# What is the base current of a class ab amplifier with darlington pair configuration

I am trying to design a 3 stage amplifier using transistors, 2 pre-amp stages using class A configuration and an output stage class AB. I have built the class ab amplifier with a darlington configuration and 4 diodes for the bias voltage and it has a gain of about 0.8v but now I am trying to calculate the input impeadance of the ac model so that I can design the 2nd stage but I am having some issues with finding the input impedance since I do not know how to calculate the transconductance with a darlington configuration.

Can you guys give me some hints on what to do ? Thanks

simulate this circuit – Schematic created using CircuitLab

• I have an issue with your circuit. Q3 and Q4 are a "normal" Darlington, OK but Q2 and Q1 aren't, I would expect Q2 to be a PNP. If Q2 was a PNP then the total voltage between the bases of Q3 and Q2 would be 4 Vbe, so 4 diodes would be needed. Indeed you have 4 diodes (D1-D4) for the DC biasing. For help on the small signal model of a Darlington transistor, have you read: coefs.uncc.edu/dlsharer/files/2012/04/F5.pdf ? – Bimpelrekkie Apr 15 at 9:22
• Both darlington pairs are normal, I just made an error in the circuit diagram – Samwel Portelli Apr 15 at 9:25
• Feel free to update and fix your schematic then. – Bimpelrekkie Apr 15 at 10:09
• Use a free simulation tool. – Andy aka Apr 15 at 11:57
• As drawn, the lower Darlington does not provide the needed polarity of current flow into base of the higher-current(output) device. Is the schematic correct? – analogsystemsrf Apr 16 at 10:36

On a single bipolar, wiggling the base by 18milliVolts will cause a 2:1 variation in the emitter (and collector) current. The GM (transconductance) is Iemitter(amps)/0.026volts. At 26milliAmps, the GM is ONE (amp/volts).

For a Darlington, the GM will be affected by resistance shunting the base-emitter of the high-current device. And the frequency (as transient charging current plays a role).

Try a 1Kohm resistor paralleling the EB junction of Q4(high current device), and the same for the PNP device. For more power, consider 100 ohms, or whatever keeps the low-current device in "relative constant current" operation.

At "relative constant current", the low-current devices have no effect on the GM. IMHO

(A) At first, here are some information on the Darlington pair only (h-parameters, small-signal operation) - consisting of two identical transistors:

1) Input resistance: rin=h11,1 + (1+h21,1)*h11,2

2) Current gain: h21=h21,1*h21,2

3) Transconductance: gm=0.5*gm,2

Comment: Of course, the parameters h11 and gm depend on the quiescent DC currents through both transistors. Remember: h11=h21/gm and gm=Ic/Vt (Vt: temperature voltage).

(B) The total small-signal input resistance must take into consideration the resistors R1 and R2 as well as the dynamic (differential) resistances of the 4 diodes (which are determined by the DC current through the diode chain). More than that, the emitter resistor R3 provides negative feedback and, hence, increases the overall input resistance of the Darlinton configuration.

• Would there be any benefit to using a Vbe multiplier instead of the 4 diodes? – Captainj2001 Nov 6 at 21:11