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by hand calculation I know that loop gain Aβ of this (current-current) circuit is 30, I tried to break this circuit and add Vtest to it , and setting input to zero but I didn't get a result that match my calculated result

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  • \$\begingroup\$ Please explain how you break the loop and add the Vtest. \$\endgroup\$ – Bimpelrekkie Dec 19 '18 at 10:49
  • \$\begingroup\$ I st the input at zero, then tried to break the loop two times one before RF and second before CF, I added Vtest with 1AC on left side and 0Vn on left side, but I didn't could find any voltage gain between these points \$\endgroup\$ – Mohan Dec 19 '18 at 11:58
  • \$\begingroup\$ You might have influenced the DC operating conditions by doing it like that. Try placing a 1 F (yes, 1 Farad) capacitor in series with the AC source, that way the DC voltage at the base of Q1 is not influenced. \$\endgroup\$ – Bimpelrekkie Dec 19 '18 at 12:34
  • \$\begingroup\$ If you download LTspice, there are two examples provided once installed. One method from Michael Tian, V. Visvanathan, Jeffrey Hantgan, and Kenneth Kundert, "Striving for Small-Signal Stability", IEEE Circuits and Devices Magazine, vol. 17, no. 1, pp. 31-41, January 2001. Another method from Middlebrook, R. D., "Measurement of Loop Gain in Feedback Systems", Int. J. Electronics, vol 38, No. 4, pp. 485-512, 1975. \$\endgroup\$ – jonk Dec 19 '18 at 15:12
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For measuring, calculating or simulating the loopgain there are three important considerations:

(It is assumed that the loop is opened at CF for injecting the test signal)

  • The signal source Vs must be set to zero;

  • The opening must not disturb the normal operational DC bias point (in your circuit, this is not a problem because of CF)

  • The opening must not change (too much) the loading at the break point. (In your case, it is required that the output resistance at the emitter of Q2 is much smaller than the input resistance at the most right node of RF). Otherwise, the load impedance must be recreated.

Comment: In your circuit diagram you have shown i(out) and i(in). I suppose you know that these two currents have nothing to do with loop gain. Loop gain is measured as a ratio of voltages at the breakpoint.

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