How to test power supply peak current of a GSM module?
I dsigned a GSM module with 8W output.
Now I used a power supply for it, i.e. 12V/0.45A, that is stably displayed on the power supply, so I thought it was average current, 0.45A.
But now I want to test the peak current?
How to test it? Because GSM is pulse signal, so the peak current should be 8*0.45A, I only used one slot.
That should be calculated in theory.
But I need to test it, so how to do ?
Best,
Tony Liu
Don't understand what the problem is.
You can run the power supply with the actual GSM module, or connect an artificial load with a current waveform according to the GSM module datasheet. In any case, you would acquire the real current and voltage waveforms with an oscilloscope.
I have an idea, pls comment.
1. Insert a 15m Ohm/2512 resistor in the power supply wire.
2. Use two Oscilloscope probes to test the voltage difference bewteen the two terminals of the 15mR resistor.
3. Use the voltage divide 15mR, get the peak current.
Because the peak current is about 3A~4A, so the volatge should be 45mV~60mV.
Some high-end Oscilloscope Manufacturers ( Tektronix, Keysight ) supply "Current Probes" for their products but it's a bit expensive.
Hi,
This is a common current measurement method. I recommend to use a 4 wired sense resistor.
Klaus
Right now I am using a 0.1 ohm (1%) resistor placed in series with Vcc to measure the current of a GSM PA.
Never use a differential probe (like the one in the link below) for a GSM application. It gives you a lot of erroneous and non-consistent results.
The best is to use the setup that you mentioned, using two scope probes on different channels, and check the voltage difference. Most of the smart scopes can do (and display) this difference automatically.
http://www.keysight.com/en/pd-231731...9&cc=RO&lc=eng