I need to implement the following circuit in LTSpice but I don't know if my schematic is ok. Also, I don't know what values for I should use for Cinf and R2.

The following constraints are imposed:

  • VCC= 10V
  • VCM = 1.5V
  • Vgis a sine wave with a frequency of 1 kHz. Its amplitude will be chosen so as to ensure minimal distortion of the output signal (i.e. v0 should also be a sine wave)
  • R1= 15kΩ.

I need to choose a value for R2 and Cinf in order to have a (theoretical)gain value Av=1500.

How can I compute the theoretical expressions of the DC collector currents of bipolar transistors Q1–Q8, considering VBE≈ 0.6V and the theoretical expression of the small-signal voltage gain Av= vo/vg?

The first picture is the schematic and the second is the LTSpice implementation.

What I found is that:

  • ICq5=ICq6=ICq7=ICq8=(Vcc - Vbe)/R2
  • ICq1=ICq2 = [ICq7 / 2 ] * [Beta/(Beta+1)]
  • ICq4=iCq2-current of Q5
  • ICq3=ICq1-current of Q3-current of Q4
  • Beta in my case is 182.1 because I'm using a BC847A transistor.
  • A=Vo/Vg = Gm_q2 * Beta_q5 * R1Gm_q2=ICq2 / Vt that is approx 40*ICq2

How can I find the value of R2? This is what I have now, an sine like wave for Vg

enter image description here enter image description here

enter image description here

  • 1
    \$\begingroup\$ Those terms are all explained in the text below the schematic. What's unclear about them? \$\endgroup\$ May 7, 2020 at 19:50
  • \$\begingroup\$ @BrianDrummond they are explained, but what should I put for Vcc? I can't just let a wire and that's it, I should put something there and idk what to put... \$\endgroup\$ May 7, 2020 at 20:33
  • \$\begingroup\$ Oh. Use a voltage source component. set it to your desired supply voltage like 5V or 12V. Ditto Vcm (use 0.5 * VCC as a starting point; vary it to see what that does to the amplifier later). Vg is an AC source; start with 1 or 10 or 100mV, 1 kHz. And Cinf ... just set it to an impedance RL/10 or less at your lowest test frequency. \$\endgroup\$ May 7, 2020 at 20:39
  • \$\begingroup\$ @BrianDrummond thanks! what values should I put for R2, C∞ in order to a(theoretical)gain value Av= 1500? \$\endgroup\$ May 8, 2020 at 6:22
  • \$\begingroup\$ I've already answered Cinf, and finding R2 is probably part of your exercise. \$\endgroup\$ May 8, 2020 at 9:07

4 Answers 4


Here, my answer is similar to Spehro's with the following differences.

  • hFE reduces sharply to 10% if Vce < ~0.5 @ 5mA due to saturation effects.
    • Vce(sat) increases with Ic depending on Q (up to 2V at Imax)
  • Vcm >=1.5V depends on the load or negative feedback, if used.

  • R1 (load) >= than R2 ( current bias)

Proof of concept with 100 uV signal

enter image description here

  • open-loop gain is large Aol ~ 33k * R1/R2 (with hFE=100)
  • 10uV Vcm offset is significant with open loop.

  • DC offset due to differences in Vcm and hFE, shown in 2nd simulation depends on hFE.
  • also, the load current is DC offset from Vcc/2 due to connection at 0V thus be aware of DC offset voltage or use neg. FB.
  • \$\begingroup\$ So where it's the theoretical gain=1500?? \$\endgroup\$ May 13, 2020 at 10:05

The parameters you mention are all external to the amplifier design.

Your SPICE schematic does not match the design. You need a Vcm voltage source and there should not be a current source connected to Q1 base.

Generally, you might start with Vcm = Vcc/2 and pick Cinf such that Xcinf << RL at the input frequency.

Eg. Vcc = 10, Vcm = 5, Vb= 100mV (AC) or whatever you want to use.

  • \$\begingroup\$ Please edit this new information into your question, not into the comments. And how are you doing with the calculation? R2 is a bias resistor so it's obviously going to affect the transconductance. \$\endgroup\$ May 7, 2020 at 20:30
  • \$\begingroup\$ I've edited the question. \$\endgroup\$ May 8, 2020 at 7:56
  • \$\begingroup\$ oops edited wrong.... one \$\endgroup\$ May 12, 2020 at 13:26
  • \$\begingroup\$ so...any answer @TonyStewartSunnyskyguyEE75 ? \$\endgroup\$ May 13, 2020 at 13:02

I'm more used to thinking with FETS, tho bipolars are my preferred technology.

Takes a while to figure out WHAT IS BEING IGNORED. Clearly the Early Voltage params are not included.

Seems like the diffpair and active_current_mirror load are simply used to dump current into the base of that top PNP amplifier Q5.

Thus the transconductance of the diffpair/active_load is scaled by BETA of Q5, before converted into OUTPUT VOLTAGE across the load resistor.


Here is my short answer:

  • The gain up to the base of Q5 is (app): A1=-(gm_q2)hie_q5

with: gm_q2=Ic_q2/VT and hie_q5=hfe_q5(VT/Ic_q5) and Ic_q5=2Ic_q2

  • The gain of the last stage is A2=-R1(gm_q5)=-R1(Ic_q5/VT)

  • With Ic_q6=2Ic_q2 the only unknown quantity in the total gain A=A1A2 is the current Ic_q6, which is determined by R2 and Vcc.

  • The above calculation (I hope, it is without errors) neglects the transistor output resistances and, of course, the common mode gain.

  • Final result: A=0.5(R1/R2) * hfe * (Vcc-0.6)/VT


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