We now can add a load resistor to the circuit: Since an ideal scope has ~infinite resistance (or ~10 megaohm in reality), so most Vth can be loaded on the scope. Now, in order to calculate Rth, we need to know the voltage of Vth - you can simply measure it with a scope at point C. Unfortunately, lots of voltage from Vth will be applied on Zout, and only a portion of the Vth can pass into the next stage.With small resistance, Zout won’t have too much voltage drop, so most of the signal from Vth can pass into the next stage.Since we are looking the black box from the point of view of output, the input signal may have been transformed by the black box and have a different wave form, so we can turn it into a Thevenin equivalent circuit:Īnd Rth will be the output impedance. Let’s put a function generator on the black box input (we still need send in some signal, otherwise we won’t see anything): (Don’t attempt to look at the input to determine the output impedance, since your black box may transform your input signal anyway) It determine how much voltage will be shared between the black box and the output load The definition of the output impedance is ” “How much impedance(resistance) from the point of view of the OUTPUT” ![]() If unsure, change the input frequency and measure the input impedance again. ![]() *Some special device are sensitive to the input frequency (like capacitors). *A good circuit should have high input impedance. Put a different test resistor, until the voltage amplitude at B is HALF of the voltage at A - implies the new test resistor value is the SAME as the ZinĬongratulation, you just measure the input impedance!.Measure the voltage amplitude on point A and B, and use the ratio to calculate Zin, or:.This is nothing but two resistors in series - a simple voltage divider. So… how to we determine the input impedance then? Since the circuit now is nothing but resistors in series, we can add a test resistor on the input:Ĭompared to the 20k test resistor, the 5 ohm internal R is almost nothing - we can safely assume that the test resistor and Zin will take most of the voltage drops. Don’t need to worry about electric bill.Most of the voltage will go into the black box, since internal resistance is small and won’t take much of voltage drop.By Ohm’s law, large resistance implies low current.Has to increase input voltage to compensate the loss - not economic.Lots of voltage will goes to the internal resistance, waste lots of power, and very few signal actually reach the black box. ![]() ![]()
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