# Current Measurement

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#### hj47

##### New Member
I am not sure how to measure the current draw on a pulsed circuit, i would like to measure the maximum current draw during each pulse.

I have attached a quick diagram of how i think it is done, but i am unsure of what numbers i use or if i have it correct?

From the diagram we have a 10v pulse & the reading across the resistor, from the example is 2v with a 5 Ohm resistor.

First up, is this where you check the current from the pulse?

The big question is how then do i get the current measurement, i get 2v across the resistor so the voltage drop from the original 10v is 8v.

Do i use the 8v figure or the 2v figure?

I=V/R = 8v/5 Ohm = 1.6A

Or is it

I=V/R = 2v/5 Ohm = .4A

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#### blackboxcommando

##### New Member
Hi HJ!
I had to consult my Ugly's Electrical Reference as I don't trust my memory on the basics.
It looks like you are trying to determine the current in a series circuit by measuring the voltage drop across a resistor of known value. So yes this is the place to take the measurement because the total current in a series circuit is equal to the current in any part of the circuit. And you use the voltage measured across the resistor in your Ohm's law formula.

I=V/R = 2v/5 Ohm = .4A

is correct.

Hope this helps!

BBC

#### hj47

##### New Member
Thanks BBC,

Yes, it does help me out a great deal.

Thank You

HJ

#### hj47

##### New Member
The reason i asked the original question is that when i place a DMM in series with a circuit the same as above to read Current i get .635 A but when i use the resistor method i get .4A?.

I would have thought the DMM would only read average current draw & this would be less than the .4A.

That's why i thought maybe you had to use the total voltage drop from 10v to 2v which is 8v & use this in the Ohms Law calculation?

A problem with the DMM maybe?

#### MikeMl

##### Well-Known Member
No, the resistance of the shunt built inside the DMM is a few mΩ, while the external resistor across which you are measuring the voltage drop is 5Ω.

The final current in the series circuit when the circuit is pulsed is determined by the inductance, the length of the pulse, and the total resistance in the loop. In one case the resistance is the DC series resistance of the inductor plus a few mΩ, while in the other case it is the inductor resistance in series with ! Why would the current not be different?

btw-to measure the current in a circuit without changing the current by inserting a current shunt resistor, the shunt resistance must be much less than the circuit resistance. In your case the DC resistance of the inductor is probably a couple of Ω. Try making your current shunt resistor 10mΩ or 100mΩ.

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#### hj47

##### New Member
Thanks MikeMl,

What Value resistor should i use to test this, the total resistance i have for the circuit is only .7Ω

#### Leftyretro

##### New Member
One thing that needs to be expressed is what measurement of current do you wish to obtain? Peak current, average current, RMS current? Due to the wave form you are wishing to measure that needs to be defined.

Lefty

#### jmaf

##### New Member
Hi hj47, you probably won't be able to match the theoretical current to the current in practice.

Your waveform, frequency and inductance on that coil will also influence the result.

The resistor method seems correct how you and blackboxcommand solved it. It's dropping 2V and so the current through it must be 400 mA as you calculated, the only problem there is that resistor probably isn't 5 ohm precisely, if you've got a precise ohmmeter do measure that resistor for a more precise calculation. Should be near .4 A though.

#### hj47

##### New Member
Thanks,

It is the Peak Current drawn for each individual pulse that i need to measure.

#### jmaf

##### New Member
Thanks,

It is the Peak Current drawn for each individual pulse that i need to measure.
What wave form is your pulse? Square?

What is the frequency?

The maximum current in DC would be 2A, so that's your absolute maximum current here. All AC waveforms and frequencies will be below that value.

#### MikeMl

##### Well-Known Member
Thanks MikeMl,

What Value resistor should i use to test this, the total resistance i have for the circuit is only .7Ω
As I said, if you are measuring current in a circuit by inserting a resistor, and then measuring the voltage drop across the resistor, in order not to modify the behavior of the circuit, the inserted shunt resistor should drop less than 1% of the total applied voltage. That implies 0.007Ω, or 7mΩ. Depending on the current range of your DMM, the internal shunt is likely 1mΩ, 10mΩ, or 100mΩ.

If you are measuring a complex, time varying voltage across the shunt resistor, most oscilloscopes have a 10mV/div scale, so with a 10mΩ shunt, your scale will be 1A/div.

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