How to operate a Variable DC source

[fog37]

TL;DR Summary

Understanding how to use the voltage and current knobs

Hello,
I am experimenting with DC variable power supply and want to make sure I am using it correctly.

My understanding is that it can output a variable DC voltage by turning the voltage know CW. The current-limit knob, as far as I know, is only there to set the maximum current that we would like the source to output. For example, once we set a certain voltage and connect a load, the load may draw a current I from the source that is above the set current limit and the source prevents that.

• In many cases, the current knob is turned completely CW to the maximum...Why?
• Given a certain load, whatever that load may be, how would we know what the safest amount of current would be for any specific voltage applied to the load itself?
• Source DC sources have three outputs: the positive +, the negative -, and a ground. What is the purpose of the ground if we only use the positive and negative outputs? Should the negative output be connected to the ground output?

Thanks!

 

[berkeman]

Good questions.

(If you can post the model number of your source, that will help in checking what I'm about to offer.)

Your understanding seems correct to me. There are some subtleties with different sources (some can be put in constant current CC mode, which is a bit different), but for a variable DC voltage source with an adjustable current limit, what you have said is correct. As long as the current drawn is below the set current limit, you will get the set output voltage.

If there is an overcurrent condition, the voltage supply can either just lower the output voltage to the level that matches the set current limit, or it can "crowbar" the output voltage until the power is cycled and the high current draw load is removed.

Typically switching power supplies limit output current using a "burp" or other transient retry mode, and typically linear power supplies use crowbar current limiting. A lab DC power supply instrument may just use a simple lowering of the output voltage until you raise the current limit to match the load.

When bringing up new circuit boards or ASICs, for example, we may set the current limit a little above the expected Idd/Icc for the device, and slowly ramp up the Vdd output of the supply while watching the current drawn. If there is a fault in the device that starts causing an overcurrent, using a slow ramp up with current limit let's us shut things off before we destroy the device. Sometimes you need an intact defective device to do failure analysis on, and just powering it up with no current limiting may not preserve the faulty first article device...

 

[berkeman]

In many cases, the current knob is turned completely CW to the maximum...Why?

You do this when you know the circuit or device will not overcurrent, and don't want to be bothered/confused by current limits lowering the output voltage.

Source DC sources have three outputs: the positive +, the negative -, and a ground. What is the purpose of the ground if we only use the positive and negative outputs? Should the negative output be connected to the ground output?

Many DC voltage sources have floating voltage outputs with respect to Earth ground. This is useful if you want to "stack" voltages from multiple sources (connect them in series), or if you don't want to have the powered device referenced to Earth ground (for some regulatory tests, for example).

 

[Tom.G]

Many DC voltage sources have floating voltage outputs with respect to Earth ground. This is useful if you want to "stack" voltages from multiple sources (connect them in series), or if you don't want to have the powered device referenced to Earth ground (for some regulatory tests, for example).

That configuration could be a dual supply, having both + and - outputs available, both being referenced to the GND connection. To determine what you have, you would have to either read the specs or measure the voltages from the + and from the - to the GND connection.

 

[sophiecentaur]

Summary:: Understanding how to use the voltage and current knobs

how would we know what the safest amount of current would be for any specific voltage applied to the load itself?

That information could well be given in the documentation that comes with the device. Most 'appliances' have a specified supply voltage and will consume a known amount of power. Otoh, 'devices' that are used within experimental circuits could be driven inappropriately 'hard' and could be forced to take more than their rated current. A current limited PSU can protect that device but that wouldn't be suitable for long term use and a finalised design should protect its own components.
Setting a maximum value for supply current may give misleading results unless the experimenter spots that the PSU has gone into limiting. The situation could be rather like a 'Brown-out' and could actually damage the circuit that's being tested.

 

[berkeman]

Setting a maximum value for supply current may give misleading results unless the experimenter spots that the PSU has gone into limiting.

That's never happened to me!

 

[fog37]

Thank you everyone!

This is the power supply:https://tacklifetools.com/products/dc-power-supply-variable-switching-dc-regulated-power-supply-with-4-digital-lcd-display-reverse-polarity-high-temperature-protection-110v-115cm-alligator-leads-included

This discussion about voltage lead me to a related thought expressed in the figure below.
When a simple load is connected to a DC voltage source, the difference in potential between the top and bottom leads of the load, $V_{top} - V_{bottom}$, is a number that change in time going zero, positive, negative. We don't know the absolute electric potential $V_{bottom}$ or $V_{top}$ of either lead, only their difference.

What happens if we connected the lower lead $V_{bottom}$ to ground, which is equivalent to attaching to planet Earth at two different points? That is not the same as connecting planet Earth in parallel with the lower wire. Or it it?
I think the electric potential $V_{bottom}$ will stay fixed, the same, while the electric potential $V_{top}$ changes going positive to negative relative to $V_{bottom}$? Why? The Earth connection to the lower wire make that section of the circuit an extremely large equipotential conductor. The load is connected to the battery + terminal through a wire while it is connected to the - battery terminal through a huge conductor (lower wire+ earth). Why would the potential $V_{bottom}$ not change much or at all?

 

[sophiecentaur]

The top left diagram is idealised. The diagram with the Planet Earth in it is 'practical' and would involve a lengths of circuit wires so not only the R_Earth in the third diagram would be involved. This sort of question, involving a hybrid situation of real and ideal values doesn't really have an answer but the R_Earth could be said to be in parallel with 0Ω - so it would have no effect.

Long connecting wires tend to be susceptible to voltages and currents, induced by EM interference so 'more information needed', I think.

 

[fog37]

thank you sophiecentaur.

In the case of a DC voltage source, things are easy to interpret: the lower branch of the circuit has an electric potential $V_{bottom}$ that is always 10V lower than $V_{top}$. We don't know what these potentials are, just their difference.

But if the voltage source was variable, the difference $V_{top} - V_{bottom}$ would change from positive to negative. When we create the connection with planet Earth, does $V_{bottom}$, which is the Earth+lower wire potential, stays fixed while $V_{top}$ moves positive and negative relative to it? But Why? Someone told me this but I cannot justify it...

I am thinking of this example to help me understand: a capacitor formed two spherical conductors A and B with radius $R_{A} >> R_{B}$. If we connected the two spheres to a DC voltage source, electric charge of opposite sign would become present on each sphere. For example, sphere A becomes positive and sphere B becomes negative. However, because $R_A >> R_B$, we would get $|Q_A >> Q_B|$ to achieve $|V_A = V_B$...

Is the condition $|V_A| = |V_B|$ correct? D

If we connected this unusual capacitor to a variable voltage source, the change in the amount of charge of the larger sphere would be smaller, hence its absolute electric potential wouldn't change much while the potential on the smaller sphere would sensibly change... Is that correct?

thanks again for the patience.

 

[eq1]

If we connected this unusual capacitor

Spherical capacitors are actually one of the standard geometries because we can actually do that integral. It if were cow shaped, that would be hard... :). But we’re getting off topic.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html

 

[fog37]

I guess I am thinking that if a small spherical conductor has higher electric potential than a larger spherical conductor it means that the smaller conductor, because of its smaller radius, has much less charge on it than the larger conductor. If the potential difference changes, the charge on the larger conductor changes but not very much, relatively speaking, keeping the large conductor potential nearly the same. This would explain why the conductor connected to Earth would keep a rather constant potential while the top wire potential changes more significantly...