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Bjt Differential Amplifier

I have the following differential amplifier and I would like to find which is the output resistance of this circuit in the cases of differential input and common mode input.I know that the input resistance in the cases of differential input signals is Ridm = 2*rπ and for common mode signals is Ricm = rπ/2 + (β+1)R3.But for output resistance which is the relations?

Also Ι simulate this differential amplifier in Pspice.The results was Rout= 100kΩ when I set a common mode input and Rout= 200kΩ when I set a differential mode input.Hοwever, I can understand the reason for this result and how I can prove through theory.Can anyone help?

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  • \$\begingroup\$ You have gone this far, why not just measure it? You can set the AC value at in+ and in- to 0. Then add a 100nA current source between out+ and out-. Divide the voltage across the current source by 1uA and you'll have it. You might want to change your labels out+ and out- so the output will be positive when you put a positive input on in+. Do you know about using half circuits to analyze this amp and Early voltage? \$\endgroup\$ – owg60 Apr 27 '17 at 18:59
  • \$\begingroup\$ Thank you very much owg60.Yes, I know the method using half circuits and also Ι simulate this differential amplifier in Pspice.However the results was Rout= 100kΩ when I set a common mode input and Rout= 200kΩ when I set a differential mode input.I can understand the reason for this different result and how I can prove through theory. \$\endgroup\$ – elecV1 Apr 27 '17 at 20:43
  • \$\begingroup\$ You call this a MOS differential amplifier, yet your schematic shows a BJT differential amplifier. Which one is it? \$\endgroup\$ – Hearth Apr 30 '17 at 17:53
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The output impedance is consists of several parts:

A perfect BJT would behave as a current source of infinite impedance. A real BJT has a collector resistance ro, which depends on the transistor model.

  • Capacitance

At high frequency, the parasitic capacitances in your transistor will appear in parallel with the output resistors. 10pF will form a pole with 100kOhm at only 150 kHz so don't expect huge bandwidth here.

  • Miller Effect

Cbc interacts with the impedance in the base of the transistor, and reduces collector impedance. Basically if you inject AC current into the collector, then its voltage will vary. Some current goes through Cbc and into the base, which then adjusts Ic to prevent Vce from moving. The effect is proportional to the impedance driving the base: your schematic will hide it since you connected the base to a voltage source without any resistor in between...

  • Self-heating

At very low frequencies, higher Vce heats the transistor, which lowers its Vbe and increases the current. This only matters if Ic is high enough to heat things up, which isn't the case here.

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