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I have the following CC-CE (Common-Collector Common Emitter) cascade amplifier:

enter image description here

I am trying to bias the amplifier by using LTSpice. Here is my schematic below:

enter image description here

Apparently, from the displayed voltages, am doing something wrong. Unfortunately I can't figure out what am doing wrong, can somebody please help me by showing me how to bias the circuit properly In LTSpice??? Thank you for your help in advance.

BTW below is the dc operating point results:

enter image description here

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  • \$\begingroup\$ That isn't a cascode amplifier. \$\endgroup\$
    – Andy aka
    Commented Jun 20, 2018 at 20:16
  • \$\begingroup\$ @AndyAka, thank you for your comment, but this is what my text book calls it \$\endgroup\$
    – JordenSH
    Commented Jun 20, 2018 at 20:53
  • \$\begingroup\$ It’s a cascade amplifier not cascode. \$\endgroup\$
    – Andy aka
    Commented Jun 20, 2018 at 23:03
  • \$\begingroup\$ ah! your right! I will fix this shortly \$\endgroup\$
    – JordenSH
    Commented Jun 21, 2018 at 17:25

2 Answers 2

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You did the same as what is in the original circuit.

However you're treating this amplifier as if it has a voltage output. You can treat it that way but you should then draw a small signal equivalent and figure out what the voltage gain will be. It will be quite large! That means that any DC offset in the circuit will be amplified a lot making the output clip to its minimum or maximum voltage. That's what happens here, it clips to the minimum.

You could tune I3 such that the output vc2 is at 2.5 Volts for example but don't be surprised if you have to make the value of I3's current a very precise value. Then change the temperature (for example) and the output clips again! Hmm, that's not very useful now is it?

OK, now what if we treat the output vc2 as if a current comes out. So fix the voltage (at 2.5 V, use a 2.5 V DC voltage source) and observe the current through that voltage source. Now you can simulate the small signal current gain Iout/Iin (Iin is the current from I3).

Here's how I would simulate this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ @Bimperlrekkie, can you elaborate more by providing a schematic that shows the biasing you mean?? I am trying to implement your biasing but am getting no luck so far. May be am doing something wrong \$\endgroup\$
    – JordenSH
    Commented Jun 21, 2018 at 17:52
  • \$\begingroup\$ I added the schematic, note how I force the output to 2.5 V DC and measure the current as the output signal is a current. I suggest that you do a DC simulation and sweep Iin between 0 and 10 uA and observe what happens to Iout. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 21, 2018 at 19:16
  • \$\begingroup\$ So I just implemented your circuit with LTSpice, I got Ib1=1.837uA, Ie1=201.94uA, Ic1=200.11uA. For Q2 I got: Ib2=1.942uA, Ic2=200.97uA. But what is confusing is that Vb2=-0.723mA, and vb1=7mV which way below the allowed threshold. \$\endgroup\$
    – JordenSH
    Commented Jun 22, 2018 at 4:27
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I will propose this circuit:

enter image description here

It should give you the correct results in AC analysis from 10Hz to 100KHz.

Or you can try this way also:

enter image description here

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  • \$\begingroup\$ thank you very much for your answer. Would you please elaborate on why you connected the emitter of Q3 to -5V, and please explain how does the circuit on the far right works? This is a very ingenious technique and would like to explore it. Again, thank you very much \$\endgroup\$
    – JordenSH
    Commented Jun 22, 2018 at 15:52
  • \$\begingroup\$ The negative supply voltage is not needed. E1 is a voltage controlled voltage source and he is used as an op-amp in integrator configuration. It provides a large negative feedback around the circuit to stabilize the DC-bias point at 2.5V. \$\endgroup\$
    – G36
    Commented Jun 22, 2018 at 16:04
  • \$\begingroup\$ Oh wow! That is advanced, am not near that stage yet. It is interesting how my book presented this problem with no information on how to bias the circuit, yet now I look at your answer and it contains advanced solution. Do you think there is a simpler solution to this problem, because in the end I want to bias the circuit correctly so I can find the input resistance \$\endgroup\$
    – JordenSH
    Commented Jun 22, 2018 at 16:08
  • \$\begingroup\$ I updated my answer. Is the method is simpler? Also, for your circuit finding Rin is a trivial task. Do you know the Rin for CE amplifier with emitter degeneration resistor? \$\endgroup\$
    – G36
    Commented Jun 22, 2018 at 16:35
  • \$\begingroup\$ My theoretical calculations say Rin=rpi1+(beta+1)rpi2. Is that correct? \$\endgroup\$
    – JordenSH
    Commented Jun 22, 2018 at 16:39

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