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I observed this when I was doing laboratory for one of my courses: when current flows back into my voltage source, its voltage output suddenly increases.

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This is my experiment circuit. I set the voltage source on the left to 5volts and then connected it to the circuit. But as the current is flowing backward into the 5volts voltage source, the reading on my voltage source says that its actually output is 5.7volts.

Could anyone tell me why?

The voltage sources are Tektronix PS280/PS283 lab power supplies.

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  • \$\begingroup\$ What is your voltage source? A battery? A linear regulator? \$\endgroup\$ – The Photon May 4 '12 at 15:49
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An ideal voltage source is is "perfectly stiff". Its voltage does not vary with load, and its impedance is zero. A non-ideal voltage source has a nonzero impedance. This will cause its voltage to vary with the current. As more and more current is drawn from the voltage source with a bigger load (smaller resistance), its voltage will drop. So, conversely, if you go the other way and push current back, the voltage likewise rises.

Why the voltage drops when you draw current from the voltage source is that your load is bringing electrons toward ground. Electrons are losing potential as they travel through the non-ideal voltage source's resistance. The voltage loss is V = IR.

When you force current backwards through the voltage source, you are driving that with a potential which is higher than that of the voltage sources. So now electrons are flowing the other way. Your voltage source is closer to the ground than the one driving it and so the current can be regarded as negative. The voltage difference is still V = IR, but the I has changed sign.

The above describes an "ideal non-ideal" voltage source: i.e. a voltage source which is non-ideal in the simplest, first order way (the inclusion of a constant ("ohmic") parasitic resistance).

A regulated power supply will not exhibit this behavior. For example, consider a simple a supply which uses a pair of transistors to create an arrangement where a feedback voltage regulates the current flowing through a pass transistor. Such an arrangement will stiffly (though not ideally) regulate the voltage from falling against increasing current demands. But it will not regulate against a reverse overvoltage. The overvoltage will give rise to a large feedback signal which will simply cut off the pass transistor. But beyond that, the current through the pass transistor will not reverse. The transistor looks like an open circuit at that point, and the supply is then just a passive Thevenin resistance made up of the internal voltage divider that provides the feedback signal.

Some basic voltage regulator circuits given here, with discussion: http://www.circuitstoday.com/controlled-transistor-series-regulator-with-overload-and-short-circuit-protection

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  • \$\begingroup\$ Thanks a lot Kaz. This is exactly what I need. Too bad I can' t vote up. My reputation is too low for now. \$\endgroup\$ – Zening Qu May 4 '12 at 6:51
  • \$\begingroup\$ I have another thing that I don't quite understand: according to your explanation when you draw current from an non-ideal voltage source the voltage should drop. But actually I haven't observed any voltage drop. Is this because there are diode-like element inside the source so that the impedance is different in two directions? I am guessing that the R in V = IR is different depending on the current's direction... \$\endgroup\$ – Zening Qu May 4 '12 at 6:56
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    \$\begingroup\$ The manual for these devices doesn't say much about the internals, other than a reference to fold-back current limiting. Brusing up on this by re-looking at some basic schematics for voltage regulation, I can easily see how it wouldn't be symmetric. The basic feedback regulator works by regulating current in a pass transistor, using another transistor which monitors a feedback voltage. The pass transistor current regulation really works in one direction only (that is your diode-like element). Excess voltage in the feedback cannot activate the pass transistor such that current flows backwards. \$\endgroup\$ – Kaz May 4 '12 at 16:46
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    \$\begingroup\$ So basically if we have a power supply is of this type (juice regulated through pass transistor) if we generate an overvoltage against the supply, that transistor can do no better than to cut off. And once it cuts off, that part of the circuit is open; the regulation is out of the picture. The voltage you see then is a combination of the internal resistors in the supply (the ones that make the voltage dividier for the feedback signal), the external ones you're using and the voltage you are driving with. \$\endgroup\$ – Kaz May 4 '12 at 17:00
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    \$\begingroup\$ @CherryQu, you should put that information in the question. The exact make and model of the supply and a photo of the whole set up could be helpful. \$\endgroup\$ – The Photon May 4 '12 at 18:11
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You probably mean that the voltage at node 4 is 5.7 V.

If there's current going right-to-left through the 150 Ohm resistor, then there must be a corresponding voltage difference between nodes 4 and 5 according to Ohm's law. So its not surprising that node 4 reads higher than 5 V.

If you are measuring 5.7 V at node 5, that would be very surprising, or your simulator is including some hidden behavior in the 5 V source that isn't shown in your drawing (i.e. an additional equivalent series resistance).

EDIT

Reading the comments and reading the question again, it appears you're asking about a physical experiment and not a simulation experiment.

You don't say what kind of voltage source you have, and that's very important to the behavior.

If your voltage source is a battery, then the "classic" "non-ideal voltage source" model, with an ideal source and a resistor in series is pretty close to accurate. That would be the case where there is an additonal resistance that you haven't shown in your drawing, and as I explained above, when you push current back through that resistor, it produces a voltage across that resistor according to Ohm's law.

If your voltage source is a linear regulator (whether a "classic" or a low-dropout type) it depends exactly which model you use. But classic linear regulators typically completely fall out of regulation when you push reverse current through them. They no longer behave like voltage sources at all. Most likely they connect the output back to whatever the input supply was through a diode.But again it depends exactly what regulator you use --- some have reverse-bias protection that would make them look like an open-circuit instead.

If your voltage source is a benchtop supply or a wall-wart supply, then again it depends exactly what model supply and what its internal circuit is. Depending on the designer's expectations for how the supply is to be used, it could continue to look like a voltage source, it could shut itself down, or it could fall out of regulation and behave erratically when you push reverse current through it.

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    \$\begingroup\$ But that's what a non-ideal voltage source is. The model of a basic non-ideal voltage source is an ideal voltage source plus a series resistance incorporated in that voltage source, right? \$\endgroup\$ – Kaz May 4 '12 at 5:15
  • \$\begingroup\$ It could be that the 5V is intended to be stiff voltage source and the 150 ohmer is part of the model of the non-ideal voltage source. (I.e. OP is taking the 0-4 to be the non-ideal voltage source, not 0-5). \$\endgroup\$ – Kaz May 4 '12 at 5:21
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    \$\begingroup\$ The voltage at node 4 will be 6.2V, not 5.7V! \$\endgroup\$ – Federico Russo May 4 '12 at 13:26
  • \$\begingroup\$ @Kaz, on your first comment, a non-ideal voltage source is anything that isn't an ideal voltage source. It could have a series resistor, it could have inductors and capacitors that reduce its ability to respond to transients. It could have nonlinear elements. In school we often study the case of a series resistor, but in a real experiment, it could be just about anything. \$\endgroup\$ – The Photon May 4 '12 at 15:43
  • \$\begingroup\$ @Kaz, on your second comment, that's the assumption I made about what the OP was asking; but I was too lazy to do the simulation to see if the reported result was consistent. \$\endgroup\$ – The Photon May 4 '12 at 15:44

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