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This is a voltage amplifier circuit I want to build and curious about the series parallel connections so to understand the working of each element. My understanding of the circuit is:

  1. 6.8k ohm resistor and 680 ohm are in series with the the diode (at collector and emitter respectively).
  2. 10nF capacitor is in series with the curcuit to its right.

How can 56k ohm and 5.6k ohm resistors be classified? Are they in series or parallel or anything else? I want to know so I can find voltage/current at the base of the diode.

circuit

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  • \$\begingroup\$ Would Electrical Engineering be a better home for this question? \$\endgroup\$ – Qmechanic Jun 16 at 17:46
  • \$\begingroup\$ That't not a diode, it's a transistor. \$\endgroup\$ – Justme Jun 16 at 18:12
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Series and parallel resistors are discussed here, Finding the equivalent resistance.

6.8k ohm resistor and 680 ohm are in series with the the diode (at collector and emitter respectively).

Not exactly as the currents through those two resistors are not the same as \$i_{\rm 680} = i_{\rm 6.8k} + i_{\rm base}\$.
However given the current gain is approximately 40 - 100 you can say that the current through both those two resistors is approximately the same and so they are in series.

10nF capacitor is in series with the curcuit to its right.

Yes, but why not say that current entering the node is the sum of the currents leaving the node?

How can 56k ohm and 5.6k ohm resistors be classified?

For dc bias conditions classified as series with, as before, the proviso that the dc base current is very much smaller than the dc current through the two resistors.
For AC, given that the \$V_{\rm cc}\$ rail is effectively grounded the two resistors are in parallel.

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