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I have a table of \$V_{out}\$ values for different source resistance values. Using this data, how can I calculate the effective input resistance of this circuit?enter image description here

My attempted solution was using KCL around the base-emitter loop: $$\frac{V_s - I_bR_s-V_{be}}{I_b}=R_{in}$$ and $$ \frac{I_c}{\beta}=I_b$$

this doesn't appear to be correct, my values are varying with \$R_s\$ and I'm even getting negative values. What am I missing here?

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  • \$\begingroup\$ Difficult to do without knowing the hFE of Q1 and Q2. Different generations of transistors will have different gain, which is to say that older generations of transistors had a relatively high leakage current, thus a lower input resistance. \$\endgroup\$ – Sparky256 Dec 9 '17 at 23:22
  • \$\begingroup\$ Do you have a way you've considered about how you might validate an answer that presents itself to you? Suppose, for example, I said "It's \$6.5\:\textrm{k}\Omega\$!" How would you test it? Would you use Spice to verify? If so, how would you approach that? Would you build it? If so, how would you approach that test? If I made up a formula, could you argue against it? And why is \$R_S\$ present? That's the source resistance, right? Not the circuit input impedance. Or? \$\endgroup\$ – jonk Dec 10 '17 at 0:45
  • \$\begingroup\$ If it isn't clear, I am trying to get you to think about what input impedance means. Your circuit has the input source actually being responsible for supplying the operating current at the base. It doesn't make as much sense, because of that. Normally, input impedance is an AC thing. \$\endgroup\$ – jonk Dec 10 '17 at 1:03
  • \$\begingroup\$ When vout declines by 6dB half amplitude Rin = RS \$\endgroup\$ – sstobbe Dec 10 '17 at 3:22
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Looking between the two bases, the small signal resistance will be

2 * beta * 1/gm

At 1milliAmp, the 1/gm will be 26 ohms.

If beta = 100, the small signal reac (from base to emitter) will be 26 * 100 = 2,600 ohms.

Measures across 2 bases, total is 5,200 ohms.

This ignores all other effects, and ignores any effects of parasitic capacitors.

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