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How would the DC equivalent circuit be constructed? I have a vague idea knowing that AC voltage sources are connected to the ground and capacitors are open circuited but what happens to the resistance \$R_B\$? I have seen some solutions where the whole branch is connected to the ground but that doesn't sem right

  • \$\begingroup\$ Your text doesn't match the diagram. What is the circuit's intended function? Or is it just a random? \$\endgroup\$ – Chu Jan 11 '19 at 18:51

For the dc equivalent the capacitor becomes an open circuit. Since Rb is in series with the capacitor then no current will flow through either component, and both of them can be removed from the schematic.

What you have left is the four resistors that set the bias point of the amplifier. It seems like you are working backwards here.

  • 1
    \$\begingroup\$ Yup. Exactly as I'd written in a comment to the OP, in an earlier question. Didn't seem to understand it then. May not, now. I suspect a detailed walk-through is required to make the point clearer, unfortunately. \$\endgroup\$ – jonk Jan 11 '19 at 17:01
  • \$\begingroup\$ I know that is what makes sense to me too, but if you see my edit to the same earlier question I only get the answer right if the \$R_B\$ is included as stated. which doesn't make much sense but it works, and not knowing why is getting me really frustrated. Or maybe the correct answer is maybe just wrong, I have no idea \$\endgroup\$ – Bidon Jan 11 '19 at 17:05
  • \$\begingroup\$ @Bidon I believe the first term in the answer you were told is "correct" is actually "wrong." \$\endgroup\$ – jonk Jan 11 '19 at 17:23
  • \$\begingroup\$ Bidon...what is the meaning of "....but it works"? To me, without any long calculation, the capacitor C allows no DC current through resistor Rb - hence, Rb must not appear in the DC equivalent diagram. \$\endgroup\$ – LvW Jan 11 '19 at 17:23
  • \$\begingroup\$ The meaning of "...but it works" is my own self doubt, because my intuition agrees with you 100% but my slim knowledge and experience of this discipline makes me doubt. But anyway, thank you for for the patience \$\endgroup\$ – Bidon Jan 11 '19 at 17:28

The trick is to find the bias (base) voltage of the circuit. For a DC voltage applied without the cap, if the input is exactly at this voltage it will produce no change in the output voltage. The output, of course, will be one diode drop lower than the input.

If you reference your DC input to this bias voltage, for small +/- changes you'll get the appropriate amplified change at the output.

More or less.

Problem is, the base-emitter voltage drop, which contributes to the bias point, will change with temperature. So using this sort of circuit for DC measurements is not a good idea, since the zero point will vary with temperature. It will also vary from transistor to transistor, since transistors vary between lots.

This is why IC op amps are the preferred approach when dealing with DC. They use intrinsically matched differential transistors on the input, and negative feedback to handle the gain. The result is excellent DC performance over temperature.

But even then, it's not perfect.


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