Simple current-controlled circuit

Thread Starter

mikla90

Joined Sep 13, 2018
11
Good morning everybody!

My boss asked me to create a very simple circuit, with 5 resistors in parallel powered by a current-controlled PSU. I need to measure the voltage drop across every single resistance once a day for testing purposes. Can I do it with all 5 connected and just measure the voltage drop on every single resistor or should I disconnect all other resistor one at a time in order to not have any false reading influenced by the other components?

Thanks!
 

crutschow

Joined Mar 14, 2008
27,736
with 5 resistors in parallel powered by a current-controlled PSU. I need to measure the voltage drop across every single resistance once a day for testing purposes.
If the resistors are in parallel, then the voltage drop is the same across all the resistors.

Connecting them in series however, would allow you to measure each resistor voltage individually to determine their resistance (if that is the purpose of the measurement).

This can be done without disconnecting as long as your measurement circuit has good common-mode voltage rejection.
 

Thread Starter

mikla90

Joined Sep 13, 2018
11
Another way could be to put all resistor in series and measure the voltage drop between the negative of the PSU and each resistor, then subtracting the precedent value. That could work, right?
 

Thread Starter

mikla90

Joined Sep 13, 2018
11
Then they must all 5 be in parallel, unless the boss has had a change of heart.
My bad, actually! The boss wants to verify how long resistance can survive under current, by measuring the voltage drop across them. Given this, can I do what I proposed?
 

AlbertHall

Joined Jun 4, 2014
11,505
A similar situation I came across was a company making automotive lamps. They had a large array of the lamps (perhaps 200) connected to a power supply running in constant current mode. They then noted how many lamps had blown over time to establish a failure rate.

This method is flawed. When a lamp blows, the set current is shared between fewer lamps and so each lamp now has an increased current and so the failure rate accelerates over time. I advised them to use the power supply as constant voltage then when a bulb blows it does not affect the other lamps.

In your case the same problem applies.
Apply a constant voltage across the five parallel resistors and then monitor the current from the power supply to determine if a resistor has blown. Say each resistor draws 1A so the total current should be 5A. If the current is just 3A then two resistors have blown.

Whie they are connected in parallel you cannot determine which resistor has blown.

If you need this information then you need to use five ammeters in series with each resistor. This coud be five small shunts connected in series with each resistor which you can monitor the voltage across to determine the current in each resistor.

Alternately you could connect the resistors in series but then when one of them blows it disconnects the current from the other four so it terminates the test at that point.
 

Alec_t

Joined Sep 17, 2013
12,206
Connect the resistors in series, as mentioned already. But as soon as one fails you will need to bypass it so that the others can continue under test. I suspect your circuit could end up more complicated than you would like.
 

Uilnaydar

Joined Jan 30, 2008
118
In that case the test is flawed and will not give you meaningful results.
If the boss says parallel and the power source is required to be constant current. The only way to measure resistance is with 4x current sensors and a voltmeter. (This assumes the constant current sources has an adequate current meter on it.... mine sure doesn't!)

Run the test and good ole V/I will give you the resistance. The 5th resistor will need to have I derived from supply current minus the other 4 measured currents.

I look at this as an excuse to get MORE test equipment.... YES! BUY BUY BUY!
 

Thread Starter

mikla90

Joined Sep 13, 2018
11
Hang on, I should have specified better the test parameters.

Mandatory is the current supply. The fact that one blows up and I need to bypass it is not much of a problem for us, as I would do it anyway.

So, as a recap: if I put them all in series with a constant current supply and then measure all voltage drops from the negative of the PSU, then just by subtracting the previous voltage I'm able to calculate the resistance of every single DUT. Right?
 

AlbertHall

Joined Jun 4, 2014
11,505
Mandatory is the current supply. The fact that one blows up and I need to bypass it is not much of a problem for us, as I would do it anyway.
Unless you use some sort of 'automatic' bypass, the current through the other four resistors is interrupted until the byapss is put in place and so any timings would be disrupted.
You could use a zener diode across each resistor with a voltage rating higher than the maximum expected voltage across each resistor and a current rating higher than your constant current.
 

Thread Starter

mikla90

Joined Sep 13, 2018
11
Well, the current is constant, so that wouldn't be a problem.
As for the bypass, the test is running only during work hours, so it's not a problem.
 

ebp

Joined Feb 8, 2018
2,332
If you use a constant current power supply and paralleled resistors, if one resistor fails open-circuit (the usual failure for a resistor) the current will be shared by the remaining resistors. This leads to a
bang...................bang..........bang...bang.bang
failure scenario, where the surviving resistors become increasingly overstressed as others fail

Measuring voltage from the "bottom" end of a string of resistors and calculating the differences is certainly a reasonable thing to do provided the meter you are using has sufficient accuracy and resolution for the requirement. High resolution (e.g. 6 digit) meters are useful for this sort of thing.
 

crutschow

Joined Mar 14, 2008
27,736
if I put them all in series with a constant current supply and then measure all voltage drops from the negative of the PSU, then just by subtracting the previous voltage I'm able to calculate the resistance of every single DUT. Right?
Yes.
But it would be more accurate to measure the voltage drop across each resistor.
 

Thread Starter

mikla90

Joined Sep 13, 2018
11
Ok, because of the type of test I'm conducting I'm more interested in the difference in resistance rather then the actual resistance itself (so if like after 100 hours resistance is less or more than when I started) so measuring from a common point looks accurate enough for me.

Thanks guys!
 

AnalogKid

Joined Aug 1, 2013
9,355
It sounds like this is some kind of burn-in or aging test, yes?

Are you testing to see if the resistors change in value with respect to each other, or with respect to their intended value? For example, if all resistors change over time from 100 ohms to 110 ohms, that is a 10% change. But if they all changed the same amount, then the differences among the group might be less than 1%. Which is more important.

Also, what is the nominal resistor value and what is the test current?

ak
 

Thread Starter

mikla90

Joined Sep 13, 2018
11
It's actually a life test. For simplicity, we make wires and have to validate how they degrade in time. Current depends on the type of wire, from 1A to 60A.

Keep in mind that every difference will be measured, as they don't go from 0.5 Ohm to 10 Ohm in a day. So, I'll be able to track every change with enough accuracy.
 
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