Engineering Calculate R3 in Transistor Circuit

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To calculate R3 in the transistor circuit, it is identified as the emitter resistance that regulates Hfe, typically set at 1V and 1mA, resulting in R3 being 1000 ohms. When Q4 connects to Q2's emitter, the voltage drops to 0.6V, reducing the current through R3 and allowing for the base current of Q4 to be calculated. The base current of Q4 is derived from the difference between the emitter current and the current through R3. Some participants suggest that R3 functions more as a pull-down resistor and that Hfe is less critical in this circuit configuration. Clarifying which transistor's emitter is referenced is also recommended for better understanding.
Harrison G
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Homework Statement



IMG_20160812_234158.png

Hello, guys! I just wanted to make sure I do this right.

Homework Equations


Forget about the rest of the circuit and focus only on Q2 and Q4. How to calculate R3?

The Attempt at a Solution


The way I do it:
R3 is basicaly the emitter resistance which purpose is to regulate Hfe. So in most cases the emitter is held at 1V at current of 1mA. We find R3 from the formula R3=1V/1mA = 1000ohms. When Q4 is connected to the emitter of Q2, the emitter will fall to 0,6V and the current through R3 will reduce, leaving room for the base current of Q4 I3=0,6/1000= 0,0006A=0,6A. I3 reduced from 0,001A to 0,0006A. Now we can find what the base current of Q 4 will be from IB=IE-I3=1mA-0,6mA=0,4mA.
 
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I think R2 should equal R3 so that there is equal and opposite drive into the push-pull pair Q3 and Q4.
 
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tech99 said:
I think R2 should equal R3 so that there is equal and opposite drive into the push-pull pair Q3 and Q4.
But the calculations for R3 are done in such a way like I did, isn't that right?
 
Harrison G said:
R3 is basicaly the emitter resistance which purpose is to regulate Hfe. So in most cases the emitter is held at 1V at current of 1mA. We find R3 from the formula R3=1V/1mA = 1000ohms. When Q4 is connected to the emitter of Q2, the emitter will fall to 0,6V and the current through R3 will reduce, leaving room for the base current of Q4 I3=0,6/1000= 0,0006A=0,6A. I3 reduced from 0,001A to 0,0006A. Now we can find what the base current of Q 4 will be from IB=IE-I3=1mA-0,6mA=0,4mA.
In order to follow your explanation, I think we need each mention of emitter to be identified as of Q2, Q3 or Q4.
 
R3 is more a pull-down resistor. In this type of a circuit we do not care about Hfe much.
If you have time try watch this lecture
 

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