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I'm trying to analyze the small signal model of this following snippet circuit (all the transistors are the same with beta = 100, and VA =100:

enter image description here

What i've come up with is:

schematic

simulate this circuit – Schematic created using CircuitLab

when the rest of the circuit is located 'above' the left part. I want to analyze what is the net resistance the the rest of the circuit 'sees', But I got stuck..

I can't figure out how to account for the current sources.

The rightmost current source, is just a resistor of 1/gm (I think), but what about the left one?

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  • \$\begingroup\$ From which collector Q15 or Q16 do you want to find Rout? Becoues fom Q16 collector it will be Rout = ro. And from Q15 colletor rout will be equal to around 1/gm \$\endgroup\$
    – G36
    Commented May 31, 2022 at 21:35

2 Answers 2

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The small-signal model will look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

Notice that:

\$R_{OUT} = \frac{V_X}{I_X}\$

Also, notice that \$V_{BE1} = V_{BE2} = 0V\$ thus, we can see that \$g_{m1}\times V_{BE1} = 0A\$.

So we are only left with \$r_{o1}\$.

Therefore

\$R_{OUT} = \frac{V_X}{I_X} = r_{o1}\$

Do you see it?

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  • \$\begingroup\$ Isn't the top side connected to ac gnd because of Vcc? \$\endgroup\$
    – Shredder
    Commented Jun 1, 2022 at 15:28
  • \$\begingroup\$ Yes, but what does that change? \$\endgroup\$
    – G36
    Commented Jun 1, 2022 at 15:30
  • \$\begingroup\$ It doesn't change the final answer, I agree it's just ro1, just commenting for accuracy's sake for OP \$\endgroup\$
    – Shredder
    Commented Jun 1, 2022 at 15:33
  • \$\begingroup\$ @Shredder I see, how would you draw it? \$\endgroup\$
    – G36
    Commented Jun 1, 2022 at 15:52
  • \$\begingroup\$ Oh wait, you know what I'm realizing is that the entire circuit is simply drawn upside down, but it's correct as is. \$\endgroup\$
    – Shredder
    Commented Jun 1, 2022 at 16:01
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If Q1=Q2, Vbe = Vce on both thus Ic are the same on both as a function of gm and Vbe, Ib and temp. Also Icb supplies equal base current to both BJT's.

Other

Essentially the improvement over just a diode mirror is the hFE value at Ic and Vcb=0 so it is not forward conducting except ro. But the diode and Vbe diverge significantly since junctions different sizes and doping characteristics unless a series R is matched to both on the P side.

When Vcb becomes forward biased which at high currents appears reverse biased from bulk Rce <2V but you goal is to match Rce/Rbe to Vcb to Vbe , but since those internal you can verify on any transistor with Ic <=10uA.

Questions?

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