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Opamp Current Analysis

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Thank you for reading my post. I'm currently studying Op-amp, and there is a homework about LH 0005 of National Semiconductor for me to do: determine the collector current through each transistor

I'm trying to find the IC2 but don't know how. Please help me with this :)

1. The problem statement, all variables and given/known data

Given: β = 100 & VBE = 0.7

2. Relevant equations


3. The attempt at a solution

I tried to analyze but I'm stuck at the moment. Here are somethings that I've figured out so far:
VC2 = VCC - 10k*IC2
VE6 = VC2 - VBE6 = VC2 - 0.7
VC5 <<<
I2 = (VE2 + VEE) / 12k
I6 = (VE6 + VCC) / 2k

I'm trying to find the IC2 but don't know how. Please help me with this :)
a1d0e395e0.png

I'm looking forward to seeing your response soon :)
 
Last edited by a moderator:

LvW

773
200
Txp2037, you have overlooked that Q6 is pnp (VBE=-0.7V).
 
Txp2037, you have overlooked that Q6 is pnp (VBE=-0.7V).
Thank you for reminding me :) I'm struggling with the problem. By any chance you can give me some hints?
 

LvW

773
200
Thank you for reminding me :) I'm struggling with the problem. By any chance you can give me some hints?
Question: Are you required to find the quiescent collector currents for the shown input voltages?
This will create problems because - assuming VBE=0.7V as mentioned - the base potentials for Q1 and Q3 are equal.
However, this is in contradiction to the given signal voltages.
It is more likely that Q3 will be completely off. However, in this case, we are not in linear mode anymore.
Perhaps you should review the task again (with the dc input voltages)?
 
Question: Are you required to find the quiescent collector currents for the shown input voltages?
This will create problems because - assuming VBE=0.7V as mentioned - the base potentials for Q1 and Q3 are equal.
However, this is in contradiction to the given signal voltages.
It is more likely that Q3 will be completely off. However, in this case, we are not in linear mode anymore.
Perhaps you should review the task again (with the dc input voltages)?
Thank you for your reply. My requirement are (and I quote): Referring to the LH0005 op-amp equivalent circuit shown in figure, determine (a) the collector current through each transistor and (b) the de voltage at the output terminal. (Assume that each transistor has βdc = βac = 100 and VBE = 0.7 V.)
I think I can sort the (b) out, but I don't know how to find the collector current. I've written down some equations so when I know the IC2, I can solve the problem. About the input voltage, there are nothing given :)
 

LvW

773
200
Oh sorry - I have misinterpreted the drawing. So "1" and "3" are node names and not two different input voltages.
Therefore, I now assume that both open base terminals are grounded, correct?

In this case, it is a simple task to find the voltage at the common emitter node (because of VBE=0.7V for each transistor). This allows to compute the current to the emitter resistor (12k) - and the current is equally splitted between both transistors Q2 and Q4.
 
Oh sorry - I have misinterpreted the drawing. So "1" and "3" are node names and not two different input voltages.
Therefore, I now assume that both open base terminals are grounded, correct?

In this case, it is a simple task to find the voltage at the common emitter node (because of VBE=0.7V for each transistor). This allows to compute the current to the emitter resistor (12k) - and the current is equally splitted between both transistors Q2 and Q4.
You gave something awesome here. So as you said, the VB1 will be 0V. Then VE1 = VB1 - VBE1 = 0 - 0.7 = -0.7V.
VE1 = VB2 = -0.7V ==> VE2 = VB2 - VBE2 = -1.4V
Then I can find the I2 = 2IE2 = (VE2 + VEE) / 12k = (-1.4 + 10) / 12k =~ 0.717mA.
Am I currently on the right way?
 
I rather think that I2=IE/2.
Thank you for your help. Under your guide, I've managed to solved the problem completely :) I was misunderstood about the "ground" voltage that should be 0V. With that information, everything was clear :) Thank you sir
 
1
0
Could you further elaborate how voltage gain, Ri etc are calculated.....
 

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