How Can I Use a Power Supply as Both a Voltage and Current Source?

[fog37]

Hello,
Certain power supplies can be both a voltage source or a current source.
a) Voltage source:
We set the voltage to a specific value we like and change load resistor. The current will vary (get smaller for larger R) but the voltage will not vary. To make the power supply a voltage source, I think we need to set the current knob to its maximum and the voltage knob to zero before turning to power supply on. Is that correct? Why?
b) Current source:
We set the voltage to a specific value we like and change load resistor. The current will vary (get smaller for larger R) but the voltage will not vary. To make the power supply a current source, I think we need to set the current knob to its minimum and the voltage knob to its max before turning to power supply on. Is that correct? Why?

One power supply I have seen has three terminals: positive, negative and ground. Which pair do we use when we want the power supply to be a voltage or a current source?

Thanks!

 

[davenn]

a) Voltage source:
We set the voltage to a specific value we like and change load resistor. The current will vary (get smaller for larger R) but the voltage will not vary. To make the power supply a voltage source, I think we need to set the current knob to its maximum and the voltage knob to zero before turning to power supply on. Is that correct? Why?
b) Current source:
We set the voltage to a specific value we like and change load resistor. The current will vary (get smaller for larger R) but the voltage will not vary. To make the power supply a current source, I think we need to set the current knob to its minimum and the voltage knob to its max before turning to power supply on. Is that correct? Why?

the first part of your 2 descriptions is identical so it doesn't really tell the difference ... can you see why ?
.
the second parts of your comments, I would not do either of those because of the risk of damaging the device

have you actually googled the definitions of voltage and current sources ??
it mite be a good idea and then come back with any questionsDave

 

[berkeman]

As Dave says, your explanations are not correct. Your Voltage Source explanation is almost correct, but it's better if you understand how the "Current Limit" feature of a Voltage Source works.

And for Current Source, you don't use a Voltage Source in Current Limit mode, usually.

One power supply I have seen has three terminals: positive, negative and ground. Which pair do we use when we want the power supply to be a voltage or a current source?

A Voltage Source Power Supply with 3 terminals like that allows you to "float" the output voltage with respect to Earth Ground (the 3rd terminal). That is useful for some situations, especially if you are "stacking" several power supply outputs to make a higher voltage or some other configuration.

 

[Babadag]

Actually, as Dave already said, both source are the same. If the source is strong enough and the voltage drop negligible you may consider a voltage source independent of load.
In order to supply a fix value current in a circuit the source impedance should be so large as the load impedance does not change it.
V=Z*I where Z=Zsystem+Zload and if Zsystem is -for instance- 100 times Zload you may neglect the load.
Positive, negative and neutral ? It could be a d.c. source with grounded middle way.
So you have “+” to neutral [in this case it is the “-“] half a voltage and the second part
the neutral will be the “+” terminal and the “-“ second and you’ll get the second half of voltage “+” to “-“.

 

[jim hardy]

The current limit adjust is there so you can protect your load against destruction by overcurrent. A big power supply can let the smoke out of an experiment that goes awry. Limiting current to less than 1/10 amp might well save an expensive or hard to find part.

Therefore you should think of it as a "Foolproofing" provision, in same spirit as a fuse . You can use it to set current as you describe but read the directions carefully, because in doing that at high current for extended time you might overheat its internal parts . Make sure to observe any precautions that are in the instruction book.

It's a foolproofing feature but not a Darn-Fool-Proofing one.

Estimate worst case power dissipated internally by (Vmax-Vout) X I max and plot with Vout on horizontal axis . That'll help you in your studies.

old jim