Solving Op-Amp Current Analysis: LH0005 National Semiconductor

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SUMMARY

The forum discussion centers on analyzing the collector current through transistors in the LH0005 op-amp circuit from National Semiconductor. The user seeks assistance in determining the collector current (IC2) given parameters such as β = 100 and VBE = 0.7V. Key insights include the realization that both open base terminals are grounded, which simplifies the analysis of the common emitter node voltage and allows for the calculation of emitter currents. The user successfully computes I2 using the formula I2 = (VE2 + VEE) / 12k, leading to a current of approximately 0.717mA.

PREREQUISITES
  • Understanding of operational amplifier (op-amp) circuits
  • Knowledge of transistor parameters including β (beta) and VBE (base-emitter voltage)
  • Familiarity with Kirchhoff's voltage law (KVL) and current calculations
  • Basic circuit analysis skills for analyzing common emitter configurations
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  • Study the characteristics and applications of the LH0005 op-amp
  • Learn about transistor biasing techniques and their impact on circuit performance
  • Explore advanced op-amp configurations and their voltage gain calculations
  • Investigate the effects of varying β and VBE on transistor operation in circuits
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Electrical engineering students, circuit designers, and anyone involved in analyzing or designing op-amp circuits will benefit from this discussion.

txp2037
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Member advised to use the homework template for posts in the homework sections of PF.
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 :)

Homework Statement


Given: β = 100 & VBE = 0.7

Homework Equations

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 :)
 
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Txp2037, you have overlooked that Q6 is pnp (VBE=-0.7V).
 
LvW said:
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?
 
txp2037 said:
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)?
 
LvW said:
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 :)
 
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.
 
LvW said:
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?
 
txp2037 said:
Then I can find the I2 = 2IE2
I rather think that I2=IE/2.
 
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LvW said:
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
 
  • #10
Could you further elaborate how voltage gain, Ri etc are calculated...
 

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