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I'm running through some old exam papers at the moment and one of the questions are:

The parameter \$h_{oe}\$ is needed to calculate the output impedance of an amplifier. Explain in detail how to measure the parameter \$h_{oe}\$ practically in totality from start to end by using a signal generator and oscilloscope.

I've founf a couple of examples showing how to meazure the output impedance of an amplifier using a signal generator and an osciloscope with a variable resistor such as this example at learnabout-electronics.

In my book they go through an equation where you can get \$h_{oe}=\frac{Z_oR_3}{R_3-Z_o}\$ (This is from a Bootstrapped common emitter ammplifier circuit.) So would you go through the process to find the output impedance abnd then plug the values into the equation above and calculate \$h_{oe}\$?

Update
\$Z_o\$ is the output impedance and \$R_3\$ is the resister between the \$Ecc\$ and the collector which is dynamically in parrallel with the \$R_L\$.

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  • \$\begingroup\$ What do Zo and R3 represent in your last paragraph? \$\endgroup\$
    – Dave Tweed
    Oct 20, 2012 at 12:12
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    \$\begingroup\$ This is your homework, so you need to think about it. Think about what exactly the "output impedance" of a signal source like a amplifier means. Consider that Thevenin may have some ideas for you. Once you have the equivalent circuit, think about how to measure the resistance. One wrinkle is that you probably want to ignore impedance of the DC bias, just measure the impedance of the AC part of the signal. Note that this may change with frequency, so the answer might not be as simple as a single number. \$\endgroup\$
    – Olin Lathrop
    Oct 20, 2012 at 13:54

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The answer to your question is yes. Measure Zo and then calculate hoe.

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  • \$\begingroup\$ This looks a bit short as far as explanation/teaching goes. \$\endgroup\$
    – Kortuk
    Oct 21, 2012 at 19:49
  • \$\begingroup\$ @Kortuk: I think the OP just needed the specific question answered so that he could proceed with solving the homework problem. I just wanted to take it one step farther than Olin did. \$\endgroup\$
    – Dave Tweed
    Oct 21, 2012 at 19:53

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