transistor

[John Popelish]

electricked wrote:

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?

Stop right there and forget about D1 opening up. It has a continuous
supply of current passing through it, regardless of the tiny voltage
the mic is producing. The diode voltage drop just bounces up and down
from the variations the mic is producing, while the current through it
is almost unvarying. That is because the mic voltage is tiny compared
to the big voltage of Vcc.

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.

Right. That means the top of the diode is .7 volts more positive than
ground.

Since we need only .7 volts at the base
to forward bias the transistor, R1 has to drop 4.3-.7=3.6V.

Wrong. The base needs .7 volts more positive than its emitter which
is at ground, in this case) to become forward biased. After all, it
is just another diode.

Vcc

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

So the current through D1 drops to about half of what it was, and the
other half detours through the BE diode. So both junctions are
forward biased at the same time. When the mic produces low level AC,
it changes the division from half and half to some other split. When
it produces a negative voltage, D1 has more drop than the BE junction,
so hogs more of the current from R1. When the mic produces positive
voltage, D1 drops less voltage than the BE junction, so it hogs less
of the current from R1 and more goes through the BE junction. In
effect, a variation in current that looks just like the mic current
gets moved over to the base junction, even though the actual charge
that is going through the transistor is coming from Vcc. Its
variation is based on the mic voltage.

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

Beta deals with the ratio of collector current compared to base
current, do first you have to calculate how much the base current will
vary when the voltage changes from .7 to .71 volts. Then you can
apply beta to that change in base current. There is also a formula
that relates collector current directly to base voltage (see
discussion going on about whether base current or voltage controls
collector current going on in the "High gain amplifier?" thread.

(snip)
--
John Popelish

 

[Dimitrij Klingbeil]

"electricked" <no_emails_please> wrote in message
news4Wdnf_ndPiF5MLdRVn-ig@comcast.com...

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?

Exactly! That's it.

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?

Look up John's explaination. D1 never really opens up. D1 and BE are in
parallel and share the voltage. It's the current that splits up.

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.

No. R1 still has to drop 4.3V. The BE junction and the diode+mic circuit are
in parallel, so their total voltage is 0.7V, -> 4.3V left for R1.

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

Beta is a current gain factor, so replace the voltages with current values.
The calculation gets awkward since a transistor junction's resistance is not
linear, neither is a diode. The only feasible way is to use a simulation
program to get the values, doing the math by hand for such a circuit without
simplifying away more than 75% of it would take aeons.

Does this seem right?

Not really, see voltage vs. current. See also the detailed explaination John
has just written.

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)

Wrong assumption, base and diode are not in series.

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?

Not applicable because of mistakes above.

I see, you get the general idea right, you only mixed up voltage vs. current
and took a parallel connection for a series one. When you get this straight,
you're done.

Let's take a look at a schematic:

Vcc = 5V
|
|
-
| | R1
| |
-
|
o-----o---- Out
| |
V D1 V D2
- -
| |
o-----o---- Gnd

For simplicity I left out the mic. It's not essential here.

D2 is actually meant to be the base-emitter junction of the transistor, the
collector being left out since it does not influence the base current.

Now you should see that R1 drops 4.3V and the diodes together 0.7V since
they form a simple parallel circuit. Their voltage does not double up. That
is Out = 0.7V. Remember the basics of parallel circuits: Each part's voltage
equals the total voltage.

Now let's turn the diode and the mic around:

Vcc = 5V
|
|
-
| | R1
| |
-
|
o---- Out 2
|
[mic]
|
0---- Out 1
|
V D1
-
|
o------ Gnd

Transistor left out for simplicity.

As you probably see, the voltage at Out1 is constant (0.7V), well, strictly
speaking, it isn't, but let's assume the variations to be very low, they
indeed are.

To this voltage the mic voltage gets added at Out2. This means Out2 = Out1 +
X, with X being the mic's voltage.

As with every series connection, it does not matter, which way we place the
diode and mic relative to each other, this circuit is still equivalent to
yours, only yours does not have a constant Out 1.

Now let's place the transistor where it belongs:

Vcc
|
|
-
| | R1
| | .--> wherever
- |
| C
o-------B
| E
[mic] |
| |
V D1 |
- |
| |
o--------o----- Gnd

It should be evident that the former Out 2 is applied to the base directly.

So far, I hope, the circuit was simple. D1 provides constant bias and mic
creates the difference to be amplified.

There' s a proverb about lightning: "It does not hurt until it hits".

Now, if you attempt the circuit above, reality is going to hit you.

You now ask yourself something like "how exaclty", right?

Well, a diode's resistance is not linear, you know. So is a transistor's.
Now assume the voltage drop on the diode is just a little lower than on BE
for a given current. Since this is a parallel circuit and a diode's
differential resistance is very low when the diode is open, the diode would
get all the current and the base would get none (simplified of course). Or
assume the opposite: The transistor's drop is lower. Then the diode would be
out of service. Unfortunately these parts are sensitive, it only tages
millivolts to get them out of balance. The mic's voltage however is very
small, so if the parts are even slightly unbalanced, it would not be enough
to drive the transistor reliably. Small changes can be made by varying R1.
This will force different currents through the mic and if the mic's
resistance is enough, one could thus adjust the circuit into equilibrium.
However this will likely not last. Semiconductor devices' properties vary
greatly with temperature. If the transistor, for example, gets a little
warmer than the diode, its drop will decrease. This will force the base
current to get higher and the transistor to heat up further, while the
unloaded diode cools off. Techs often call it "thermal runaway". (Sorry,
John, for such an overstatement, I only meant the general idea. The circuit
will of course have enough stability with small signal transistors and small
loads. I doubt it however slightly for an 8ohm speaker connected directly,
here a small transistor may well overheat.)

Now let's not get too deep into the details. The description above is just
to give you a general idea about what I meant by "hard to trim". It's too
much simplified anyway. Actually there is a whole range besides the exact
equilibrium that is still usable, but even this range is hard to keep the
circuit within in a stable way.

That's why I suggested that second diode and RC circuit, simply to stabilize
the amp. I won't explain that in detail just now, please ask some specific
questions. I'll try to answer them later.

This is my last reply for today, here in central Europe, it's past two in
the night, and I really need some sleep.

Dimitrij

 

[John Popelish]

electricked wrote:

I think I found my mistake. I was mixing up the voltage and current
properties of the diode.

So in effect the diode is like a variable resistor when it comes to current,
but it has a constant voltage drop if enough voltage is applied. Basically,
the diode acts as a constant resistance once it reaches it's forward biased
voltage, is that it?

Not quite. Diodes (any PN junction) have a nonlinear voltage current
relationship. Increase the current through them and the voltage drop
across them increases, just not in proportion to the current.

Take a look at the garden variety 1 amp rectifier diode data sheet:
http://www.fairchildsemi.com/ds/1N/1N4004.pdf

Focus on the forward characteristics graph on page 2. Note that the
voltage axis along the bottom is linear, ranging from .6 to 1.4 volts,
while the current axis is logarithmic, ranging from .01 amp to 20 amps
( a ratio of 2000 to 1). If the diode were perfect, the line would
not be a curve, but would just continue up at the slope it has at the
lower left corner. The sweep to the right in the right corner is
there because there is a bit of ohmic resistance in series with the
ideal diode. Draw the straight line over the curve and you can
estimate the extra voltage caused by the resistance and calculate
about what that resistance is to cause that extra voltage drop.

Anyway, this is representative of how all diodes act in forward bias.

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

So the current through D1 drops to about half of what it was, and the
other half detours through the BE diode. So both junctions are
forward biased at the same time. When the mic produces low level AC,
it changes the division from half and half to some other split. When
it produces a negative voltage, D1 has more drop than the BE junction,
so hogs more of the current from R1. When the mic produces positive
voltage, D1 drops less voltage than the BE junction, so it hogs less
of the current from R1 and more goes through the BE junction. In
effect, a variation in current that looks just like the mic current
gets moved over to the base junction, even though the actual charge
that is going through the transistor is coming from Vcc. Its
variation is based on the mic voltage.

Got a question. What if the mic produced .10 volts. How will the situation
changed around the diode? It will be reverse biased since its cathode is
more positive than it's anode, no?

If the diodes start with .7 volts across each, and you push one up by
..01 volt, that voltage will be divided roughly equally across both
junctions. So the drop across D1 will be reduced by about .005 volts,
and the drop across the BE junction will increase by about .005
volts. Both will still be somewhat forward biased. If the mic
produced +1 volt, then this description falls down, and the transistor
base would be a bit above .7 volts (sucking up all the current that
falls through R1) and D1 would have about .3 volts of reverse bias
across it and be essentially non conducting. So this signal coupling
method has a limited peak input current - that being all the current
that falls through R1 with Vcc - a diode drop across it.

So if the mic produces -.5 volts then D1 would have to drop .7-(-.5)=1.2V? I
think is right since the top of the diode is at .7
and the bottom is -.5 so
that makes a difference of 1.2 volts which is more than enough to forward
bias the diode, thus it will drive more current compared to D2, which is at
.7

No. The diode D1 current would increase to about twice what it had
been (Look at that diode curve to estimate what doubling the current
does to the forward drop. Hint - it ain't much.) because it would
suck up all the current from R1, while the BE junction would fall far
enough below .7 volts that it would conduct almost no current.


By the way, I am thoroughly enjoying this discussion.
--
John Popelish

 

[electricked]

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?

Stop right there and forget about D1 opening up. It has a continuous
supply of current passing through it, regardless of the tiny voltage
the mic is producing. The diode voltage drop just bounces up and down
from the variations the mic is producing, while the current through it
is almost unvarying. That is because the mic voltage is tiny compared
to the big voltage of Vcc.

I think I found my mistake. I was mixing up the voltage and current
properties of the diode.

So in effect the diode is like a variable resistor when it comes to current,
but it has a constant voltage drop if enough voltage is applied. Basically,
the diode acts as a constant resistance once it reaches it's forward biased
voltage, is that it?

Diode properties: diode will drop any voltage up to .7, is this right? If
the voltage accross it was .4 it will drop it but the current will increase
more linearly once the diode is forward-biased. I don't know if this is
right but that's how I'm looking at it now. Does it make sense?

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.

Right. That means the top of the diode is .7 volts more positive than
ground.

Since we need only .7 volts at the base
to forward bias the transistor, R1 has to drop 4.3-.7=3.6V.

Wrong. The base needs .7 volts more positive than its emitter which
is at ground, in this case) to become forward biased. After all, it
is just another diode.


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

So the current through D1 drops to about half of what it was, and the
other half detours through the BE diode. So both junctions are
forward biased at the same time. When the mic produces low level AC,
it changes the division from half and half to some other split. When
it produces a negative voltage, D1 has more drop than the BE junction,
so hogs more of the current from R1. When the mic produces positive
voltage, D1 drops less voltage than the BE junction, so it hogs less
of the current from R1 and more goes through the BE junction. In
effect, a variation in current that looks just like the mic current
gets moved over to the base junction, even though the actual charge
that is going through the transistor is coming from Vcc. Its
variation is based on the mic voltage.

Got a question. What if the mic produced .10 volts. How will the situation
changed around the diode? It will be reverse biased since its cathode is
more positive than it's anode, no?

So if the mic produces -.5 volts then D1 would have to drop .7-(-.5)=1.2V? I
think is right since the top of the diode is at .7 and the bottom is -.5 so
that makes a difference of 1.2 volts which is more than enough to forward
bias the diode, thus it will drive more current compared to D2, which is at
..7

It's starting to make more and more sense... I think...

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

Beta deals with the ratio of collector current compared to base
current, do first you have to calculate how much the base current will
vary when the voltage changes from .7 to .71 volts. Then you can
apply beta to that change in base current. There is also a formula
that relates collector current directly to base voltage (see
discussion going on about whether base current or voltage controls
collector current going on in the "High gain amplifier?" thread.

(snip)
--
John Popelish

Gotcha! Beta is for the current ratios.

--Viktor

 

[electricked]

"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:405f8fac_2@news.arcor-ip.de...

(snip)

Not applicable because of mistakes above.

I see, you get the general idea right, you only mixed up voltage vs.
current
and took a parallel connection for a series one. When you get this
straight,
you're done.

I see my mistakes now and it's more logical now.

Let's take a look at a schematic:

Vcc = 5V
|
|
-
| | R1
| |
-
|
o-----o---- Out
| |
V D1 V D2
- -
| |
o-----o---- Gnd

For simplicity I left out the mic. It's not essential here.

D2 is actually meant to be the base-emitter junction of the transistor,
the
collector being left out since it does not influence the base current.

Now you should see that R1 drops 4.3V and the diodes together 0.7V since
they form a simple parallel circuit. Their voltage does not double up.
That
is Out = 0.7V. Remember the basics of parallel circuits: Each part's
voltage
equals the total voltage.

R1 drops 4.3 and the diodes drop .7 because we set it up like that, right?
For example if we make R1 to drop 4, then 1V drops across both diodes, and
more current will flow through them. Same if R1 dropped 4.5, the diodes will
drop .5 and much less current will flow since they are not fully "open" at
that point. I see it now. This is the "AHA!" moment I've been hoping for.
I'm glad it came sooner than later, thanks to you and John,

Now let's turn the diode and the mic around:

Vcc = 5V
|
|
-
| | R1
| |
-
|
o---- Out 2
|
[mic]
|
0---- Out 1
|
V D1
-
|
o------ Gnd

Transistor left out for simplicity.

As you probably see, the voltage at Out1 is constant (0.7V), well,
strictly
speaking, it isn't, but let's assume the variations to be very low, they
indeed are.

So Out1 is .7 since no matter how high the voltage is (as long as it's
higher or equal to .7) the diode will drop only .7V (assuming the mic is 0),
correct?

So if the mic produces .3 volts, is that measured between Out2 and Out1 or
Out2 and ground?
What if the mic voltage is -.3? Say the mic is producing -.3 volts, and Out2
is .7. Does that mean that .7-.3=.4V are dropped across D1?

Basically, my question is, when you say that the mic is producing voltage,
between which two points does that voltage measure AND if the voltage the
mic produces is negative, how is it measured/related to the different points
in the circuit above?

To this voltage the mic voltage gets added at Out2. This means Out2 = Out1
+
X, with X being the mic's voltage.

Is it the same when the mic voltage is negative? Wouldn't then the voltage
at Out2 = Out1 - X? I'm still trying to grasp potential differences,
especially when negative voltages are involved. Can you give me some analogy
to give me a good grounding on this topic?

As with every series connection, it does not matter, which way we place
the
diode and mic relative to each other, this circuit is still equivalent to
yours, only yours does not have a constant Out 1.

Now let's place the transistor where it belongs:

Vcc
|
|
-
| | R1
| | .--> wherever
- |
| C
o-------B
| E
[mic] |
| |
V D1 |
- |
| |
o--------o----- Gnd

It should be evident that the former Out 2 is applied to the base
directly.

So far, I hope, the circuit was simple. D1 provides constant bias and mic
creates the difference to be amplified.

There' s a proverb about lightning: "It does not hurt until it hits".

Now, if you attempt the circuit above, reality is going to hit you.

You now ask yourself something like "how exaclty", right?

Well, a diode's resistance is not linear, you know. So is a transistor's.
Now assume the voltage drop on the diode is just a little lower than on BE
for a given current. Since this is a parallel circuit and a diode's
differential resistance is very low when the diode is open, the diode
would
get all the current and the base would get none (simplified of course). Or
assume the opposite: The transistor's drop is lower. Then the diode would
be
out of service.

Shouldn't this be the other way around? If diode, D1, is dropping less than
another diode, D2, connected in parallel with D1, then wouldn't more current
go through D2?

Say D1 drops .4 and D2 drops .7, then D2 would allow higher current to pass
so D1 is left with almost nothing.

Unfortunately these parts are sensitive, it only tages
millivolts to get them out of balance. The mic's voltage however is very
small, so if the parts are even slightly unbalanced, it would not be
enough
to drive the transistor reliably. Small changes can be made by varying R1.
This will force different currents through the mic and if the mic's
resistance is enough, one could thus adjust the circuit into equilibrium.
However this will likely not last. Semiconductor devices' properties vary
greatly with temperature. If the transistor, for example, gets a little
warmer than the diode, its drop will decrease. This will force the base
current to get higher and the transistor to heat up further, while the
unloaded diode cools off. Techs often call it "thermal runaway". (Sorry,
John, for such an overstatement, I only meant the general idea. The
circuit
will of course have enough stability with small signal transistors and
small
loads. I doubt it however slightly for an 8ohm speaker connected directly,
here a small transistor may well overheat.)

Now let's not get too deep into the details. The description above is just
to give you a general idea about what I meant by "hard to trim". It's too
much simplified anyway. Actually there is a whole range besides the exact
equilibrium that is still usable, but even this range is hard to keep the
circuit within in a stable way.

That's why I suggested that second diode and RC circuit, simply to
stabilize
the amp. I won't explain that in detail just now, please ask some specific
questions. I'll try to answer them later.

This is my last reply for today, here in central Europe, it's past two in
the night, and I really need some sleep.

Dimitrij

Hey Dimitrij, have a good night and thanks for all the great help you've
provided. I'm sure I'll have a lot of questions along the way, so if you
feel like answering to any of them, be my guest

--Viktor

 

[electricked]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405FC3B8.ABA72839@rica.net...

electricked wrote:

I think I found my mistake. I was mixing up the voltage and current
properties of the diode.

So in effect the diode is like a variable resistor when it comes to
current,
but it has a constant voltage drop if enough voltage is applied.
Basically,
the diode acts as a constant resistance once it reaches it's forward
biased
voltage, is that it?

Not quite. Diodes (any PN junction) have a nonlinear voltage current
relationship. Increase the current through them and the voltage drop
across them increases, just not in proportion to the current.

I was looking at the following page when I wrote that:
http://www.phys.ualberta.ca/~gingrich/phys395/notes/node58.html#SECTION00521000000000000000

which I guess is a simplified version and since you mentioned earlier the
resistance is non-linear I should've known better than that. Anyhow...

Take a look at the garden variety 1 amp rectifier diode data sheet:
http://www.fairchildsemi.com/ds/1N/1N4004.pdf

Focus on the forward characteristics graph on page 2. Note that the
voltage axis along the bottom is linear, ranging from .6 to 1.4 volts,
while the current axis is logarithmic, ranging from .01 amp to 20 amps
( a ratio of 2000 to 1). If the diode were perfect, the line would
not be a curve, but would just continue up at the slope it has at the
lower left corner. The sweep to the right in the right corner is
there because there is a bit of ohmic resistance in series with the
ideal diode. Draw the straight line over the curve and you can
estimate the extra voltage caused by the resistance and calculate
about what that resistance is to cause that extra voltage drop.

If I draw a straight line at the slope of the lower left side I get about
1.1V at 20amps.

So the extra voltage needed at that current due to the resistance is
1.4-1.1=.3volts.

What do you mean by "what that resistance is to cause that extra voltage
drop?"

How do I calculate that? I'm not sure what I should be looking at.

What I gather from that graph is that the resistance decreases with increase
in voltage.

What about the current through the diode from 0 to 0.6? Is it virtually 0
amps there?

So all this means that I can't use OHM's law to calculate the relationships
between changing current and voltage but have to look at a datasheet. Great!

Anyway, this is representative of how all diodes act in forward bias.


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

So the current through D1 drops to about half of what it was, and the
other half detours through the BE diode. So both junctions are
forward biased at the same time. When the mic produces low level AC,
it changes the division from half and half to some other split. When
it produces a negative voltage, D1 has more drop than the BE junction,
so hogs more of the current from R1. When the mic produces positive
voltage, D1 drops less voltage than the BE junction, so it hogs less
of the current from R1 and more goes through the BE junction. In
effect, a variation in current that looks just like the mic current
gets moved over to the base junction, even though the actual charge
that is going through the transistor is coming from Vcc. Its
variation is based on the mic voltage.

Got a question. What if the mic produced .10 volts. How will the
situation
changed around the diode? It will be reverse biased since its cathode is
more positive than it's anode, no?

If the diodes start with .7 volts across each, and you push one up by
.01 volt, that voltage will be divided roughly equally across both
junctions. So the drop across D1 will be reduced by about .005 volts,
and the drop across the BE junction will increase by about .005
volts. Both will still be somewhat forward biased. If the mic
produced +1 volt, then this description falls down, and the transistor
base would be a bit above .7 volts (sucking up all the current that
falls through R1) and D1 would have about .3 volts of reverse bias
across it and be essentially non conducting. So this signal coupling
method has a limited peak input current - that being all the current
that falls through R1 with Vcc - a diode drop across it.

How do you calculate the .3 volts of reverse bias on D1?

So if the mic produces -.5 volts then D1 would have to drop
..7-(-.5)=1.2V? I
think is right since the top of the diode is at .7
and the bottom is -.5 so
that makes a difference of 1.2 volts which is more than enough to
forward
bias the diode, thus it will drive more current compared to D2, which is
at
.7

No. The diode D1 current would increase to about twice what it had
been (Look at that diode curve to estimate what doubling the current
does to the forward drop. Hint - it ain't much.) because it would
suck up all the current from R1, while the BE junction would fall far
enough below .7 volts that it would conduct almost no current.

I think I'm not understanding how this works exactly. How is the mic voltage
measured? Do I hook up the top of mic to ground, or top of mic to bottom of
mic? If the mic was a pump that pushes up or down depending on voltage
polarity, how would you desribe its behavior in accordance to the rest of
the circuit?

By the way, I am thoroughly enjoying this discussion.

Are you serious? If I knew as much as you do, I'd be bored out of my mind
;-)

--
John Popelish

Thanks John!

--Viktor

 

[John Popelish]

electricked wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:405FC3B8.ABA72839@rica.net...

Not quite. Diodes (any PN junction) have a nonlinear voltage current
relationship. Increase the current through them and the voltage drop
across them increases, just not in proportion to the current.

I was looking at the following page when I wrote that:
http://www.phys.ualberta.ca/~gingrich/phys395/notes/node58.html#SECTION00521000000000000000

which I guess is a simplified version and since you mentioned earlier the
resistance is non-linear I should've known better than that. Anyhow...

Take a look at the garden variety 1 amp rectifier diode data sheet:
http://www.fairchildsemi.com/ds/1N/1N4004.pdf

Focus on the forward characteristics graph on page 2. Note that the
voltage axis along the bottom is linear, ranging from .6 to 1.4 volts,
while the current axis is logarithmic, ranging from .01 amp to 20 amps
( a ratio of 2000 to 1). If the diode were perfect, the line would
not be a curve, but would just continue up at the slope it has at the
lower left corner. The sweep to the right in the right corner is
there because there is a bit of ohmic resistance in series with the
ideal diode. Draw the straight line over the curve and you can
estimate the extra voltage caused by the resistance and calculate
about what that resistance is to cause that extra voltage drop.

If I draw a straight line at the slope of the lower left side I get about
1.1V at 20amps.

So the extra voltage needed at that current due to the resistance is
1.4-1.1=.3volts.

So what resistance drops .3 volts with 20 amps going through it?

What do you mean by "what that resistance is to cause that extra voltage
drop?"

How do I calculate that? I'm not sure what I should be looking at.

What I gather from that graph is that the resistance decreases with increase
in voltage.

What about the current through the diode from 0 to 0.6? Is it virtually 0
amps there?

Just extend the slope of the lower left end beyond the lower left
corner of the graph.

I estimate between .4 and .5 volts for .001 amp.

So all this means that I can't use OHM's law to calculate the relationships
between changing current and voltage but have to look at a datasheet. Great!

There is a different formula for diodes. If you start with the
current, you use a log function to find the voltage. If you start
with voltage you need an exponential function to solve for current.
Once you get used to it, it isn't much harder than ohm's law for
resistors.

(snip)

If the diodes start with .7 volts across each, and you push one up by
.01 volt, that voltage will be divided roughly equally across both
junctions. So the drop across D1 will be reduced by about .005 volts,
and the drop across the BE junction will increase by about .005
volts. Both will still be somewhat forward biased. If the mic
produced +1 volt, then this description falls down, and the transistor
base would be a bit above .7 volts (sucking up all the current that
falls through R1) and D1 would have about .3 volts of reverse bias
across it and be essentially non conducting. So this signal coupling
method has a limited peak input current - that being all the current
that falls through R1 with Vcc - a diode drop across it.

How do you calculate the .3 volts of reverse bias on D1?

The BE junction sucks up all the current from R1 while its forward
voltage rises only slightly (that nonlinear logarithmic function) But
the mic keeps raising the voltage on the bottom of D1, till the
voltage across the diode reverses (0.7 on the top and + 1 volt on the
bottom) so the bottom is more positive than the top. Reverse biased.

(snip)

No. The diode D1 current would increase to about twice what it had
been (Look at that diode curve to estimate what doubling the current
does to the forward drop. Hint - it ain't much.) because it would
suck up all the current from R1, while the BE junction would fall far
enough below .7 volts that it would conduct almost no current.

I think I'm not understanding how this works exactly. How is the mic voltage
measured? Do I hook up the top of mic to ground, or top of mic to bottom of
mic?

You connect an instantaneous reading volt meter across the mic. An
oscilloscope is an example of such a volt meter.

If the mic was a pump that pushes up or down depending on voltage
polarity, how would you desribe its behavior in accordance to the rest of
the circuit?

By the way, I am thoroughly enjoying this discussion.

Are you serious? If I knew as much as you do, I'd be bored out of my mind
;-)

Lets go back to your model of a diode as an overflowing tank, except
lets cut a V notch in the side of the can to produce the current curve
versus level like a diode. The two diodes in parallel are like two
tanks with a connecting pipe that lets current inflow go to either or
both tanks, depending on their levels (lower level hogs more of the
inflow).

flow from R1
v
__ __ | | __ __
| \ / | | | | \ / |
|~~V~~|--- ---|~~V~~|
| |---------| |
|_____| _|_____|__
| | | table |
|jack | | |
_____|(mic)|________| |______
ground

Now, what will happen to the divided flow from R1 if you raise or
lower the jack (add positive or negative voltage)? The table is just
to balance the height of the drawing of the jack. The mic has no
height (zero volts, average) and pushes above or lowers below ground,
but that is hard to draw.

The idea is to vary the flow to the right tank as the jack goes up and
down, since height corresponds to voltage in this model.

--
John Popelish

 

[electricked]

"John Popelish" <jpopelish@rica.net> wrote in message
news:405FE2B3.5BDA7736@rica.net...

electricked wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:405FC3B8.ABA72839@rica.net...

Not quite. Diodes (any PN junction) have a nonlinear voltage current
relationship. Increase the current through them and the voltage drop
across them increases, just not in proportion to the current.

I was looking at the following page when I wrote that:

http://www.phys.ualberta.ca/~gingrich/phys395/notes/node58.html#SECTION00521000000000000000

which I guess is a simplified version and since you mentioned earlier
the
resistance is non-linear I should've known better than that. Anyhow...

Take a look at the garden variety 1 amp rectifier diode data sheet:
http://www.fairchildsemi.com/ds/1N/1N4004.pdf

Focus on the forward characteristics graph on page 2. Note that the
voltage axis along the bottom is linear, ranging from .6 to 1.4 volts,
while the current axis is logarithmic, ranging from .01 amp to 20 amps
( a ratio of 2000 to 1). If the diode were perfect, the line would
not be a curve, but would just continue up at the slope it has at the
lower left corner. The sweep to the right in the right corner is
there because there is a bit of ohmic resistance in series with the
ideal diode. Draw the straight line over the curve and you can
estimate the extra voltage caused by the resistance and calculate
about what that resistance is to cause that extra voltage drop.

If I draw a straight line at the slope of the lower left side I get
about
1.1V at 20amps.

So the extra voltage needed at that current due to the resistance is
1.4-1.1=.3volts.

So what resistance drops .3 volts with 20 amps going through it?

What do you mean by "what that resistance is to cause that extra voltage
drop?"

How do I calculate that? I'm not sure what I should be looking at.

What I gather from that graph is that the resistance decreases with
increase
in voltage.

What about the current through the diode from 0 to 0.6? Is it virtually
0
amps there?

Just extend the slope of the lower left end beyond the lower left
corner of the graph.

I estimate between .4 and .5 volts for .001 amp.

So all this means that I can't use OHM's law to calculate the
relationships
between changing current and voltage but have to look at a datasheet.
Great!


There is a different formula for diodes. If you start with the
current, you use a log function to find the voltage. If you start
with voltage you need an exponential function to solve for current.
Once you get used to it, it isn't much harder than ohm's law for
resistors.

(snip)

If the diodes start with .7 volts across each, and you push one up by
.01 volt, that voltage will be divided roughly equally across both
junctions. So the drop across D1 will be reduced by about .005 volts,
and the drop across the BE junction will increase by about .005
volts. Both will still be somewhat forward biased. If the mic
produced +1 volt, then this description falls down, and the transistor
base would be a bit above .7 volts (sucking up all the current that
falls through R1) and D1 would have about .3 volts of reverse bias
across it and be essentially non conducting. So this signal coupling
method has a limited peak input current - that being all the current
that falls through R1 with Vcc - a diode drop across it.

How do you calculate the .3 volts of reverse bias on D1?

The BE junction sucks up all the current from R1 while its forward
voltage rises only slightly (that nonlinear logarithmic function) But
the mic keeps raising the voltage on the bottom of D1, till the
voltage across the diode reverses (0.7 on the top and + 1 volt on the
bottom) so the bottom is more positive than the top. Reverse biased.

(snip)

No. The diode D1 current would increase to about twice what it had
been (Look at that diode curve to estimate what doubling the current
does to the forward drop. Hint - it ain't much.) because it would
suck up all the current from R1, while the BE junction would fall far
enough below .7 volts that it would conduct almost no current.

I think I'm not understanding how this works exactly. How is the mic
voltage
measured? Do I hook up the top of mic to ground, or top of mic to bottom
of
mic?

You connect an instantaneous reading volt meter across the mic. An
oscilloscope is an example of such a volt meter.

If the mic was a pump that pushes up or down depending on voltage
polarity, how would you desribe its behavior in accordance to the rest
of
the circuit?

By the way, I am thoroughly enjoying this discussion.

Are you serious? If I knew as much as you do, I'd be bored out of my
mind
;-)

Lets go back to your model of a diode as an overflowing tank, except
lets cut a V notch in the side of the can to produce the current curve
versus level like a diode. The two diodes in parallel are like two
tanks with a connecting pipe that lets current inflow go to either or
both tanks, depending on their levels (lower level hogs more of the
inflow).

flow from R1
v
__ __ | | __ __
| \ / | | | | \ / |
|~~V~~|--- ---|~~V~~|
| |---------| |
|_____| _|_____|__
| | | table |
|jack | | |
_____|(mic)|________| |______
ground

Now, what will happen to the divided flow from R1 if you raise or
lower the jack (add positive or negative voltage)? The table is just
to balance the height of the drawing of the jack. The mic has no
height (zero volts, average) and pushes above or lowers below ground,
but that is hard to draw.

The idea is to vary the flow to the right tank as the jack goes up and
down, since height corresponds to voltage in this model.

--
John Popelish

Thanks John!

I'll keep working at it.

--Viktor