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741 Op Amp Circuitry Question

Take a look at the schematic of the 741 on Wikipedia:

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Here's my question. Let's say the rails are at +15 and -15 volts and the output of the differential amplifier (blue box in the figure) is

+12 volts. That would mean that the base of Q15 would be at 12 volts, the base of Q19 would be at about 11.4 volts, and the emitter of Q19 would be at about 10.8 volts. That would result in 25.8 volts across the 50 ohm resistor, requiring 516mA of current.

However, using the assumed values at the rails, the Q12/Q13 current mirror would only be able to provide a maximum of 738uA, which is 700 times less current than would be required by the emitter output of Q19.

How does this circuit work? What am I missing?

Reply to
nealkendrick
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Won't happen. The output of the diff amp is *current*, and the voltage at that node will be in the neighborhood of 1 Volt.

Reply to
Bill S.

Ok, if you apply 1 Volt at that node, you'll have -0.2V at the emitter of Q19, which will try to draw 296mA (more than is available) through Q19, correct? That would be 14.8 volts across the 50 ohm resistor.

Reply to
nealkendrick

Q15/Q19 are Darlington connected to provide hi Z input load for Q4/Q6 high impedance current output. This is where most of the single pole (with the 30pf cap) gain is achieved. Under op amp closed loop conditions, Q15+Q19 current exactly matches the Q12/Q13 mirror current to produce zero op amp output voltage. This would not turn Q22 on at all. The Q22/50ohm combo is there as a current limit for op amp output negative excursion short circuit current limiting. Note that positive excursion output current limiting is via Q17.

Dick Ballard snipped-for-privacy@att.net

Reply to
Dick Ballard

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