Getting confused with the basics and using my power supply

[HobGoblyn]

I'm getting a little confused with the basics, more the equipment rather than the result. 

I've managed to blow the mA fuse in both of my multimeters, but according to my working out, it should have been fine. 

I'm also getting a little confused as to how to set up my power supply (which might be the cause of most of my problems).

I simply wanted to set the power supply to replicate say 2 x 1.5v batteries.


Doing other experiments, I found setting the amps to 0.5 seemed to be a good place, as my experiments only seem to draw as many amps as they need.  So I set it like




and as soon as I turn the output on with nothing connected, the amps drop to 0




I create the simplest of circuits, 3v, 470 ohm resistor (measures at 469.2)




I turn the power supply output on and the amps say 0.004, I take that to read 4mA ?



So to find the actual current, 3v / 469.2 ohms = 0.0063, so the current is 6mA

I measure the current by putting the meter in series




I have to use the Amp input on my meter as I've blown the fuse on the mA

This shows as 0.006 which is what ohms law says it should be.




While I ended up with the correct result, I'm not sure why I blew up both mA fuses, and I'm not sure why my power supply only shows 4mA.  I just want to make sure I fully understand everything that's going on before I go onto the more complex stuff.

OK, I'm initially setting my power supply at 500mA, but the second I turn the output on, it drops to zero, and when  measuring the current, it was  zero until I touched the leads, then it went to 4mA.

One meter has a 250mA fuse, so if it somehow got the 500mA, that explains how it fused, but I don't understand how it would have got it, the other meter has a 1 amp fuse which shouldn't have blown, even if it did get the 500mA.

This was something I was reading in a book that I thought I'd spend 5 mins trying as I hadn't measured current before. Two hours of confusion, then I check fuses in both meters

[rdl]

The current control on the power supply is a current limit control, the load is what determines how much current actually flows. If you set 500 mA, the the power supply will deliver up to that amount, but not more.

Multimeters are essentially a short circuit when measuring current. This is why the 250 mA fuses blew with a current limit setting of 500 mA.

[ataradov]

Also, power supply measurement is just an indicator of a rough value. It is not a measurement tool. It is not accurate at all. So your results are pretty well aligned with what they should be.

And yes, when the output is off, you are setting the limit. When you turn the output on, it switches to showing the actual supplied current. So with no load it would be 0.

And 1 A fast blow fuse may blow before power supply would react to the over-current condition.

The limit you set is a safety feature designed to potentially limit the damage, not fully prevent it. Especially for sensitive electronics.

[tunk]

A 2mA error on a 5A scale is a 0.04% error.
Also make this thought experiment: What would happen
if you instead used a (high-wattage) 2ohm resistor.

[ArthurDent]

Note that when replacing the fuses you use exactly the same type and rating of ceramic body fuses as the original to maintain the DMM's safety rating. 

[TimFox]

Setting the voltage and current controls on a lab PS determines the maximum value of each.  So, as you decrease the load resistance from a high value, the voltage starts constant, then decreases for low resistance values to keep the current at the set value.  Some power supplies have “fold-back” limiting, where the current limit decreases the current limit to protect the supply against short circuits.

[rstofer]

When you are setting the current, the output is shorted (internally) and REAL current is flowing, the set amount of current.

Maybe, when you switch the output on, the internal short is removed but the current still wants to flow (inductance?) and it spikes the output.  This is the kind of thing you can only see with a scope.  I'm just guessing but, if I had this as a problem, I would be going after it with a scope.  There just shouldn't be any glitches at turn-on.  It would be very bad for most circuits.

So, make the connections more 'permanent' in the sense that everything isn't a bodge and place a shorting jumper across the test leads using a breadboard jumper.  Neatness counts on breadboards! Start up the power supply, verify that the current is reasonable and then remove the short across the probes.

Since you now the maximum current is < 10 mA, why set the limit at 500 mA?  How about 10 mA?

[newbrain]

One meter has a 250mA fuse, so if it somehow got the 500mA, that explains how it fused, but I don't understand how it would have got it, the other meter has a 1 amp fuse which shouldn't have blown, even if it did get the 500mA.

Very often power supplies have a capacitor (some microfarad) directly in parallel with their output.
The current limiting circuitry cannot act on the current provided by this cap, and the energy contained in it is enough to blow a fuse as the current will momentarily spike, limited only by the internal equivalent resistance of the capacitor (ESR), the leads and contacts resistance and the resistance of the fuse itself: all very low, possibly < 1 total.

[HobGoblyn]

Thanks all

Since you now the maximum current is < 10 mA, why set the limit at 500 mA?  How about 10 mA?

I will make sure in future I work out the current and set appropriately.

I set it that way as I had a few experiments that needed 3v, thought I might as well use my power supply rather than a couple of batteries, and wrongly hoped that I could set the power supply once, to imitate a couple of 1.5v batteries, then forget about it while doing these very simple experiments. 

Many thanks

[ataradov]

It does not really matter. For smaller currents it makes no sense to set accurate limit. I don't think I ever set limit lower than 100 mA in my entire life.

Power supply will not protect your circuit. As newbrain mentioned, there is an output capacitor that will discharge into your circuit no matter what you do.

Also, setting lower current may trigger CC mode during initial transient consumption. And that can cause a whole slew of issues on its own.

[HwAoRrDk]

The current readout on these Tenma (aka Korad) power supplies isn't very accurate below 10 mA. I have one too and it's the same. In a case like this, trust your meter rather than the PSU.

[HobGoblyn]

I've replaced the fuse in my UT61E, this time I set the amps to 10mA and it worked perfectly.

As I bought a pack of 10 fuses, I tried setting it to 100mA as per ataradov's post, this also worked perfectly, so I will leave it set at 100mA for all my small experiments.

Many thanks for all your advice

[HobGoblyn]

I'm doing (or thinking) something very very stupid, I thought I'd grasped this.

I repeated the experiment in my OP but using 1.5v and 10.02 ohms



The circuits the same, just different value resistor and voltage




1.5v / 10.2 = 0.147A  = 147mA

So I set the amps of the power supply to 160mA



Turned it on.  Amps on power supply showed 144mA  which is about what my working out says.



I measured the voltage across the resistor





This was 1.46v

So far all looks fine.

Then I put my amp meter in series



This shows 74.23mA, the power supply shows 70mA





These appear to be the correct figure but divided by 2, and I can't work out why?

I've probably been at it too long and need a break

[ArthurDent]

One thing to keep in mind if you are using your DMM in series with your power supply and your load to measure the current drawn is that your DMM is now part of the circuit and it will have a voltage drop so your load will see less voltage than your power supply indicates. This voltage drop caused by an ammeter in series with a load is called 'burden voltage' and varies with the range you're using and your load. The lower the voltage and the higher the current you use can cause the error caused by this burden voltage to be quite high. If you use both your DMMs in your circuit, one set to MA and in series, and the other set to voltage directly across the load, you can see how this burden voltage affects your readings. Here is a good video that shows the effects quite well.

[HobGoblyn]

Thanks.  I was only using one meter at a time.

It gets even weirder.  If I use the Amp rather than mA part of my meter, it correctly shows 0.143A  and my power supply shows 144,





 but as soon as I go back to the mA, I get the wrong result

[Caliaxy]

Can you use your other DMM to measure the voltage across the resistor (while your Uni-T Ameter is still in series with the resistor)? What happens to the current value (as displayed on the power supply meter and on the Uni-T meter if you short the Uni-T meter (while in series with the resistor)? Try it with the Uni-T meter set to mA then on A.

[HobGoblyn]

Can you use your other DMM to measure the voltage across the resistor (while your Uni-T Ameter is still in series with the resistor)? What happens to the current value (as displayed on the power supply meter and on the Uni-T meter if you short the Uni-T meter (while in series with the resistor)? Try it with the Uni-T meter set to mA then on A.

Sorry for being thick, I presume when you say "what happens ....if  you short the Uni-T meter ....", by short,  you are saying measure the current (as if I understand things correctly, that is in effect shorting)?

When the Uni-T is set to A, it measures 0.145A, if I measure the voltage across the resistor (at the same time as  measuring the amps) , it's 1.45v,  the power supply says 144mA

If I change the Uni-T to mA, it measures 74.50 mA,  the voltage across the resistor is 74.28 and the power supply says 73mA

[ArthurDent]

Saying you're getting the 'wrong' results is misleading. The meters are accurately displaying what they are measuring but you haven't understood what they are actually measuring. You are using a 10 ohm resistor and when your DMM is on the higher current range the burden resistor is a very small resistance so the drop across the meter is very small-close to  zero volts. Once you switch to the MA range you change the value of the burden resistor from almost nothing to 10 ohms. Now instead of having a 10 ohm resistor across the supply you have a 20 ohm load and the current will be 1/2 what it was on the A range. The values of current you got were about 150 and 75 which verifies that the MA range indeed has a 10 ohm burden resistor equal to your load resistor and thereby giving you 1/2 the current.
 

[HobGoblyn]

Saying you're getting the 'wrong' results is misleading. The meters are accurately displaying what they are measuring but you haven't understood what they are actually measuring. You are using a 10 ohm resistor and when your DMM is on the higher current range the burden resistor is a very small resistance so the drop across the meter is very small-close to  zero volts. Once you switch to the MA range you change the value of the burden resistor from almost nothing to 10 ohms. Now instead of having a 10 ohm resistor across the supply you have a 20 ohm load and the current will be 1/2 what it was on the A range. The values of current you got were about 150 and 75 which verifies that the MA range indeed has a 10 ohm burden resistor equal to your load resistor and thereby giving you 1/2 the current.

Sorry, I misread your previous post with the vid, to be saying the problem was using two meters at once.  hence my reply saying I'm only using one meter at a time.

After reading this post, I re-read the previous post and watched the vid, and it makes a lot of sense and explains a lot, but as is always the case, the more I learn, the more I get confused

I understand everything that the vid (and you) are saying.  But that opens up a dilemma for me, when going through a beginners book.  I was expecting to simply do a few experiments to prove to myself I understand whats happening. And it's been harder than I thought. Then again, I have learnt a lot by them giving unexpected (to me)  results.   

I sort of feel that while starting out (in that I want to learn electronics properly from the beginning), having a mA or even uA setting on my DMM is almost pointless.

But I'm learning a lot, thats the main thing.

[Caliaxy]

As ArthurDent says.  A-meters don't measure the current directly, they actually measure the voltage drop across a small internal resistor (shunt or burden resistor, r) they put in series with the circuit. They display the results as current (using Ohm's law, I=u/r, where you measure u and know r). 

The current through the circuit changes when you insert the A-meter, from I=U/R to I=U/(R+r), where U is the voltage of the power supply and R is your external resistor (10 ohms). Ideally, the internal shunt resistor r has a very small value, much smaller than that of the resistor R under test (through which you want to measure the current), so (R+r) can be approximated as R. The bigger the shunt (r), the bigger the error. From your measurements, r is ~10 ohms on the mA scale and much smaller (negligible) on the A scale, hence the "instrument error". In the experiment I was suggesting, shorting r (forcing it to be 0 ohms) brings the current back to its original value and the voltage drop over r (as displayed by the A-meter) becomes, of course, 0V (displayed as 0 A).

[HobGoblyn]

I've done a few tests as I like to have proof    Same simple circuit, 1.5v,  each time a different resistor



1)  20.2 ohm   1.5/20.2 = 74mA  Actual amp meter reading = 49mA

Meter on mA has internal resistance of 10 ohm

20.2 + 10 = 30.2 ohm    1.5/30.2 = 49mA. Same as amp meter



2)  5.1 ohm (this was five 1 ohm resistors in series)

1.5/5.1= 294mA  Actual amp meter reading 114

5.1 + 10 = 15.1      1.5/15.1 = 99mA  close to amp meter  (only test that wasn't spot on)



3) 47.1 ohm        1.5/47.1 = 31mA   Actual amp meter reading = 26mA

47.1+ 10 = 57.1 ohm.    1.5/57.1 =  26mA  same as amp meter



4) 100.3 ohm    1.5/100.3 = 15mA   Actual amp meter reading 13.5ma

100.3 + 10 = 110.3 ohm    1.5/110.3 = 13.5mA  same as amp meter



5)  219.7 ohm     1.5/219.7= 7mA   Actual amp meter reading = 6.5mA

219.7 + 10 = 229.7 ohm.   1.5/229.7 = 6.5mA



6) 989.7 ohm     1.5/989.7 =  1.5mA  Actual amp meter reading 1.5

989.7 + 10 = 999.7 ohm    1.5/999.7 = 1.5ma


So it proves there is a 10 ohm resistance when meter is set to mA.  The only one that wasn't spot on was test no2, that was the only test that didn't use just 1 resistor, it had 5 in series.


If I set meter to uA  for test 6, instead of 1.5mA amp meter shows 877.u

This must mean the meter when set to uA has a 720 ohm resistance?

989.7 + 720 = 1709.7 ohm.  1.5/1709.7 = 0.000877 which I presume is 877uA ?

[ArthurDent]

I think you've got it figured out. 

[Siwastaja]

Right, and do note that a 10 ohm burden resistor (also called "current measurement resistor", or "shunt resistor") for the mA range is just huge. Sure, they (multimeter designers) get a larger voltage drop over this resistor, making it easier to measure accurately, but such high value disrupts many actual circuits (as you have found out).

A higher-quality meter would use a smaller burden resistor (say, 1 ohm), and a better amplifying circuit (higher gain, lower offset, lower drift, lower noise...) inside to still measure the current accurately, even though it has to work with a smaller voltage drop over the measurement resistor.

Though, the principle stays the same: even if the burden resistor was just 1 ohm, you would still see a difference between your expectations and reality.

Voltage drops happen in almost all "high current" circuits, in wiring, contacts, current measurement resistors, etc. Hence, having another multimeter available just to read voltage at the point of interest, bypassing wiring losses etc., is a good idea.

[AVGresponding]

I prefer to make my own shunts for some measurements, as I can make them 0.1 ohm or 0.01 ohm for example, and using many paralleled resistors also gives you greater current measuring capacity.

You do need a quite high precision mV reading meter to make use of these at lower currents, but they do have the advantage of having a low enough burden voltage not to interfere with most circuits operation.

[Fungus]

...my experiments only seem to draw as many amps as they need.

Ohms law.