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For a voltage source (except RF circuits), the output impedance of the source signal being coupled must be much smaller than the input impedance of the load. This is first of all done to prevent loading. Also even better to set Zin>>Zout to prevent the voltage divider non-linearity effects.

But why is that the other way around when it comes to a current source?

Why should Zout>>Zin? What is the logic behind it?

  • \$\begingroup\$ Because the current source's impedance shunts the supply rather than being in series. And you'd be concerned about the current divider effect rather than voltage divider. \$\endgroup\$ – The Photon Dec 23 '15 at 17:41
  • \$\begingroup\$ do you mean Zin and Zout are in parallel in current source case? \$\endgroup\$ – user16307 Dec 23 '15 at 17:43
  • \$\begingroup\$ Because in this case, the total current is MAINLY determined by the value of Zout. Hence, the total current is - more or less - independent on the load (it is constant).. Hence, we have a "constant" current source. \$\endgroup\$ – LvW Dec 23 '15 at 17:47
  • \$\begingroup\$ You are assuming output voltage (for a voltage source) or output current (for a current source) should minimally be affected by the load. However maximum transferred power is an equally valid assumption to make, with different results for the load. In other words, there is no such "should", it all boils down to what you want to accomplish. \$\endgroup\$ – jippie Dec 23 '15 at 18:10
  • \$\begingroup\$ Another thing to note about real-world current sources is that--like you can't have infinite amperage coming out of a [real-world] voltage source--you can't have infinite voltage coming out of a [real-world] current source. The voltage limit for the latter is sometimes called compliance voltage (range). This limit, which is the open-circuit voltage for a current source, is obviously related to the internal impedance, just like the short-circuit current is for a voltage. \$\endgroup\$ – Fizz Dec 23 '15 at 18:33

The output impedance shunts the source in a current source:


simulate this circuit – Schematic created using CircuitLab

If the source impedance were in series, it would have no effect on the source's behavior, because the current source would simply compensate for its presence to produce the required output current.

Since the source impedance is shunting the source, it must have a high value to avoid drawing away source current and reducing the current delivered to the load.

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  • \$\begingroup\$ nice but just one question about this: so if one increases Zsrc it seems it is getting better since the current divider effect starts to disappear. and if we bring Zsrc to extreme case for example if we replace the Zscr with a dry wood or something has extreme low conductivity. Or even if we remove Zsrc means Zsrc goes to infinity and still we should be happy about it. But when I think this way it seems Im making a logical mistake. What do you think? \$\endgroup\$ – user16307 Dec 23 '15 at 17:58
  • \$\begingroup\$ Or Zsrc is a Thevenin equivalent which "has to be some value"? \$\endgroup\$ – user16307 Dec 23 '15 at 18:04
  • \$\begingroup\$ Why do you think you would make a LOGICAL mistake? \$\endgroup\$ – LvW Dec 23 '15 at 18:04
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    \$\begingroup\$ Yes, the power delivered to the load increases toward the maximum as \$Z_{src}\to\infty\$. But you'll get almost all the improvement as long as you have \$Z_{src}\gg{}Z_{load}\$. Increases beyond, say \$Z_{src}=10(Z_{load})\$ will have very little effect. \$\endgroup\$ – The Photon Dec 23 '15 at 18:05
  • \$\begingroup\$ Is there a current source in the world which has infinite Zsrc? I think Zsrc is not something like a one piece 1/4 watt actual resistor component. It is Thevenin equivalent right? \$\endgroup\$ – user16307 Dec 23 '15 at 18:07

A 1V voltage source with a series resistance of 1 ohm can produce 1 amp into a short circuit. The terminal impedance looking into the voltage source and resistor is 1 ohm.

If you converted this circuit to a current source, your source would be 1A with a parallel resistance of 1 ohm. If you were to calculate the terminal impedance of that set-up it would be 1 ohm.

Both these circuits provide identical performance and functionality - if both were placed inside a box with their terminals coming out and you were asked to probe the terminals and measure this (and that) you would find zero electrical difference.

This means that the current source without the parallel resistor MUST have infinite impedance and, as is more commonly known, the voltage source (without the series resistor) has zero impedance.

Another way to look at a current source is as an infinite voltage source in series with an infinite resistance. Despite both voltage and resistor being infinite they can still have a ratio and if this ratio is 1 then the current source is 1 amp and no matter how little or much resistance you connected to the output terminals, 1 amp would always flow.

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  • \$\begingroup\$ The resistor in parallel with a current source is the norton equivalent of the series resistance of the voltage source and, the series resistance of the voltage source is the thevenin equivalent of that of the current source. \$\endgroup\$ – Andy aka Dec 23 '15 at 18:20

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