0
\$\begingroup\$

When calculating the input impedance of a voltage divider biased BJT common emitter amplifier, we calculate the parallel resistance of the bias resistors and the emitter resistance. But why do we consider both bias resistors? Looking from the amplifier input (the transistor base), the other resistor leads to the voltage source, while the other leads to ground. Shouldn't we only consider the one that leads to ground? Why would any of the input current go to the voltage source/battery?

\$\endgroup\$
  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/298560/… \$\endgroup\$ – G36 Jan 4 at 18:52
  • \$\begingroup\$ Because in"small signal analysis" or AC analysis, both supplies are considered 0V. So they ARE in parallel. In practice. the big supply decoupling capacitor guarantees that Vs has an AC component of 0V. \$\endgroup\$ – Brian Drummond Jan 4 at 21:19
0
\$\begingroup\$

When you determine an input impedance (which is a transfer function), you install a test generator (the stimulus) \$I_T\$ across the connecting ports and you determine the response voltage \$V_T\$ collected across the current source terminals. This configuration is called a DPI or driving point impedance as described by the fast analytical circuits techniques or FACTs. The circuit below shows the configuration for the bipolar amplifier you described:

enter image description here

When you do that, you imply that the source supplying your circuit is uncorrelated with the stimulus. It means that if you sweep the input impedance with \$I_T\$, you won't see any ripple over the \$V_{cc}\$ rail supplied by a perfect dc source (0 output impedance). The small-signal amplitude of the \$V_{cc}\$ rail is thus 0 V during the analysis and the source can be replaced by a short circuit for the dynamic analysis. You can also imagine an infinite decoupling capacitor from the \$V_{cc}\$ rail to ground perfectly decoupling the rail during the analysis. As the drawing shows, when the source is shorted, the biasing resistances naturally come in parallel.

\$\endgroup\$
  • \$\begingroup\$ Thank you, accepted! \$\endgroup\$ – S. Rotos Jan 5 at 1:01
0
\$\begingroup\$

It would be better if you provide the schematic drawing.

When calculating the input resistance, all independent source must be turned off. So, the voltage source is connected to the ground. The resistor that connected to the voltage source is grounded now. That is the reason that you have to consider both bias resistors. Now, those resistors are in parallel.

\$\endgroup\$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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