transistor base input

L

lerameur

Guest
Hi all

I have an old analog alarm system. I am hooking up the bell output to
a microcontroller.
Although the output voltage for the bell fluctuates around 6 to 9
volts. I need either a steady 0 or 5v for the input of the
controller.
I made a circuit with 2 2n222 transistor, acting as an inverting
gate. The problem is that the spec sheet of these and most transistor
have a Veb of 5 volt. my Ve is at ground therefor my Veb exceeds
this. any ideas on how to get pass this issue.
Thought about a voltage divider at the output of th bell , but I am
scared that if voltage drops a bit below 6v, the inverter will not
pick it up as positive voltage.
thanks
k
 
lerameur wrote:
Hi all

I have an old analog alarm system. I am hooking up the bell output to
a microcontroller.
Although the output voltage for the bell fluctuates around 6 to 9
volts. I need either a steady 0 or 5v for the input of the
controller.
I made a circuit with 2 2n222 transistor, acting as an inverting
gate. The problem is that the spec sheet of these and most transistor
have a Veb of 5 volt. my Ve is at ground therefor my Veb exceeds
this. any ideas on how to get pass this issue.
Thought about a voltage divider at the output of th bell , but I am
scared that if voltage drops a bit below 6v, the inverter will not
pick it up as positive voltage.
thanks
Click to expand...
The Veb spec is a limit on how much reverse voltage that
junction can stand, not any normal operating voltage. If
the emitter is at zero volts, the transistor will switch on
when the base voltage is about 0.7 volts more positive than
ground, and the base-emitter junction becomes forward biased
enough for the base current to rise to about 1/20th of the
collector load current.

However, using a transistor as a switch will do nothing to
regulate the voltage of the collector load.

It would help us help you if we could see a schematic of
your circuit. Either you could post a link to your circuit,
draw one with text characters (and a fixed width font, (like
Courier) and post it here (this is made easy with a little
chunk of software:
http://www.tech-chat.de/ascii-circuits.html )
or post a normal graphic file representation on the
alt.binaries.schematics.electronic group, where attachments
are allowed.

--
Regards,

John Popelish
 
lerameur wrote:

ok good thanks for the link. below is my circuit . I have a 7805 that
regulates the 5v. The ground is common to all. The source is anywhere
between 6 to 9 volts. The source as mentionned in the circuit is the
output of the analog alarm system going to the siren. I will hooking
up the 5v regulator most probably on the batteries (12v) actin g as
backup and being recharge by the analog circuit.
Hope that helps you help me.
Click to expand...
Is the voltage swing labeled as "source, below also limited
to swing between +5 volts and ground? I guess I don't
understand why you need this circuit, at all. What prevents
you from connecting the "source" directly to the microprocessor?

VCC 5V
+
VCC 5v |
+ |
| .-.
| | |
.-. | |
| | '-'
| | | ___
'-' -------|___|-- to
microcontroller
| |
| ___ |/
|----|___|--|
| |
___ |/ |
Source--|___|- --------| |
|> |
|
| GND
|
GND
Click to expand...
Once I understand exactly what the purpose of this circuit
is, I might have more suggestions.


--
Regards,

John Popelish
 
lerameur wrote:
On Apr 30, 4:09 pm, John Popelish <jpopel...@rica.net> wrote:

Is the voltage swing labeled as "source, below also limited
to swing between +5 volts and ground? I guess I don't
understand why you need this circuit, at all. What prevents
you from connecting the "source" directly to the microprocessor?
(snip)
Well section 13.9 of the pic16F88 datasheet is copied below:
From what I can read, if I supply 5v to the chip, my input to the
microcontroller has to be between 4.4v and 5.6v.
So the circuit is there to have a steady output of 5v into the
microcontroller for any voltage between 6 to 9 v off alarm system
(source)
Click to expand...
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.

A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.
Click to expand...
Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.

The analog input, therefore, must be between
VSS and VDD.
Click to expand...
Yes.

If the input voltage deviates from this
range by more than 0.6V in either direction, one of the
diodes is forward biased and a latch-up condition may
occur.
Click to expand...
Yes.

A maximum source impedance of 10 k? is rec-
ommended for the analog sources.
Click to expand...
Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?

Any external com-
ponent connected to an analog input pin, such as a
capacitor or a Zener diode, should have very little
leakage current.
Click to expand...
Of else, what?

--
Regards,

John Popelish
 
John Popelish wrote:
lerameur wrote:
(snip)
Well section 13.9 of the pic16F88 datasheet is copied below:
(snip)
Click to expand...

http://ww1.microchip.com/downloads/en/DeviceDoc/30487c.pdf

--
Regards,

John Popelish
 
lerameur wrote:
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.

A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.
Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.
Because the input is from the analog alarm system, which varies
between 6 to 9 volts. I cannot use that as an input.


A maximum source impedance of 10 k? is rec-
ommended for the analog sources.
Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?
Just high or low, alarm on or off...
Click to expand...
So you need to distinguish between an output of 9 volts and
an output of 6 volts, with some decision voltage between those?

I suggest you use a two resistor divider to lower the input
range to about 1/3rd (3 to 2 volts) and do a one bit analog
to digital conversion of this range with one of the inputs
configured as a comparator. The comparators include an
internal reference voltage derived from the PIC supply
voltage, or can use a second input as the reference voltage
input (the decision voltage) so you could use an external
pot to set the detection point.

See section 13.0 for comparator setup and Tables 18.1 and
18.2 for the comparator DC specs. Note that the comparator
input voltage range is 0 to Vcc-1.5 volts. Note, also that
only the decision voltage must be within this range, since
the measured input will still give the correct result if it
goes more positive than this limit. So you may get by with
a 1/2 input voltage division (4.5 to 3 volts), or some
ration between 1/3rd and 1/2, as long as the decision
voltage is below 3.5 volts.

If the microprocessor might not be powered when the input is
active, you might want to add a Schottky diode from the
comparator input to Vcc to limit the input voltage to about
Vcc+0.3 volts.

--
Regards,

John Popelish
 
lerameur wrote:

[snip]

just as 5 volt.

When the voltage from my source is between 6v and 9 volts, my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high
Click to expand...
Just use your micro as a digital input; no analog required.

6-9vdc in --10K-----+------ micro
|
/\ 5v zener diode
|
Gnd

HTH
 
The Veb spec is a limit on how much reverse voltage that
junction can stand, not any normal operating voltage. If
the emitter is at zero volts, the transistor will switch on
when the base voltage is about 0.7 volts more positive than
ground, and the base-emitter junction becomes forward biased
enough for the base current to rise to about 1/20th of the
collector load current.

However, using a transistor as a switch will do nothing to
regulate the voltage of the collector load.

It would help us help you if we could see a schematic of
your circuit. Either you could post a link to your circuit,
draw one with text characters (and a fixed width font, (like
Courier) and post it here (this is made easy with a little
chunk of software:http://www.tech-chat.de/ascii-circuits.html)
or post a normal graphic file representation on the
alt.binaries.schematics.electronic group, where attachments
are allowed.

--
Regards,

John Popelish
Click to expand...
ok good thanks for the link. below is my circuit . I have a 7805 that
regulates the 5v. The ground is common to all. The source is anywhere
between 6 to 9 volts. The source as mentionned in the circuit is the
output of the analog alarm system going to the siren. I will hooking
up the 5v regulator most probably on the batteries (12v) actin g as
backup and being recharge by the analog circuit.
Hope that helps you help me.


VCC 5V
+
VCC 5v |
+ |
| .-.
| | |
.-. | |
| | '-'
| | | ___
'-' -------|___|-- to
microcontroller
| |
| ___ |/
|----|___|--|
| |>
___ |/ |
Source--|___|- --------| |
|> |
|
| GND
|
GND
 
lerameur wrote:
On Apr 30, 5:29 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.
A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.
Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.
Because the input is from the analog alarm system, which varies
between 6 to 9 volts. I cannot use that as an input.
A maximum source impedance of 10 k? is rec-
ommended for the analog sources.
Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?
Just high or low, alarm on or off...
So you need to distinguish between an output of 9 volts and
an output of 6 volts, with some decision voltage between those?
no

just as 5 volt.

When the voltage from my source is between 6v and 9 volts, my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high
Click to expand...
Man, it took you enough posts to make that simple
explanation of what you want. All you need is a series
resistor to limit the current from the source and a means to
clamp the input voltage to no more than Vcc +0.3 volts. You
could do that with a 10k to 100k series resistor and a 5.1
volt zener or a Schottky diode to Vcc. Set the PIC pin as a
digital input. If you want the low state to be more noise
immune, use a 2:1 voltage divider for the input, since the
decision voltage is between 0.8 volts and 2.0 volts (see
18.4, DC characteristics.)


--
Regards,

John Popelish
 
On Apr 30, 4:09 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
ok good thanks for the link. below is my circuit . I have a 7805 that
regulates the 5v. The ground is common to all. The source is anywhere
between 6 to 9 volts. The source as mentionned in the circuit is the
output of the analog alarm system going to the siren. I will hooking
up the 5v regulator most probably on the batteries (12v) actin g as
backup and being recharge by the analog circuit.
Hope that helps you help me.

Is the voltage swing labeled as "source, below also limited
to swing between +5 volts and ground? I guess I don't
understand why you need this circuit, at all. What prevents
you from connecting the "source" directly to the microprocessor?





VCC 5V
+
VCC 5v |
+ |
| .-.
| | |
.-. | |
| | '-'
| | | ___
'-' -------|___|-- to
microcontroller
| |
| ___ |/
|----|___|--|
| |
___ |/ |
Source--|___|- --------| |
|> |
|
| GND
|
GND

Once I understand exactly what the purpose of this circuit
is, I might have more suggestions.

--
Regards,

John Popelish
Click to expand...
Well section 13.9 of the pic16F88 datasheet is copied below:
From what I can read, if I supply 5v to the chip, my input to the
microcontroller has to be between 4.4v and 5.6v.
So the circuit is there to have a steady output of 5v into the
microcontroller for any voltage between 6 to 9 v off alarm system
(source)

A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS. The analog input, therefore, must be between
VSS and VDD. If the input voltage deviates from this
range by more than 0.6V in either direction, one of the
diodes is forward biased and a latch-up condition may
occur. A maximum source impedance of 10 k? is rec-
ommended for the analog sources. Any external com-
ponent connected to an analog input pin, such as a
capacitor or a Zener diode, should have very little
leakage current.
 
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.

A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.

Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.
Because the input is from the analog alarm system, which varies
Click to expand...
between 6 to 9 volts. I cannot use that as an input.


A maximum source impedance of 10 k? is rec-
ommended for the analog sources.

Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?
Just high or low, alarm on or off...
Click to expand...

k
 
lerameur wrote:

I think we have it now, thats sounds pretty good. Sorry for the bad
explanation, I thought I had it good enough.
Click to expand...
If it makes sense to you, then that's fine. If not, we'll
hit it some more.

As to "good enough", I probably have a lot more
possibilities floating around in my imagination than you
have, so you don't understand what I have to eliminate to
get to your case.

--
Regards,

John Popelish
 
On Apr 30, 5:29 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.

A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.
Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.
Because the input is from the analog alarm system, which varies
between 6 to 9 volts. I cannot use that as an input.

A maximum source impedance of 10 k? is rec-
ommended for the analog sources.
Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?
Just high or low, alarm on or off...

So you need to distinguish between an output of 9 volts and
an output of 6 volts, with some decision voltage between those?
no
Click to expand...

just as 5 volt.

When the voltage from my source is between 6v and 9 volts, my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high

ken
 
On Apr 30, 6:23 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
On Apr 30, 5:29 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.
A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.
Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.
Because the input is from the analog alarm system, which varies
between 6 to 9 volts. I cannot use that as an input.
A maximum source impedance of 10 k? is rec-
ommended for the analog sources.
Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?
Just high or low, alarm on or off...
So you need to distinguish between an output of 9 volts and
an output of 6 volts, with some decision voltage between those?
no

just as 5 volt.

When the voltage from my source is between 6v and 9 volts, my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high

Man, it took you enough posts to make that simple
explanation of what you want. All you need is a series
resistor to limit the current from the source and a means to
clamp the input voltage to no more than Vcc +0.3 volts. You
could do that with a 10k to 100k series resistor and a 5.1
volt zener or a Schottky diode to Vcc. Set the PIC pin as a
digital input. If you want the low state to be more noise
immune, use a 2:1 voltage divider for the input, since the
decision voltage is between 0.8 volts and 2.0 volts (see
18.4, DC characteristics.)
Click to expand...

I think we have it now, thats sounds pretty good. Sorry for the bad
explanation, I thought I had it good enough.

k
 
On Apr 30, 6:23 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
On Apr 30, 5:29 pm, John Popelish <jpopel...@rica.net> wrote:
lerameur wrote:
I assume this is a PIC input configured as a digital input,
based on your two transistor interface circuit.
A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.
Why would you use the input configured as an analog input,
instead of digital? Your circuit will output only two
states-- high and low.
Because the input is from the analog alarm system, which varies
between 6 to 9 volts. I cannot use that as an input.
A maximum source impedance of 10 k? is rec-
ommended for the analog sources.
Only if you need the full analog to digital converter
resolution over the full temperature range. What, exactly
do you want the microprocessor to know about the input signal?
Just high or low, alarm on or off...
So you need to distinguish between an output of 9 volts and
an output of 6 volts, with some decision voltage between those?
no

just as 5 volt.

When the voltage from my source is between 6v and 9 volts, my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high

Man, it took you enough posts to make that simple
explanation of what you want. All you need is a series
resistor to limit the current from the source and a means to
clamp the input voltage to no more than Vcc +0.3 volts. You
could do that with a 10k to 100k series resistor and a 5.1
volt zener or a Schottky diode to Vcc. Set the PIC pin as a
digital input. If you want the low state to be more noise
immune, use a 2:1 voltage divider for the input, since the
decision voltage is between 0.8 volts and 2.0 volts (see
18.4, DC characteristics.)
Click to expand...

I think we have it now, thats sounds pretty good. Sorry for the bad
explanation, I thought I had it good enough.

k
 
On Wed, 30 Apr 2008 21:55:53 GMT, Randy Day <randy.day@shaw.cax>
wrote:

lerameur wrote:

[snip]

just as 5 volt.

When the voltage from my source is between 6v and 9 volts, my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high

Just use your micro as a digital input; no analog required.

6-9vdc in --10K-----+------ micro
|
/\ 5v zener diode
|
Gnd

HTH
Click to expand...

Another thought - perhaps the 6 to 9 volts going to the siren is
actually AC or pulsed DC, rather than plain DC. In that case, you
would want a diode in series with the 10K resistor in the above
circuit, cathode towards the Zener, and a capacitor (0.1 uF or so,
perhaps) and a 100K resistor, both in parallel with the zener. This
would give a steady DC voltage to the microcontroller input while the
siren is operating.

(The "jumps anywhere between 6 to 9 volts" bit makes me think that the
siren drive is not DC.)


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
 
Peter Bennett wrote:

Just use your micro as a digital input; no analog required.

6-9vdc in --10K-----+------ micro
|
/\ 5v zener diode
|
Gnd

HTH

Another thought - perhaps the 6 to 9 volts going to the siren is
actually AC or pulsed DC, rather than plain DC. In that case, you
would want a diode in series with the 10K resistor in the above
circuit, cathode towards the Zener, and a capacitor (0.1 uF or so,
perhaps) and a 100K resistor, both in parallel with the zener. This
would give a steady DC voltage to the microcontroller input while the
siren is operating.

(The "jumps anywhere between 6 to 9 volts" bit makes me think that the
siren drive is not DC.)
Click to expand...
Good point.
 
On Apr 30, 2:55 pm, Randy Day <randy....@shaw.cax> wrote:
lerameur wrote:

[snip]

just as 5 volt.

When the  voltage from my source is between 6v and 9 volts,  my chip
reads high (or 5v)
otherwise the chip reads low or zero.
like I said, the source is the voltage going to the siren, so it is
almost always ZERO.
When siren goes off, the voltage jumps anywhere between 6 to 9 volts.
Then the chip should read high

Just use your micro as a digital input; no analog required.

6-9vdc in --10K-----+------ micro
                    |
                   /\ 5v zener diode
                    |
                   Gnd

HTH
Click to expand...
You should rethink the 10k resistor.
You chose to make your input signal the max specified impedance for
the chip -- without even allowing for the additional impedance of the
source signal, or just plain old wiggle room; which you can easily
afford. You should also consider zener current. A smaller resistor
is in order, maybe by an order of magnitude.
 
gearhead wrote:
On Apr 30, 2:55 pm, Randy Day <randy....@shaw.cax> wrote:

Just use your micro as a digital input; no analog required.

6-9vdc in --10K-----+------ micro
|
/\ 5v zener diode
|
Gnd

HTH

You should rethink the 10k resistor.
You chose to make your input signal the max specified impedance for
the chip -- without even allowing for the additional impedance of the
source signal, or just plain old wiggle room; which you can easily
afford. You should also consider zener current. A smaller resistor
is in order, maybe by an order of magnitude.
Click to expand...
Why?
The lower the resistor, the more the source will be loaded.

--
Regards,

John Popelish
 
On May 1, 2:42 pm, John Popelish <jpopel...@rica.net> wrote:
gearhead wrote:
On Apr 30, 2:55 pm, Randy Day <randy....@shaw.cax> wrote:
Just use your micro as a digital input; no analog required.

6-9vdc in --10K-----+------ micro
                    |
                   /\ 5v zener diode
                    |
                   Gnd

HTH

You should rethink the 10k resistor.
You chose to make your input signal the max specified impedance for
the chip -- without even allowing for the additional impedance of the
source signal, or just plain old wiggle room; which you can easily
afford.  You should also consider zener current.  A smaller resistor
is in order, maybe by an order of magnitude.

Why?
The lower the resistor, the more the source will be loaded.

--
Regards,

John Popelish- Hide quoted text -

- Show quoted text -
Click to expand...
Well, it's a bell output from an alarm. I guess I was operating on
the assumption that it has the power to drive a physical bell, so it
will hardly be able to tell the difference between 200uA or 2mA;
either way you'd draw magnitudes less than the alarm's potential
output, so it hardly matters.
But if he feeds a maximum specified impedance to the micro input, he
risks having the circuit not work.
But the OP knows more about his alarm than I do. Let him work it out.
 
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