Transistor amp question ---- not homework

Discussion in 'Electrical Engineering' started by Peter, Sep 5, 2005.

1. PeterGuest

I've posted a circuit on my webpage for everyone to view since I don't
think I should upload pictures here.

http://home.comcast.net/~bostonpeter73/

I'm trying to understand how to find Vout. Last night I played around with
loop equations, assumptions, and after hours of trying I had no luck.

The biggest part that is killing me is that 68ohm resistor up top because
you can't just say 20 volts goes to the left part and the right part of the
circuit. I know approx. what I should get because I plugged this into
MultiSim, but that's cheating in my book, I should know how to calculate
this circuit.

If anyone knows the best way to come up with the answer, please let me
know.

Peter, Sep 5, 2005

2. ehsjrGuest

Peter wrote:
> I've posted a circuit on my webpage for everyone to view since I don't
> think I should upload pictures here.
>
> http://home.comcast.net/~bostonpeter73/
>
>
> I'm trying to understand how to find Vout. Last night I played around with
> loop equations, assumptions, and after hours of trying I had no luck.
>
> The biggest part that is killing me is that 68ohm resistor up top because
> you can't just say 20 volts goes to the left part and the right part of the
> circuit. I know approx. what I should get because I plugged this into
> MultiSim, but that's cheating in my book, I should know how to calculate
> this circuit.
>
>
> If anyone knows the best way to come up with the answer, please let me
> know.
>
>
>

Here's how to "sketch" it in a post, and a hint
at the bottom:

o +12
|
[R1] 68
|
+------------+
| |
[R2] 8.25K [R5] 150
| |
| e/
+-----------| Q1 2N3906
| \
| |
[R3] 36.4K +---- Vout
| |
| /
+-----------| Q2 2N3904
| e\
| |
[R4] 8.25K [R6] 150
| |
+------------+
|
Gnd

Hint: You were concerned about the left hand side and the
68 ohm resistor. What is the lowest possible voltage,
with respect to ground, at the bottom of the 68 ohm
resistor? Can you analyze the circuit, if the 68 ohm
resistor was replaced by a jumper wire?

Ed

ehsjr, Sep 5, 2005

3. PeterGuest

>
> Hint: You were concerned about the left hand side and the
> 68 ohm resistor. What is the lowest possible voltage,
> with respect to ground, at the bottom of the 68 ohm
> resistor? Can you analyze the circuit, if the 68 ohm
> resistor was replaced by a jumper wire?
>
> Ed
>

I believe it can be done. you pretty much ONLY have I1 on the left and I2
on the right because the emitter current of Q2 is the collector current of
Q1 and the emitter current of Q1. So it's almost a parallel circuit.

Peter, Sep 5, 2005
4. ehsjrGuest

Peter wrote:
>>Hint: You were concerned about the left hand side and the
>>68 ohm resistor. What is the lowest possible voltage,
>>with respect to ground, at the bottom of the 68 ohm
>>resistor? Can you analyze the circuit, if the 68 ohm
>>resistor was replaced by a jumper wire?
>>
>>Ed
>>

>
>
> I believe it can be done. you pretty much ONLY have I1 on the left and I2
> on the right because the emitter current of Q2 is the collector current of
> Q1 and the emitter current of Q1. So it's almost a parallel circuit.

Ok, so what do you get (voltages) at the base of the transistors
that way? Will B-E current flow? What will happen in terms
of C-E current?

Ed

ehsjr, Sep 6, 2005
5. PeterGuest

ehsjr <> wrote in news:fDnTe.1204\$d6.317@trndny05:

> Peter wrote:
>>>Hint: You were concerned about the left hand side and the
>>>68 ohm resistor. What is the lowest possible voltage,
>>>with respect to ground, at the bottom of the 68 ohm
>>>resistor? Can you analyze the circuit, if the 68 ohm
>>>resistor was replaced by a jumper wire?
>>>
>>>Ed
>>>

>>
>>
>> I believe it can be done. you pretty much ONLY have I1 on the left
>> and I2 on the right because the emitter current of Q2 is the
>> collector current of Q1 and the emitter current of Q1. So it's almost
>> a parallel circuit.

>
> Ok, so what do you get (voltages) at the base of the transistors
> that way? Will B-E current flow? What will happen in terms
> of C-E current?
>
> Ed
>

I think I need a few loop equations and not just pulling the 68ohm resistor
out of the circuit to make assumptions. There should be a scientific
approach to this. In any case, if I don't hear from someone soon, I'll pull
the circuit off the webpage.

Peter, Sep 7, 2005