Help to measure inrush current with an oscilloscope

[JPU]

Hi All

Thanks for looking.

I need to measure the inrush current to my circuit when it is switched on. I have a 24V Lifepo4 battery connected to a brush less controller circuit with a Picaxe controlled PCB dealing with user inputs.

Please can someone tell me if there is a simple way to do this with my scope. I haven't even tried yet as I fear damaging the oscilloscope. The average current of the circuit is 2.0 - 2.5 amps. I do not have access to a meter which has this facility and I only need to do this once so want to avoid having to buy a meter which has a feature to do this.

Thanks for your help in advance

JPU

 

[Colin Mitchell]

Place a one ohm resistor or even 0.1 ohm resistor in the positive lead and measure the voltage across the resistor.
For a one ohm resistor, each mV of voltage will be equal to 1mA of current.

 

[BobK]

But, unless you have a Digital Storage Scope, you better read it fast!

If you do have a DSO, do what Colin says, put in single triggered mode, and set the trigger level at some small voltage. Set the timing for the appropriately, maybe 10 to 100mS. Then turn the device on and you should capture a trace of the current used in that first 10 to 100 mS.

Bob

 

[JPU]

Thanks for your help.

Maybe a daft question, but can I use a 10K resistor?

 

[garublador]

Thanks for your help.

Maybe a daft question, but can I use a 10K resistor?

Unfortunately, no. The 10k resistor will severely limit the current the battery will be able to provide. I = V/R so the max current possible will be 24V/10,000 ohms or 2.4mA. You'll need a very low value resistor to limit the drop across the resistor which will limit the effect the added resistor has on your system. However, you'll want it high enough to give a voltage drop that's measurable by your equipment, which is why you need a resistor and (probably) can't just measure across a PCB trace or wire.

Also, make sure to pick a resistor that will be able to handle the power generated. At 2A, a 1 ohm resistor will dissipate 4 watts. A 0.1 ohm resistor will dissipate 0.4 watts.

 

[JPU]

Unfortunately, no. The 10k resistor will severely limit the current the battery will be able to provide. I = V/R so the max current possible will be 24V/10,000 ohms or 2.4mA. You'll need a very low value resistor to limit the drop across the resistor which will limit the effect the added resistor has on your system. However, you'll want it high enough to give a voltage drop that's measurable by your equipment, which is why you need a resistor and (probably) can't just measure across a PCB trace or wire.

Also, make sure to pick a resistor that will be able to handle the power generated. At 2A, a 1 ohm resistor will dissipate 4 watts. A 0.1 ohm resistor will dissipate 0.4 watts.

Thanks for explaining this to me.