I have circuit drawn in AC, they were 2 cascaded BJT transistors, and I have changed them to the hybrid pi model. I wish to remove the source and load in order to write a general gain equation for each hyrbid pi model itself, so I can re-draw them both as amplifier models.

In order to do this, I need to know across which element is Vout for transistor 1?

The upper image shows the original circuit, I know that the DC VCC is now set to zero therefore ground.

I know that in a figure "Vout" will be where ever they have indicated the name with respect to ground. It appears to me that Q1 Vout would be zero if Vout is from the collector node to the ground as they are the same node

1) When cascading is the second transistors Vinput = the first transistors Vout? (this is my hunch but I am not convinced)

2) How do I find Vout in this example?

enter image description here

  • 1
    \$\begingroup\$ Would be better to use the tool and draw a schematic, it is hard to see your circuit \$\endgroup\$ – laptop2d Jul 11 '18 at 21:01
  • \$\begingroup\$ Unless you are positive that the values of C2 and C3 are not going to affect your band of interest (or were otherwise instructed to ignore them for this problem), I think you have to include them in your small signal analysis. I am not seeing them in your small signal equivalent circuit. \$\endgroup\$ – esilk Jul 11 '18 at 21:09
  • \$\begingroup\$ The gain of each stage Is Zc_equiv/ Rbe and Rbe is derived from Ib (DC) for small signal linear operation \$\endgroup\$ – Sunnyskyguy EE75 Jul 12 '18 at 0:06
  • \$\begingroup\$ I just want to see the OP calculate the DC operating point. That alone will produce interesting results/questions. \$\endgroup\$ – jonk Jul 12 '18 at 0:59
  • \$\begingroup\$ In your diagram Q2 (PNP) is working in CE configuration. But in your small-signal mode, the Q2 is CC amplifier (emitter follower). Why is so? \$\endgroup\$ – G36 Jul 12 '18 at 15:46

@esilk, my error was indeed to wrongly assume that "all capacitors are shorts in AC" I have since learned that, in the pass band, not all capacitors are shorts. Had C1, C2, and C3 been shorts, then indeed the gain would be zero.

Below is a sketch, the circuit is a high pass response, low pass and a final high pass, which using straight line approximations looks something like the below. ( finding Thevinin impeadance seen by the capacitor seems too much work to illustrate the point to get actual corners frequencies)

C1 and C3 short but C2 open

Then I was able to do the AC Analysis, thank you G36, I did make an error regarding Q2, and resolved it enter image description here

so to make sure I answer my question

1)when cascading is Q2's vin = Q1 vout - Yes, in the pass band

2) how do I find vout - understand C1 and C3 are shorts but C2 is an open, then do regular circuit analysis

Lastly Jonk wanted to see me find DC operating point. Below is my DC analysis, I did verify its fwd active mode, and I do understand that to find Q point of one transistor it is the mid point of Vbe's active region, but Im unsure how to handle the "Q point of a circuit", but I would like to learn. enter image description here


You are nearly there if you have a small signal equivalent circuit and are confident in its accuracy for your application. Each node will have an equation that describes it, describing all nodes will give you a system of equations that can be solved for your variable of interest. Based upon your given small signal equivalent, I see 6 nodes, which means you will 6 equations. One of these nodes, Vin, is known. Its just algebra from there to solve for Vout.

Edit to address clarification in comments: Vin is not always Vbe, and output is not always Vce. It depends on your topology. In a common emitter with degeneration (a resistor placed from emitter to ground) Vo is the sum of Vce and Vre. In a common base amplifier, input is Veb, output is Vcb. In a common collector amplifier, while input is still Vbe, output is Vre.

  • \$\begingroup\$ A better way to ask my question is, "By definition, is Vin to an NPN transistor VBE, and Vout VCE?" \$\endgroup\$ – user65669 Jul 11 '18 at 22:44
  • \$\begingroup\$ Ahhh ok. No, it is not. It depends on its mode of operation. A common-base amplifier has input applied to the emitter, and output taken across the collector and base. Similarly, a common collector has its output taken at the emitter across whatever biasing is in use (normally just a resistor). \$\endgroup\$ – esilk Jul 12 '18 at 15:17
  • \$\begingroup\$ Oh, and: in a common emitter with degeneration, output is taken across the collector and ground, so Vo = Vce + Vre, where Re is a small emitter resistor (and often a large capacitor). \$\endgroup\$ – esilk Jul 12 '18 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.