This image is taken from eCircuit Center, available in the link. In a specific section, the author discusses the function of RC1, mentioning that RC1 helps set Ic1 equal to Ic2. I initially thought that RC1's role was to convert current to voltage for the following stage.
Can anyone explain why?

Finally, why include RC1? For DC bias purposes, RC1 helps set Ic1 equal to Ic2. Your initial estimate for this value is RC1 = vbe3 / Ic1. For instance, if the desired Ic1 is 0.5 mA, then RC1 = 0.7V / 0.5 mA = 1.4 k. You might need to adjust this value to achieve Ic1 approximately equal to Ic2. While RC1 assists in balancing collector currents, it also diminishes signal gain by diverting some of Ic1 away from the base of Q3.

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  • \$\begingroup\$ RC3 is perhaps the most glaring design flaw in that circuit. (To me.) \$\endgroup\$ Commented Jan 28 at 20:43
  • \$\begingroup\$ @periblepsis could you explain why? \$\endgroup\$
    – hana
    Commented Feb 5 at 9:22
  • 1
    \$\begingroup\$ It should be obvious. What happens when vo is at its most positive point in a cycle? Follow from vo, through Q11's emitter to base, then work out the voltage across RC3. What are the implications? (If it doesn't dawn on you from this description, let me know so I can be a little more pointed about it.) \$\endgroup\$ Commented Feb 5 at 21:40

2 Answers 2


I initially thought that RC1's role was to convert current to voltage for the following stage

RC1 can be removed and Q1's collector current flows entirely into Q3's base. In other words you don't need to have RC1 but it certainly helps stabilize the amplifier.

RC1 helps set Ic1 equal to Ic2

That's not particularly true or useful to say.

RC1 sets the small signal output impedance of Q1's collector and, because Q3 is acting as a low-pass filter (with CC and RC1), you need a defined resistor value to set the scene correctly. So, the circuit around Q3 lowers the gain at above-audio frequencies and, prevents HF oscillations when negative feedback is used (via RF2).


[...] the author discusses the function of RC1, mentioning that RC1 helps set Ic1 equal to Ic2.

Why is it important to equalize Ic1 with Ic2?
This is a matter of DC bias stability. With no input signal, we should design for output voltage of zero volts. To achieve this, the PNP differential pair should share equal current, splitting DC current from RE evenly between Q1 and Q2.
In this way, Q1's base current flowing through RIN1 will create a voltage drop that is equal to voltage drop across RF2, caused by base current of Q2.

Yes, a significant portion of Q1's collector current flows into RC1 rather than into base of Q3. This simply means that closed-loop voltage gain of Q1 is reduced. Most voltage gain in this loop belongs to Q3. This voltage gain is usually dialed-back by the feedback resistor ratio of RF2/RF1. Dialing-back open loop gain this way flattens frequency response, and reduces distortion.

It might be a good exercise to find values for RC1, RE, RC3 that maximize open-loop gain, and still yield near-zero resting output voltage. This max gain solution might not be optimum for a particular design...for example one may also wish for a large output voltage swing into a small load (RL)...that will require modifying choice of these three resistors, generally making them smaller than the max-gain solution.
It is often this trade-off that forces more of Q1's collector current into RC1 instead of Q3's base.

  • \$\begingroup\$ I think reducing gain makes sense, but isn't there a Miller cap to reduce gain already? \$\endgroup\$
    – hana
    Commented Feb 5 at 9:24
  • \$\begingroup\$ Does your "Miller cap" refer to Cc? or CF1? Cc is included only to kill ultrasonic oscillation...opamps do something similar - Cc has a small-enough value that it can be integrated onto a chip. CF1 sets in-band (audio) closed-loop gain and is much larger than Cc. Neither are considered in the design of DC biasing. \$\endgroup\$
    – glen_geek
    Commented Feb 5 at 15:01

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