transistor

[electricked]

Hi everyone,

You've been very helpful thus far in my pursuation of electronics as a hobby
and I thank you for it. Now I stumbled once again.

I found this NPN transistor today BFQ225 made by philips
(http://tinyurl.com/2uhqj). It's supposedly for cascode output and buffer
stages in high resolution monitors (whatever that means). Seems like it's
got a beta of 20 (or am I reading the datasheet wrong?)

Is this going to work for amplifying a mic to drive a speaker?

I read a website on how to test transistors for proper functioning and it
seems like the transistor is dead (basically connected DMM positive to base
and negative to collector and then postitive to base and negative to emitter
and both displayed infinite resistance). From my understanding, if it's an
NPN transistor, it has to show a low resistance when I connect base to
emitter in the way I desribed above, but it didn't. Is the transistor dead?
Or I'm thinking the DMM I'm using doesn't provide enough voltage when
testing the resistance so the base-emitter junction stays reverse biased. Am
I right about this? Is there an ultimate test that tells you if the
transistor is working properly?

Anyway, assuming the transistor is working, here's a scheme of what I'd like
to accomplish.

_R_ 9V|
---|___|---||---------|
| | |
| |25ohms 0.3W
| |
| | __ /|
| -| | |
old phone mic c| -|__| |
b |/ | \|
| / \------------| BFQ225 |
|(Mic) |> e |
| \_/-| | |
| | |
------------------------------------

R=100ohms

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de



Basically, I'm using an old phone mic to drive the base of the transistor.
I've heard that I might have to place a cap between the mic and the base of
the transistor so it converts to voltage to DC otherwise the base-emitter
junction will be reverse-biased with negative voltages. Does that have any
effect? Also, I'm thinking that I might need a higher resistance at the
collector. I'm using only 100ohms but I don't think that's enough. So if R
is 100 and the speaker is 25 then we have a total resistance of 125. Then
the current would be 9/125=72mA. So the power dissipation of the speaker
would be P=9*72mA=648mW, or approximately, 0.7V which is more than double
the max power of the speaker (0.3W). I actually built and tested the above
circuit, and didn't hear anything from the speaker. I might have damaged the
transistor or speaker in some way.

I'd like to hear your opinions, and educated guesses/suggestions why this
circuit wouldn't work.

THanks!

--Viktor

 

[John Popelish]

electricked wrote:

Hi everyone,

You've been very helpful thus far in my pursuation of electronics as a hobby
and I thank you for it. Now I stumbled once again.

I found this NPN transistor today BFQ225 made by philips
(http://tinyurl.com/2uhqj). It's supposedly for cascode output and buffer
stages in high resolution monitors (whatever that means).

It is essentially a small transistor mounted in a big case. The small
chip size keeps the internal capacitance down, to allow it to be fast,
while the big case allows it to run at a high energy density without
melting.

Seems like it's
got a beta of 20 (or am I reading the datasheet wrong?)

At least 20 with 25 mA of collector current and 10 volts collector to
emitter.

Is this going to work for amplifying a mic to drive a speaker?

I read a website on how to test transistors for proper functioning and it
seems like the transistor is dead (basically connected DMM positive to base
and negative to collector and then postitive to base and negative to emitter
and both displayed infinite resistance). From my understanding, if it's an
NPN transistor, it has to show a low resistance when I connect base to
emitter in the way I desribed above, but it didn't. Is the transistor dead?
Or I'm thinking the DMM I'm using doesn't provide enough voltage when
testing the resistance so the base-emitter junction stays reverse biased. Am
I right about this? Is there an ultimate test that tells you if the
transistor is working properly?

Many digital DVMs put out no more than .1 volt on the ohm ranges, so
you can measure resistor values while they are interconnected with PN
junctions. But those meters usually also have a diode symbol on one
range indicating a junction test range. This setting applies a fixed
current ot the leads, and displays the resultant voltage. So a good
silicon junction will display something around .6 volts.

Anyway, assuming the transistor is working, here's a scheme of what I'd like
to accomplish.

_R_ 9V|
---|___|---||---------|
| | |
| |25ohms 0.3W
| |
| | __ /|
| -| | |
old phone mic c| -|__| |
b |/ | \|
| / \------------| BFQ225 |
|(Mic) |> e |
| \_/-| | |
| | |
------------------------------------

R=100ohms

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Basically, I'm using an old phone mic to drive the base of the transistor.
I've heard that I might have to place a cap between the mic and the base of
the transistor so it converts to voltage to DC otherwise the base-emitter
junction will be reverse-biased with negative voltages. Does that have any
effect? Also, I'm thinking that I might need a higher resistance at the
collector. I'm using only 100ohms but I don't think that's enough. So if R
is 100 and the speaker is 25 then we have a total resistance of 125. Then
the current would be 9/125=72mA. So the power dissipation of the speaker
would be P=9*72mA=648mW, or approximately, 0.7V which is more than double
the max power of the speaker (0.3W). I actually built and tested the above
circuit, and didn't hear anything from the speaker. I might have damaged the
transistor or speaker in some way.

I'd like to hear your opinions, and educated guesses/suggestions why this
circuit wouldn't work.

THanks!

--Viktor

Your circuit sort of assumes that with no sound, the transistor is
conducting a little collector current, and any sound varies that
collector current up and down from that operating point, so that the
sound waveform gets amplified. However, a junction transistor does
not begin to conduct significant collector current till the base to
emitter voltage approaches .6 volts. Achieving that initial operating
point that allows class A operation (conduction by varying degree over
the entire signal wave) is called biasing or bias.

First of all, to make it simpler to keep track of voltages with
respect to the emitter, (since this is a common emitter amplifier
configuration) move the speaker to the other side of the battery so
that the negative terminal of the battery connects to the emitter.

Now, the simplest bias arrangement is a resistor from the positive
side of the battery to the base, to bring the base voltage up to about
.6 volts. But the microphone will be in parallel with the base
emitter junction draining current from that resistor to the emitter,
making this difficult. So you need a capacitor between the mic and
base to force all that resistor's current to pass through the base.

This isn't a good way to achieve class A bias, but it is an
improvement on what you have now, and should get something to come out
of the speaker.

A whole course on transistors can be built upon this simple circuit.
Many such courses are available on the web, such as:
http://www.csse.monash.edu.au/~nandita/cse1112/tut8.pdf
http://www.tpub.com/neets/book7/25c.htm
Pick some key words from these, and google will find lots more.

--
John Popelish

 

[electricked]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405D8A7A.8831B3D7@rica.net...
electricked wrote:

Hi everyone,

You've been very helpful thus far in my pursuation of electronics as a
hobby
and I thank you for it. Now I stumbled once again.

I found this NPN transistor today BFQ225 made by philips
(http://tinyurl.com/2uhqj). It's supposedly for cascode output and buffer
stages in high resolution monitors (whatever that means).

It is essentially a small transistor mounted in a big case. The small
chip size keeps the internal capacitance down, to allow it to be fast,
while the big case allows it to run at a high energy density without
melting.

Seems like it's
got a beta of 20 (or am I reading the datasheet wrong?)

At least 20 with 25 mA of collector current and 10 volts collector to
emitter.

What if I provide less voltage/current than the requirement asks for?

Is this going to work for amplifying a mic to drive a speaker?

I read a website on how to test transistors for proper functioning and it
seems like the transistor is dead (basically connected DMM positive to
base
and negative to collector and then postitive to base and negative to
emitter
and both displayed infinite resistance). From my understanding, if it's
an
NPN transistor, it has to show a low resistance when I connect base to
emitter in the way I desribed above, but it didn't. Is the transistor
dead?
Or I'm thinking the DMM I'm using doesn't provide enough voltage when
testing the resistance so the base-emitter junction stays reverse biased.
Am
I right about this? Is there an ultimate test that tells you if the
transistor is working properly?

Many digital DVMs put out no more than .1 volt on the ohm ranges, so
you can measure resistor values while they are interconnected with PN
junctions. But those meters usually also have a diode symbol on one
range indicating a junction test range. This setting applies a fixed
current ot the leads, and displays the resultant voltage. So a good
silicon junction will display something around .6 volts.

Ok, I did the test with the diode test option. I placed the positive lead at
the base and the negative lead at the collector which gave me .706 volts.
When I switched the leads from the DMM around on the same transistor leads,
it read 0V, in which case the base-collector is reverse biased and no
current can be transmitted.

I did the same test with the base and emitter junction and I got a reading
of .724 with + at base and - at emitter. When I switched the DMM leads
around, it read 0V which I think is correct.

So I guess the transistor is working fine. If it didn't work, would all
voltages read 0 or would they simply be much lower than .7?

Anyway, assuming the transistor is working, here's a scheme of what I'd
like
to accomplish.

_R_ 9V|
---|___|---||---------|
| | |
| |25ohms 0.3W
| |
| | __ /|
| -| | |
old phone mic c| -|__| |
b |/ | \|
| / \------------| BFQ225 |
|(Mic) |> e |
| \_/-| | |
| | |
------------------------------------

R=100ohms

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Basically, I'm using an old phone mic to drive the base of the
transistor.
I've heard that I might have to place a cap between the mic and the base
of
the transistor so it converts to voltage to DC otherwise the base-emitter

junction will be reverse-biased with negative voltages. Does that have
any
effect? Also, I'm thinking that I might need a higher resistance at the
collector. I'm using only 100ohms but I don't think that's enough. So if
R
is 100 and the speaker is 25 then we have a total resistance of 125. Then
the current would be 9/125=72mA. So the power dissipation of the speaker
would be P=9*72mA=648mW, or approximately, 0.7V which is more than double
the max power of the speaker (0.3W). I actually built and tested the
above
circuit, and didn't hear anything from the speaker. I might have damaged
the
transistor or speaker in some way.

I'd like to hear your opinions, and educated guesses/suggestions why this
circuit wouldn't work.

THanks!

--Viktor

Your circuit sort of assumes that with no sound, the transistor is
conducting a little collector current, and any sound varies that
collector current up and down from that operating point, so that the
sound waveform gets amplified. However, a junction transistor does
not begin to conduct significant collector current till the base to
emitter voltage approaches .6 volts. Achieving that initial operating
point that allows class A operation (conduction by varying degree over
the entire signal wave) is called biasing or bias.

So, I have to get voltage from the battery and drop it through a resistor
down to about .6 volts. Then when sound is made it will produce about 5mV
which will switch the transistor and the current will be amplified in the
collector-emitter circuit. I understand now.

First of all, to make it simpler to keep track of voltages with
respect to the emitter, (since this is a common emitter amplifier
configuration) move the speaker to the other side of the battery so
that the negative terminal of the battery connects to the emitter.

That's how I connected my original circuit but I mapped the circuit
differently for convenience. I'll try to reflect the real circuit in my
schemes from now on.

Now, the simplest bias arrangement is a resistor from the positive
side of the battery to the base, to bring the base voltage up to about
.6 volts. But the microphone will be in parallel with the base
emitter junction draining current from that resistor to the emitter,
making this difficult. So you need a capacitor between the mic and
base to force all that resistor's current to pass through the base.

Hmm. How would the mic "drain" current from the resistor to emitter (which
resistor are we talking about, the biasing one?) Can't we replaced the cap
with a diode that points toward the transistor? I would think this will
allow the mic current to go forward, but won't let the current coming from
Rb go through the mic, but will follow the base. Will this work? (Oh, I
think it won't since the drop accross the diode will be about .7V but the
mic can provide much less in general, about 5mV in general) so that won't
have enough power to switch the diode).

Also, for a NPN transistor, if we use a cap between the mic and Rb at the
base, shouldn't the cap be facing the opposite direction (i'm looking at the
diagrams you provided)? I thought the curved part should face negative
(base-emitter junction is negative) and straight plate should face positive
(mic).

Could you explain this part about the cap and the mic draining current in
more detail?

This isn't a good way to achieve class A bias, but it is an
improvement on what you have now, and should get something to come out
of the speaker.

That's what I want as a start -- to get something out of the speaker that
resembles my voice, even if vaguely.

A whole course on transistors can be built upon this simple circuit.
Many such courses are available on the web, such as:
http://www.csse.monash.edu.au/~nandita/cse1112/tut8.pdf
http://www.tpub.com/neets/book7/25c.htm
Pick some key words from these, and google will find lots more.

I'm lookin'...

--
John Popelish

Thank you, John!

--Viktor

 

[Dimitrij Klingbeil]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405D8A7A.8831B3D7@rica.net...

Now, the simplest bias arrangement is a resistor from the positive
side of the battery to the base, to bring the base voltage up to about
.6 volts. But the microphone will be in parallel with the base
emitter junction draining current from that resistor to the emitter,
making this difficult. So you need a capacitor between the mic and
base to force all that resistor's current to pass through the base.

This isn't a good way to achieve class A bias, but it is an
improvement on what you have now, and should get something to come out
of the speaker.

Since the OP described the mic as an "old phone" one, this might mean that
it is the "very old" style carbon mic, that is essentialy just a variable
resistor. In this case, a voltage would need to be applied to it to make it
work. With a bit of luck, one could also use a mic+resistor voltage divider
to provide both the signal and the necessary bias voltage to the base, this
would however require an additional resistor in the emitter circuit to make
the whole thing work in a more or less stable way. Just an idea ...

Dimitrij

 

[John Popelish]

electricked wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:405D8A7A.8831B3D7@rica.net...
electricked wrote:
(snip)
Ok, I did the test with the diode test option. I placed the positive lead at
the base and the negative lead at the collector which gave me .706 volts.
When I switched the leads from the DMM around on the same transistor leads,
it read 0V, in which case the base-collector is reverse biased and no
current can be transmitted.

I did the same test with the base and emitter junction and I got a reading
of .724 with + at base and - at emitter. When I switched the DMM leads
around, it read 0V which I think is correct.

The meter is displaying 0 any time no current passes through its leads
(rather than displaying some open circuit maximum voltage). So this
is telling you that the transistor is an NPN silicon type.

So I guess the transistor is working fine. If it didn't work, would all
voltages read 0 or would they simply be much lower than .7?

It tells you there are two PN junctions with P in the middle. It
tells you nothing about the current gain of the transistor.

Your circuit sort of assumes that with no sound, the transistor is
conducting a little collector current, and any sound varies that
collector current up and down from that operating point, so that the
sound waveform gets amplified. However, a junction transistor does
not begin to conduct significant collector current till the base to
emitter voltage approaches .6 volts. Achieving that initial operating
point that allows class A operation (conduction by varying degree over
the entire signal wave) is called biasing or bias.

So, I have to get voltage from the battery and drop it through a resistor
down to about .6 volts. Then when sound is made it will produce about 5mV
which will switch the transistor and the current will be amplified in the
collector-emitter circuit. I understand now.

The base to emitter diode will start conducting current as soon as the
voltage across that junction gets to about .6 volts. The idea is to
bias the base to emitter voltage just high enough to have some current
passing through the collector, but a middle value that can be altered
by the addition of the signal from the mic added or subtracted from
the bias situation.

First of all, to make it simpler to keep track of voltages with
respect to the emitter, (since this is a common emitter amplifier
configuration) move the speaker to the other side of the battery so
that the negative terminal of the battery connects to the emitter.

That's how I connected my original circuit but I mapped the circuit
differently for convenience. I'll try to reflect the real circuit in my
schemes from now on.

Now, the simplest bias arrangement is a resistor from the positive
side of the battery to the base, to bring the base voltage up to about
.6 volts. But the microphone will be in parallel with the base
emitter junction draining current from that resistor to the emitter,
making this difficult. So you need a capacitor between the mic and
base to force all that resistor's current to pass through the base.

Hmm. How would the mic "drain" current from the resistor to emitter (which
resistor are we talking about, the biasing one?) Can't we replaced the cap
with a diode that points toward the transistor?

I am assuming your mic is a dynamic construction that generates AC
voltage as the diapraghm vibrates (like a small speaker). These
devices have a low DC resistance (check yours with your ohm meter).
This low resistance connected directly from base to emitter will pass
lots of current from the bias resistor that should have gone through
the base. If it is a carbon mic (variable resistor) then a different
arrangement is needed.

I would think this will
allow the mic current to go forward, but won't let the current coming from
Rb go through the mic, but will follow the base. Will this work? (Oh, I
think it won't since the drop accross the diode will be about .7V but the
mic can provide much less in general, about 5mV in general) so that won't
have enough power to switch the diode).

It might work after a fashion, but a capacitor simple blocks all DC
current but allows any AC from the mic to be added directly to the
bias current and alter the average collector current.

Also, for a NPN transistor, if we use a cap between the mic and Rb at the
base, shouldn't the cap be facing the opposite direction (i'm looking at the
diagrams you provided)?

If it is a polarized (electrolytic) capacitor with a + or - sign on
the leads, the + end should be connected to the base, since it has +.6
volts compared to the emitter. The mic will operate at emitter
voltage.
(snip)

--
John Popelish

 

[electricked]

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405e3c3c_2@news.arcor-ip.de...

"John Popelish" <jpopelish@rica.net> wrote in message
news:405D8A7A.8831B3D7@rica.net...

Now, the simplest bias arrangement is a resistor from the positive
side of the battery to the base, to bring the base voltage up to about
.6 volts. But the microphone will be in parallel with the base
emitter junction draining current from that resistor to the emitter,
making this difficult. So you need a capacitor between the mic and
base to force all that resistor's current to pass through the base.

This isn't a good way to achieve class A bias, but it is an
improvement on what you have now, and should get something to come out
of the speaker.

Since the OP described the mic as an "old phone" one, this might mean that
it is the "very old" style carbon mic, that is essentialy just a variable
resistor. In this case, a voltage would need to be applied to it to make
it
work. With a bit of luck, one could also use a mic+resistor voltage
divider
to provide both the signal and the necessary bias voltage to the base,
this
would however require an additional resistor in the emitter circuit to
make
the whole thing work in a more or less stable way. Just an idea ...

Dimitrij

Yes, I got the mic from an older phone. It's basically a big white box
circular box. About 1" in diameter and about half inch high. I'm not sure if
it's a carbon mic or not. What I did was turn on my DMM to AC voltage
measurement, and placed the positive lead to one lead on the mic and
negative lead on the other lead of the mic. When I blew air in the mic, it
produced between 0-25mV depending on how hard I blew in it. Can you make up
as to what kind of mic this is from the info I've given?

If so, how should I use this mic? I have another mic that's much smaller,
1/4" (don't think it's carbon), and it doesn't produce any voltage when I
blow in the mic, no matter how hard I blow. But the one I'm using for this
experiment does.

--Viktor

 

[electricked]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405E6687.E37BC3B6@rica.net...

electricked wrote:

"John Popelish" <jpopelish@rica.net> wrote in message

Ok, so then this can't be used to test if the transistor is damaged?
That
was my main question actually, how do I test if the transistor is
damaged
(my DMM has the following options: VDC, VAC, diode test, resistance
test,
continuity test).

You can test if the transistor has an open lead or has been overheated
so badly that one or both junctions have been destroyed, but you
cannot easily test the current gain.

(snip)

The base to emitter diode will start conducting current as soon as the
voltage across that junction gets to about .6 volts. The idea is to
bias the base to emitter voltage just high enough to have some current
passing through the collector, but a middle value that can be altered
by the addition of the signal from the mic added or subtracted from
the bias situation.

I think I understand now. The way I visualize it is as follows: A big
tank
of water, we have top it off just enough so a few drops of water are
coming
out the tank. When we add the base current, it overflows and the current
starts flowing. The way I had it set up, the tank was empty, and even if
I
added a little bit of current, it wasn't enough to overflow the tank so
no
current was passing through the base to emitter. Does it sound like
what's
really happening? Or is it a crappy analogy?

I am not sure I follow your analogy. But a class A amplifier has to
be conducting somewhat so that the signal can either increase or
decrease that conduction to amplify the whole waveform. The bias
arrangement sets the transistor up to be conducting some collector
current, but not as much as the load will allow. In other words, the
transistor is acting like a variable resistor, not an open or closed
switch.

First tank, filled with water up to the top so if you add any water to it,
the water will overflow and drain on the ground (analogy comparison to the
base at .7 volts so current is flowing from base to emitter).

|/\/\/|
| |
| |
|_____|

If there's less than .7 volts at the base of the transistor, it's analogous
to a tank that is not full with water (when it's full with water it is
analogous to .7V of base). So basically, if you add a little bit of water,
no water will flow out of the tank since it's not full and can't overflow
(again full tank is at .7 volts). But if you keep adding water and the tank
is full (.7V) then any more water you add will start dripping on the floor
(analogous to the transistor forward biasing the base-emitter junction).

| |
|/\/\/|
| |
|_____|

The wavey lines represent water level in tank.

Hope it makes sense and I think it's the same think as your explanation.

(snip)

I am assuming your mic is a dynamic construction that generates AC
voltage as the diapraghm vibrates (like a small speaker). These
devices have a low DC resistance (check yours with your ohm meter).
This low resistance connected directly from base to emitter will pass
lots of current from the bias resistor that should have gone through
the base. If it is a carbon mic (variable resistor) then a different
arrangement is needed.

This is starting to make sense now. Yes, I think the mic is a dynamic
type
since it produces voltage when I blow in it (unlike the other mic I
have,
which I think is carbon type...it doesn't produce any voltage when I
blow at
it). Is it a problem if the voltage that is generated by the dynamic mic
is
positive? It ranges between 0 and 25mV if I blow really hard, average
about
5mV. I never see a negative voltage while I'm measuring. Does this sound
right?

Ac voltage always shows up as a positive magnitude on a meter. The
meter rectifies the wave and shows a value related to the average of
that rectified version of the wave. The actual signal is alternating
between positive and negative output.

I was visualizing the circuit a little differently, thus, my confusion.
I
was visualizing the biasing resistor coming on the right of the cap at
the
junction of the mic.

-------------Vcc
|
\
/R cap
|----)|-----base
|
[ ]mic
|
|
---
-

Since the dynamic mic generates voltage, it needs no connection to
Vcc.

It is the base that needs the current from Vcc to get it turned part
way on.

The above is what I was thinking of. But if we the biasing resistor goes
between the cap and the transistor base, then it would work as you
described, and I'm understanding that part of it.

Would removing the cap and placing a diode on top of the mic work if I
used
the above diagram?

Basically, this:

------------------Vcc
| |
| \
| /R1
| |---------base
\ |
/R2 - diode
| ^
|----|
[ ]mic
|
|
---
-

As I said, if the mic is dynamic, it has no use for current from Vcc.
Without R2, you could turn the diode around and the current through R1
would split with part of it going through the diode and the mic and
the rest going through the base emitter junction. Since the base
emitter junction is also a diode, it will pass a current similarly to
that which passes through the diode. This could work, but R1 will
have to be a lower value, since only part of the current going through
it acts to bias the transistor on.

Okay, makes sense. If I reverse the diode, wouldn't that stop the voltage
coming from the mic? The voltage from the mic is about 5mV, and the diode
needs .7V which is 700mV so the diode will be reverse biased, and will not
allow the diode to conduct, right? I see how R1 would have to be lowered to
increase the current so when it is split up between the base and mic, it
will have enough current at the base to forward bias it.

Looking at the following diagram:
http://www.tpub.com/neets/book7/25c.htm

Isn't positive voltage from mic, going at negative side of the cap?
Wouldn't
this damaged a polarized cap? I see how the positive side faces the
positive
side of the base. Or is the only requirement, that the positive side of
the
cap should be more positive than the negative side? If that's the case,
I
see how this could be.
I don't think they were careful about which side of the caps point
which way in that lesson, but the mic voltage will be zero volts plus
or minus a few millivolts, while the base voltage (biased into
conduction) will be about +0.6 volts. So the base is the more
positive of the two.

Okay, this far it makes sense. I can see this happening so long as we have
direct current, even if it's rippled; i.e. the current might change but it's
not reversing directions/polarity.

As you can see, I'm still having problem understand how components
related
to the different voltages. When I see + and - on a component (be it
battery,
cap, diode, or whatever) should I interpret it that the positive side
should
be more positive than the negative?

That is what it means. With diodes, the band shows the end that must
be the more negative for the diode to conduct.

Fair enough. I get it.

Even if it's a DC voltage. Can polarized
caps be used with AV/AC? (Logic says no, but I want to be sure).

As long as all the variations do not pass through zero and reverse,
the cap will be fine, unless the charge and discharge current
overheats the cap (ripple current rating).

Does this confirm what I said just above? If it's DC current, even if it's
rippled, a polarized cap will pass it, so long as the positive side is more
positive than the negative, BUT! if it's AC current (that is, the reference
point is being crossed) then the cap will get damaged. Does this make sense?

The cap I am suggesting between the mic and the base will have very
little voltage variation across it. As the mic produces AC, that AC
will also appear almost without reduction on the other end and alter
the base voltage, while the difference between the ends of the cap
remains at almost a constant 0.6 volts. The definition of a coupling
capacitor is that it blocks a constant DC but passes AC with little
attenuation. However, if the cap is way too small, then the signal
current it passes to the base will charge it up and down during the
signal swing, and lots of the signal voltage will be lost across the
capacitor. This is called a high pass filter.

Can I use a bipolar cap for this situation? Would it work?

All this is based on the mathematical description of capacitance:
I=C*dv/dt
(in words) current through a capacitor is proportional to the time
rate of change of voltage across it, and the proportionality factor is
the capacitance. So if the current through the cap is small enough
(which is to say, if the capacitance is high enough) then there is
very little change in the voltage across it in a short enough period
of time.

--
John Popelish

To repeat and make sure I understand this, if the voltage/current change
across the cap is small enough compared to it's capacitance, then it will be
passed, otherwise it will be blocked since the cap would think it's DC
voltage. So the high/low pass filter is relative only to the cap's
capacitance. So, if it's a small cap it would identify lower frequencies as
DC voltage, compared to a bigger cap, which would identify higher
frequencies as DC voltage(compared to smaller cap), and anything above that
as high frequencies.

For a bipolar capacitor, is there a difference between AC current, and DC
rippled current in the way low frequencies are blocked, and high frequencies
are passed?

What's the formula for deciding what size cap I need between the mic and the
base-emitter junction? The cap should be pretty small since anything that's
constant is passed and anything that has a little ripple in it has to be
passed through (the frequencies generated by the mic should be 20Hz-20Khz or
so, right?)

Thanks John!

--Viktor

 

[John Popelish]

electricked wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:405E6687.E37BC3B6@rica.net...

I am not sure I follow your analogy. But a class A amplifier has to
be conducting somewhat so that the signal can either increase or
decrease that conduction to amplify the whole waveform. The bias
arrangement sets the transistor up to be conducting some collector
current, but not as much as the load will allow. In other words, the
transistor is acting like a variable resistor, not an open or closed
switch.

First tank, filled with water up to the top so if you add any water to it,
the water will overflow and drain on the ground (analogy comparison to the
base at .7 volts so current is flowing from base to emitter).

|/\/\/|
| |
| |
|_____|

If there's less than .7 volts at the base of the transistor, it's analogous
to a tank that is not full with water (when it's full with water it is
analogous to .7V of base). So basically, if you add a little bit of water,
no water will flow out of the tank since it's not full and can't overflow
(again full tank is at .7 volts). But if you keep adding water and the tank
is full (.7V) then any more water you add will start dripping on the floor
(analogous to the transistor forward biasing the base-emitter junction).

| |
|/\/\/|
| |
|_____|

The wavey lines represent water level in tank.

Hope it makes sense and I think it's the same think as your explanation.

Now I see what you are saying, and it is a pretty fair analogy of the
nonlinearity of the conduction through the base emitter junction. If
you cut a narrow V notch in the edge of the tank, you could make it a
pretty accurate analogy of the diode character.

As I said, if the mic is dynamic, it has no use for current from Vcc.
Without R2, you could turn the diode around and the current through R1
would split with part of it going through the diode and the mic and
the rest going through the base emitter junction. Since the base
emitter junction is also a diode, it will pass a current similarly to
that which passes through the diode. This could work, but R1 will
have to be a lower value, since only part of the current going through
it acts to bias the transistor on.

Okay, makes sense. If I reverse the diode, wouldn't that stop the voltage
coming from the mic? The voltage from the mic is about 5mV, and the diode
needs .7V which is 700mV so the diode will be reverse biased, and will not
allow the diode to conduct, right? I see how R1 would have to be lowered to
increase the current so when it is split up between the base and mic, it
will have enough current at the base to forward bias it.

Looking at the following diagram:
http://www.tpub.com/neets/book7/25c.htm

Isn't positive voltage from mic, going at negative side of the cap?
Wouldn't
this damaged a polarized cap? I see how the positive side faces the
positive
side of the base. Or is the only requirement, that the positive side of
the
cap should be more positive than the negative side? If that's the case,
I
see how this could be.
I don't think they were careful about which side of the caps point
which way in that lesson, but the mic voltage will be zero volts plus
or minus a few millivolts, while the base voltage (biased into
conduction) will be about +0.6 volts. So the base is the more
positive of the two.

Okay, this far it makes sense. I can see this happening so long as we have
direct current, even if it's rippled; i.e. the current might change but it's
not reversing directions/polarity.

Be clear on one thing. The DC rating of electrolytic caps refers to
the voltage across them, not the current through them. You can slosh
charge into and back out of a cap, but if the total charge never goes
through zero, the voltage will always have the same polarity.

As long as all the variations do not pass through zero and reverse,
the cap will be fine, unless the charge and discharge current
overheats the cap (ripple current rating).

Does this confirm what I said just above? If it's DC current, even if it's
rippled, a polarized cap will pass it, so long as the positive side is more
positive than the negative, BUT! if it's AC current (that is, the reference
point is being crossed) then the cap will get damaged. Does this make sense?

Not quite. Any capacitor cannot pass continuous DC, since the process
of passing current through it must change the voltage across it and
that can't continue long before you exceed the voltage rating of the
cap. The cap passes AC current, while having a DC voltage across it.
The current ramps the DC voltage up when it goes one way, and ramps it
down it while it goes the other way. but if it reverses often enough
or the current is low enough, the voltage need not reverse.

The cap I am suggesting between the mic and the base will have very
little voltage variation across it. As the mic produces AC, that AC
will also appear almost without reduction on the other end and alter
the base voltage, while the difference between the ends of the cap
remains at almost a constant 0.6 volts. The definition of a coupling
capacitor is that it blocks a constant DC but passes AC with little
attenuation. However, if the cap is way too small, then the signal
current it passes to the base will charge it up and down during the
signal swing, and lots of the signal voltage will be lost across the
capacitor. This is called a high pass filter.

Can I use a bipolar cap for this situation? Would it work?

Sure. Since none of the cps in the tutorial we were talking about
have + signs on them, you can assume they were thinking of non polar
(bipolar) caps. But your circuit will apply voltage in only one
direction to them.

All this is based on the mathematical description of capacitance:
I=C*dv/dt
(in words) current through a capacitor is proportional to the time
rate of change of voltage across it, and the proportionality factor is
the capacitance. So if the current through the cap is small enough
(which is to say, if the capacitance is high enough) then there is
very little change in the voltage across it in a short enough period
of time.

To repeat and make sure I understand this, if the voltage/current change
across the cap is small enough compared to it's capacitance, then it will be
passed, otherwise it will be blocked since the cap would think it's DC
voltage. So the high/low pass filter is relative only to the cap's
capacitance. So, if it's a small cap it would identify lower frequencies as
DC voltage, compared to a bigger cap, which would identify higher
frequencies as DC voltage(compared to smaller cap), and anything above that
as high frequencies.

Pretty good. At some low enough voltage, the cap will charge up to a
voltage equal to the applied signal and cancel it out (or nearly so).
At high enough frequencies, it will charge up and down an
insignificant fraction of the signal swing, so not cancel much of it
out. If you can figure out the total resistance in series with the
signal, you can calculate at what frequency the signal will be reduced
to half energy. 1/(2*pi*f*)=R*C where R is the total resistance in
series with the cap (the mic resistance in series with the parallel
combination of the bias resistor and the base to emitter effective
junction resistance at its bias point) in ohms, C is the capacitance
in farads and f is the frequency in hertz. We can get back to this
later. But if your amplifier seems to be amplifying only the treble
parts of speech, using a bigger cap should give more base. You might
experiment with several values, but having a signal generator or some
kind of oscillator as input would make this more quantitative.

For a bipolar capacitor, is there a difference between AC current, and DC
rippled current in the way low frequencies are blocked, and high frequencies
are passed?

The formula is the same. It is up to you to make sure the circuit
design does not apply reverse voltage to the cap under any operating
conditions.

What's the formula for deciding what size cap I need between the mic and the
base-emitter junction? The cap should be pretty small since anything that's
constant is passed and anything that has a little ripple in it has to be
passed through (the frequencies generated by the mic should be 20Hz-20Khz or
so, right?)

Okay, lets plug some numbers into the above formula. Lets say the mic
resistance is very low, so we neglect it. The base bias resistor is
very high, so lets neglect that also. So the input resistance of the
common emitter stage is the dominant resistance limiting the signal
current driven into the amplifier by the mic. I am not going to go
into how you calculate that value right now but just throw out a
hypothetical. The actual input resistance is a function of the bias
point. The more the base current, the lower the input resistance. It
is like a tank that is soaking up that base swing with that overflow,
after all. A bigger overflow makes it harder to bounce the level up
and down with the same variation in input flow (signal current).

Lets assume the base input resistance is about 1000 ohms. If the cap
is to let at least half of the energy at 20 Hz to be applied to this
resistance, the equation is 1/(2*pi*20)=1000*C, so C has to be at
least
1/(1000*2*pi*20)=.00000795 farads or about 8 microfarads.

A .8 uF coupling cap would roll off the response below 200 Hz, a .08
uF below 2000 Hz, etc.

--
John Popelish

 

[Dimitrij Klingbeil]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405E6687.E37BC3B6@rica.net...

------------------Vcc
| |
| \
| /R1
| |---------base
\ |
/R2 - diode
| ^
|----|
[ ]mic
|
|
---
-

As I said, if the mic is dynamic, it has no use for current from Vcc.
Without R2, you could turn the diode around and the current through R1
would split with part of it going through the diode and the mic and
the rest going through the base emitter junction. Since the base
emitter junction is also a diode, it will pass a current similarly to
that which passes through the diode. This could work, but R1 will
have to be a lower value, since only part of the current going through
it acts to bias the transistor on.

John, that's what one calls imagination! That circuit's functionality will
probably rely on yesterday's weather, , but nevertheless I rather like
the idea. At least I think, with some changes, it will be worth testing.
However one should make it a little easier for the OP to get working or at
least facilitate the troubleshooting. What about using one more diode and a
capacitor, e.g. like this:

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

Simpler version: remove C2 and replace R2 with speaker.

Dimitrij

 

[John Popelish]

Dimitrij Klingbeil wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:405E6687.E37BC3B6@rica.net...


------------------Vcc
| |
| \
| /R1
| |---------base
\ |
/R2 - diode
| ^
|----|
[ ]mic
|
|
---
-

As I said, if the mic is dynamic, it has no use for current from Vcc.
Without R2, you could turn the diode around and the current through R1
would split with part of it going through the diode and the mic and
the rest going through the base emitter junction. Since the base
emitter junction is also a diode, it will pass a current similarly to
that which passes through the diode. This could work, but R1 will
have to be a lower value, since only part of the current going through
it acts to bias the transistor on.

John, that's what one calls imagination! That circuit's functionality will
probably rely on yesterday's weather, , but nevertheless I rather like
the idea. At least I think, with some changes, it will be worth testing.
However one should make it a little easier for the OP to get working or at
least facilitate the troubleshooting. What about using one more diode and a
capacitor, e.g. like this:

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

Simpler version: remove C2 and replace R2 with speaker.

Dimitrij

I think 1 diode would be the best. The more are in series, the higher
impedance they will have, because their forward current will approach
zero. They need current through them to turn them on enough to pass
the mic signal.
I have never seen such a circuit but it is really sort of similar to
one than has a PNP emitter follower preceding the NPN common emitter
to eliminate most of the bias voltage on the NPN, except that the
current gain of this 'follower' is zero.

Or you can look at it as a current mirror with signal injection.

--
John Popelish

 

[Dimitrij Klingbeil]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405F0DF7.1C688434@rica.net...

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

I think 1 diode would be the best. The more are in series, the higher
impedance they will have, because their forward current will approach
zero. They need current through them to turn them on enough to pass
the mic signal.
I have never seen such a circuit but it is really sort of similar to
one than has a PNP emitter follower preceding the NPN common emitter
to eliminate most of the bias voltage on the NPN, except that the
current gain of this 'follower' is zero.

Or you can look at it as a current mirror with signal injection.

John, I understand your reasoning well, but that is not why I proposed to
add one more diode. The reason for it was that the voltage drop on a diode
may be less that the one on the BE junction. In this case no current (or
almost none anyway) would flow through the transistor and it would not work.
In the opposite case, all current would flow through the transistor and none
through the diode. This won't work either. And since it would be a hell of a
job to match them exactly enough, there is a need to make the whole thing
more stable. Using 2 diodes one raises the base voltage by additional half a
volt, but has to compensate for this by adding resistance in the emitter
circuit (R3). The intention was to let D2 and R3 have (almost) the same
voltage drop. The resistors are to be chosen as to allow most of the current
(which still has to be kept small) to flow through the diodes and mic, and
only a small part of it through the base. A possible R1 value would be
around 1K to 4K. R3 is meant to have a 4 to 10 times lower resistance than
R2, but that is not critical as C1 will lower the AC impedance anyway. As
far as the impedance of the diodes is concerned, the DC resistance will be
pretty much constant and probably low enough and the AC impedance can be
lowered by adding a cap in parallel (not that it was necessary anyway).
Note: The explaination above assumes that the resistor values are chosen so
that the DC resistance of the transistor circuit's input is high and (in DC
mode) it is operating partially like an emitter follower. For the AC
component however the RC total impedance will be low and the transistor will
amplify it mostly as in common-emitter mode.

I did not test this circuit in practice, but something tells me that it
should work (although not efficiently). If not, please let me know.

Dimitrij

 

[John Popelish]

Dimitrij Klingbeil wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:405F0DF7.1C688434@rica.net...

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

I think 1 diode would be the best. The more are in series, the higher
impedance they will have, because their forward current will approach
zero. They need current through them to turn them on enough to pass
the mic signal.
I have never seen such a circuit but it is really sort of similar to
one than has a PNP emitter follower preceding the NPN common emitter
to eliminate most of the bias voltage on the NPN, except that the
current gain of this 'follower' is zero.

Or you can look at it as a current mirror with signal injection.

John, I understand your reasoning well, but that is not why I proposed to
add one more diode. The reason for it was that the voltage drop on a diode
may be less that the one on the BE junction. In this case no current (or
almost none anyway) would flow through the transistor and it would not work.
In the opposite case, all current would flow through the transistor and none
through the diode. This won't work either. And since it would be a hell of a
job to match them exactly enough, there is a need to make the whole thing
more stable. Using 2 diodes one raises the base voltage by additional half a
volt, but has to compensate for this by adding resistance in the emitter
circuit (R3). The intention was to let D2 and R3 have (almost) the same
voltage drop. The resistors are to be chosen as to allow most of the current
(which still has to be kept small) to flow through the diodes and mic, and
only a small part of it through the base. A possible R1 value would be
around 1K to 4K. R3 is meant to have a 4 to 10 times lower resistance than
R2, but that is not critical as C1 will lower the AC impedance anyway. As
far as the impedance of the diodes is concerned, the DC resistance will be
pretty much constant and probably low enough and the AC impedance can be
lowered by adding a cap in parallel (not that it was necessary anyway).
Note: The explaination above assumes that the resistor values are chosen so
that the DC resistance of the transistor circuit's input is high and (in DC
mode) it is operating partially like an emitter follower. For the AC
component however the RC total impedance will be low and the transistor will
amplify it mostly as in common-emitter mode.

I did not test this circuit in practice, but something tells me that it
should work (although not efficiently). If not, please let me know.

I paid no attention to the fact that you added an emitter resistor and
bypass capacitor. That, of course, changes everything. I still like
the elegance of the single diode to elevate the signal voltage by a
forward biased diode.

--
John Popelish

 

[Dimitrij Klingbeil]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405F207F.2D5B9797@rica.net...

I paid no attention to the fact that you added an emitter resistor and
bypass capacitor. That, of course, changes everything. I still like
the elegance of the single diode to elevate the signal voltage by a
forward biased diode.

No idea how I did not think of that instead of writing that lengthy piece of
text. In the meantime however I have thrown together the circuit using the
2-diode schematic and tested it, just to know what else can go wrong. The
whole thingy had an overall gain of 10 (with load connected), so letting it
drive a speaker directly turned out to be nonsesnse. Maybe some tinkering
with the resistor values could add another 10 or so, but it likely won't do
what was intended. It worked however reliably with a stronger input signal
(about 10 mV). The output impedance was a little too high for an average
speaker, in any case, a 100R speaker worked better than a 8R one.

As for the elegance of using a diode for raising a level, I fully agree. I
sometimes use the trick to replace a low-v zener with diodes and/or LEDs.

A notice to the OP: For mic amps, the idea of using only one transistor is
best left alone after all. Even if you get the maximal possible gain out of
an average transistor, at least 2 stages for dynamic mics will likely be a
sine qua non. Don't however put together a 3-stage class A oscillator
instead.

Dimitrij

 

[electricked]

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f0725_2@news.arcor-ip.de...

"John Popelish" <jpopelish@rica.net> wrote in message
news:405E6687.E37BC3B6@rica.net...


------------------Vcc
| |
| \
| /R1
| |---------base
\ |
/R2 - diode
| ^
|----|
[ ]mic
|
|
---
-

As I said, if the mic is dynamic, it has no use for current from Vcc.
Without R2, you could turn the diode around and the current through R1
would split with part of it going through the diode and the mic and
the rest going through the base emitter junction. Since the base
emitter junction is also a diode, it will pass a current similarly to
that which passes through the diode. This could work, but R1 will
have to be a lower value, since only part of the current going through
it acts to bias the transistor on.

John, that's what one calls imagination! That circuit's functionality will
probably rely on yesterday's weather, , but nevertheless I rather like
the idea. At least I think, with some changes, it will be worth testing.
However one should make it a little easier for the OP to get working or at
least facilitate the troubleshooting. What about using one more diode and
a
capacitor, e.g. like this:

Ok, I have a few questions. Remember, my understand of electronics is much
limited compared to yours so let's get to it.

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

Hmm... Why did you add the second diode? What's the point of the diode? If
it were me, I would use just one diode, and reverse it so it allows mic
current to flow and blocks the current coming through R1 so it goes straight
to base. Even in this case, how will a current so small pass through the
diode when it's reverse-biased? I thought you needed at least .6/7 volts in
order for the diode to start conducting, and the mic certainly cannot
provide that much (25mV at best).

One more question, why do you have R3 and C1 connected in parallel at the
ground? What does that accomplish? Is there a relationship between R3 and C1
in the way this parallel circuit works? What is it?

One more, what's the use for C2? What's it supposed to do? Does it have to
smooth the frequency or what?

Simpler version: remove C2 and replace R2 with speaker.

Ok, this makes more sense. So why did you need C2 and R2 in the first place,
especially C2? What's it do?

Dimitrij

--Viktor

 

[Dimitrij Klingbeil]

"electricked" <no_emails_please> wrote in message
news:TvKdneNlPpFcoMLdRVn-hA@comcast.com...

John, that's what one calls imagination! That circuit's functionality
will
probably rely on yesterday's weather, , but nevertheless I rather
like
the idea. At least I think, with some changes, it will be worth testing.
However one should make it a little easier for the OP to get working or
at
least facilitate the troubleshooting. What about using one more diode
and
a
capacitor, e.g. like this:

Ok, I have a few questions. Remember, my understand of electronics is much
limited compared to yours so let's get to it.

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

I'll try to answer your questions, but please do also look at my reply to
John about testing the thing and what gain it has. Looks like it's not
really worth constucting unless multiple stages are to be used. Please note
also that the above schematic was rather an intentional example of violating
all good design practices and although it works, it's still an example of
how not to design a circuit. If you get that thing working for you, my
respect.

Hmm... Why did you add the second diode? What's the point of the diode? If
it were me, I would use just one diode, and reverse it so it allows mic
current to flow and blocks the current coming through R1 so it goes
straight
to base. Even in this case, how will a current so small pass through the
diode when it's reverse-biased? I thought you needed at least .6/7 volts
in
order for the diode to start conducting, and the mic certainly cannot
provide that much (25mV at best).

You've probably got a wrong idea. The diode is NOT meant to rectify the
mic's output. It just can't and you have already answered that yourself (mic
< .6V). The diode was meant to provide a voltage drop to pull up the
potential at the base to a level that is just enough to make sure the
transistor works. The reason for this is that the BE-junction of a
transistor is a diode itself and also has a voltage drop of about .6V. A mic
can't provide that. Try to think of R1 as a pull-up resistor and the diodes
and mic together as a pull-down resistor in a voltage divider. The voltage
divider, as the name implies, will have a fraction of the total voltage at
its output (base) and if that fraction happens just enough to bias the
transistor without overloading it, the transistor will act like an amp.
That's the intention of the original circuit with one diode. The diode can't
be placed otherwise because then the R1 will pull up the base potential and
the mic would have no way to influence it. It won't, as you say, "allow the
mic current to flow and block Vcc" because current only flows between
different potentials, and the direction of this difference would, here, be
constant. After all the mic's voltage can't outperform the R1-limited Vcc.
As for the second diode, the reason for it is that it's hard to match two
diode voltage drops exactly. Doing the same thing with a resistor (R3) is
much easier. Besides, one shouold take the circuit's overall thermostability
into account, and that one would be miserable in the one diode version. With
2 diodes, the base potential is a little too high and the transistor acts
like an emitter follower with R3 being the load (at least for DC where C1
plays no part). That stabilizes the amp at a given and stable working point.
However because the AC gain would be low, C1 is used to shunt R3 for AC,
thus making the amp work like common emitter for AC while still being
DC-stabilized.

One more question, why do you have R3 and C1 connected in parallel at the
ground? What does that accomplish? Is there a relationship between R3 and
C1
in the way this parallel circuit works? What is it?

See above. There is no direct relationship, C1 is chosen based on the
frequency range to provide a low impedance (bypass cap).

One more, what's the use for C2? What's it supposed to do? Does it have to
smooth the frequency or what?

No. It does not deal with frequencies, it's just a decoupling capacitor used
to prevent DC from entering the speaker and the speaker from shunting R2. It
provides a low impedance for the AC component and blocks DC, that's all.

Simpler version: remove C2 and replace R2 with speaker.

Ok, this makes more sense. So why did you need C2 and R2 in the first
place,
especially C2? What's it do?

It only apparently makes sense. R2 should be higher than R3, but a speaker's
impedance is usually low. When connected directly, it might overload the amp
and let the gain sink below unity, in which case the amp would become an
active attenuator. In any case, the speaker should be high-impedance. The
lower its impedance is, the worse the results. Tip: Use headphones.

I hope, the above did not completely mess up your
understanding of electronics, sorry if otherwise.

Dimitrij

 

[Dimitrij Klingbeil]

"electricked" <no_emails_please> wrote in message
news:GfSdnd6f6Kl4z8LdRVn-vw@comcast.com...

Just one question, how does the current generated by the mic get to the
base
of the transistor?

Just imagine the diode as a nonlinear resistor with a fixed voltage drop, a
medium DC resistance, and a very low differential resistance. Why should it
not pass the mic signal along to the base, provided it's biased correctly.

Dimitrij

 

[electricked]

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f4acb_1@news.arcor-ip.de...

"electricked" <no_emails_please> wrote in message
news:TvKdneNlPpFcoMLdRVn-hA@comcast.com...

John, that's what one calls imagination! That circuit's functionality
will
probably rely on yesterday's weather, , but nevertheless I rather
like
the idea. At least I think, with some changes, it will be worth
testing.
However one should make it a little easier for the OP to get working
or
at
least facilitate the troubleshooting. What about using one more diode
and
a
capacitor, e.g. like this:

Ok, I have a few questions. Remember, my understand of electronics is
much
limited compared to yours so let's get to it.

+VCC ----x----------.
| |
-R1 -R2
| | | |
| | | |
- -
| |
x---. |
| | |
---D1| | +||C2
\|/ | x---||---> to spkr
--- | | ||
| | |
---D2| /'
\|/ '----| T1-NPN
--- >.
| |
--- .--x--.
[mic] | |
--- -R3 |C1
| | | ---
| | | ---
| - |
| | |
GND -----x-------x-----x-----> to spkr

I'll try to answer your questions, but please do also look at my reply to
John about testing the thing and what gain it has. Looks like it's not
really worth constucting unless multiple stages are to be used. Please
note
also that the above schematic was rather an intentional example of
violating
all good design practices and although it works, it's still an example of
how not to design a circuit. If you get that thing working for you, my
respect.

Hmm... Why did you add the second diode? What's the point of the diode?
If
it were me, I would use just one diode, and reverse it so it allows mic
current to flow and blocks the current coming through R1 so it goes
straight
to base. Even in this case, how will a current so small pass through the
diode when it's reverse-biased? I thought you needed at least .6/7 volts
in
order for the diode to start conducting, and the mic certainly cannot
provide that much (25mV at best).

You've probably got a wrong idea. The diode is NOT meant to rectify the
mic's output. It just can't and you have already answered that yourself
(mic
.6V). The diode was meant to provide a voltage drop to pull up the
potential at the base to a level that is just enough to make sure the
transistor works. The reason for this is that the BE-junction of a
transistor is a diode itself and also has a voltage drop of about .6V. A
mic
can't provide that. Try to think of R1 as a pull-up resistor and the
diodes
and mic together as a pull-down resistor in a voltage divider. The voltage
divider, as the name implies, will have a fraction of the total voltage at
its output (base) and if that fraction happens just enough to bias the
transistor without overloading it, the transistor will act like an amp.
That's the intention of the original circuit with one diode. The diode
can't
be placed otherwise because then the R1 will pull up the base potential
and
the mic would have no way to influence it. It won't, as you say, "allow
the
mic current to flow and block Vcc" because current only flows between
different potentials, and the direction of this difference would, here, be
constant. After all the mic's voltage can't outperform the R1-limited Vcc.
As for the second diode, the reason for it is that it's hard to match two
diode voltage drops exactly. Doing the same thing with a resistor (R3) is
much easier. Besides, one shouold take the circuit's overall
thermostability
into account, and that one would be miserable in the one diode version.
With
2 diodes, the base potential is a little too high and the transistor acts
like an emitter follower with R3 being the load (at least for DC where C1
plays no part). That stabilizes the amp at a given and stable working
point.
However because the AC gain would be low, C1 is used to shunt R3 for AC,
thus making the amp work like common emitter for AC while still being
DC-stabilized.

One more question, why do you have R3 and C1 connected in parallel at
the
ground? What does that accomplish? Is there a relationship between R3
and
C1
in the way this parallel circuit works? What is it?

See above. There is no direct relationship, C1 is chosen based on the
frequency range to provide a low impedance (bypass cap).

One more, what's the use for C2? What's it supposed to do? Does it have
to
smooth the frequency or what?

No. It does not deal with frequencies, it's just a decoupling capacitor
used
to prevent DC from entering the speaker and the speaker from shunting R2.
It
provides a low impedance for the AC component and blocks DC, that's all.

Simpler version: remove C2 and replace R2 with speaker.

Ok, this makes more sense. So why did you need C2 and R2 in the first
place,
especially C2? What's it do?

It only apparently makes sense. R2 should be higher than R3, but a
speaker's
impedance is usually low. When connected directly, it might overload the
amp
and let the gain sink below unity, in which case the amp would become an
active attenuator. In any case, the speaker should be high-impedance. The
lower its impedance is, the worse the results. Tip: Use headphones.

I hope, the above did not completely mess up your
understanding of electronics, sorry if otherwise.

Dimitrij

Just one question, how does the current generated by the mic get to the base
of the transistor?

Thanks Dimitrij, I have a long way to go in this hobby

--Viktor

 

[electricked]

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f5890_1@news.arcor-ip.de...

"electricked" <no_emails_please> wrote in message
news:GfSdnd6f6Kl4z8LdRVn-vw@comcast.com...

Just one question, how does the current generated by the mic get to the
base
of the transistor?


Just imagine the diode as a nonlinear resistor with a fixed voltage drop,
a
medium DC resistance, and a very low differential resistance. Why should
it
not pass the mic signal along to the base, provided it's biased correctly.

Dimitrij

Because from my understanding, the signal coming from the mic is too weak to
forward bias it. Plus, it's reverse-biased with the mic and takes much more
to conduct. Isn't this true? Or is everything I've been reading about diodes
up to this point all BS?

--Viktor

 

[Dimitrij Klingbeil]

"electricked" <no_emails_please> wrote in message
news:VeqdnTj_vtHC_8LdRVn-hg@comcast.com...

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f5890_1@news.arcor-ip.de...

"electricked" <no_emails_please> wrote in message
news:GfSdnd6f6Kl4z8LdRVn-vw@comcast.com...

Just one question, how does the current generated by the mic get to
the
base
of the transistor?


Just imagine the diode as a nonlinear resistor with a fixed voltage
drop,
a
medium DC resistance, and a very low differential resistance. Why should
it
not pass the mic signal along to the base, provided it's biased
correctly.

Dimitrij

Because from my understanding, the signal coming from the mic is too weak
to
forward bias it. Plus, it's reverse-biased with the mic and takes much
more
to conduct. Isn't this true? Or is everything I've been reading about
diodes
up to this point all BS?

No, what you assume isn't true, but what you've read about diodes isn't "BS"
either. It's rather perfectly correct (I assume). The fact is just that it's
not the mic that provides the bias voltage, it's the battery (or whatever
Vcc is). Besides, the diode is forward, not reverse biased. When biasing a
diode, the total voltage in the diode circuit counts, and in this case it
includes a part of Vcc, which is much higher than the mic signal and biases
the diode forward. The mic only provides a voltage difference, it only
influences the existing part of Vcc, it does not bias the diode itself, it
isn't meant to do so either. Because the diode's differential resistance is
low in this case, you can think of it like a voltage source. Maybe this way
it is easier. Imagine the diode as a constant zero-resistance voltage source
that passes the mic signal (as a voltage difference) along to the base. If
you connect a capacitor in parallel to the diode, maybe this behaviour will
be more evident.

Apart from that, the use of a diode as stated above is a rather nonstandard
"trick", it may be not very well comprehensive. Don't despair if it turns
out a little hard to understand to you. After all, it was John's idea, and
he is an engineer with years of experience. With all respect to him, I must
admit, it's not the best starting example for novices. My reply suggesting 2
diodes and the RC thingy was rather addressed to John as a word of caution:
not to leave to you building a circuit that is hard to trim. It was not
meant as an easy version either. It needs less trimming, but is more
complicated from the theory. Sorry for messing your understanding up with
this.

Dimitrij

 

[electricked]

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f6fdc_2@news.arcor-ip.de...

"electricked" <no_emails_please> wrote in message
news:VeqdnTj_vtHC_8LdRVn-hg@comcast.com...

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f5890_1@news.arcor-ip.de...

"electricked" <no_emails_please> wrote in message
news:GfSdnd6f6Kl4z8LdRVn-vw@comcast.com...

Just one question, how does the current generated by the mic get to
the
base
of the transistor?


Just imagine the diode as a nonlinear resistor with a fixed voltage
drop,
a
medium DC resistance, and a very low differential resistance. Why
should
it
not pass the mic signal along to the base, provided it's biased
correctly.

Dimitrij

Because from my understanding, the signal coming from the mic is too
weak
to
forward bias it. Plus, it's reverse-biased with the mic and takes much
more
to conduct. Isn't this true? Or is everything I've been reading about
diodes
up to this point all BS?

No, what you assume isn't true, but what you've read about diodes isn't
"BS"
either. It's rather perfectly correct (I assume). The fact is just that
it's
not the mic that provides the bias voltage, it's the battery (or whatever
Vcc is). Besides, the diode is forward, not reverse biased. When biasing a
diode, the total voltage in the diode circuit counts, and in this case it
includes a part of Vcc, which is much higher than the mic signal and
biases
the diode forward. The mic only provides a voltage difference, it only
influences the existing part of Vcc, it does not bias the diode itself, it
isn't meant to do so either. Because the diode's differential resistance
is
low in this case, you can think of it like a voltage source. Maybe this
way
it is easier. Imagine the diode as a constant zero-resistance voltage
source
that passes the mic signal (as a voltage difference) along to the base. If
you connect a capacitor in parallel to the diode, maybe this behaviour
will
be more evident.

Apart from that, the use of a diode as stated above is a rather
nonstandard
"trick", it may be not very well comprehensive. Don't despair if it turns
out a little hard to understand to you. After all, it was John's idea, and
he is an engineer with years of experience. With all respect to him, I
must
admit, it's not the best starting example for novices. My reply suggesting
2
diodes and the RC thingy was rather addressed to John as a word of
caution:
not to leave to you building a circuit that is hard to trim. It was not
meant as an easy version either. It needs less trimming, but is more
complicated from the theory. Sorry for messing your understanding up with
this.

Dimitrij

Hey, no problem, I'm sure that learning this earlier will help me learn
other things along the way much faster.

Anyway, let's say we have the following circuit:

Vcc
|
|
-
| | R1
| |
-
|
o-------------------------base
|
V D1
-
|
|
|
[mic]
|
|
--- ground
-

What you are saying is that we look at the whole diode circuit, which would
be vertically from Vcc to ground, right? Let's say Vcc is 5 volts. Let's say
the mic has no resistance. So this creates a voltage divider (R1 and
D1+mic). R1 will drop constant voltage and D1 will drop .7 + Vmic, is this
right? So basically, the bottom voltage drop will vary (in effect a variable
resistance), so the voltage at base will vary as well. Is this what you were
getting at?

So basically, D1 is forward biased since top part of voltage divider is
higher than the bottom portion of D1 (basically the mic). So when D1 opens
up, the mic voltage will be subtracted from the voltage right before D1
(basically the voltage at the base) and that voltage will go to base. Is
this how it works?

So say mic is not generating any voltage at the moment (0V). So D1 drops .7
volts, and R1 will drop 5-.7=4.3V. Since we need only .7 volts at the base
to forward bias the transistor, R1 has to drop 4.3-.7=3.6V. So if there's
nothing coming from the mic we have .7 volts at the base. Let's say the mic
generates 10mV so that the voltage at the base will be .7+.010=.71V So if
the transistor has a beta of 20, then the voltage drop at the collector to
emitter will be .71-.7=.010*20=.20V

Does this seem right?

Hmm.. I'm pretty close to getting it, but I can't see how the diode works
its magic exactly. If after the drop at R1 the voltage is 1.4V (.7 to base +
..7 to drop accross D1) then let's say the mic generates .3 volts, then the
difference at the diode would be .7-.3=.4 which is not enough to forward
bias the diode and let current go through it, so .7+.7+.4 volts would go to
the base. Hmm... I start with hmm I end up with hmm... interesting. Can you
provide an analogy that I can follow to understand this concept?

Thanks Dimitrij!

--Viktor