0
\$\begingroup\$

I am having this Current probe

I am using this to measure the inductor peak current.

Inductor is rated at 2.05A.

Schematics :

enter image description here

Input Voltage to the converter during measurement is 7.5V

I am setting this current probe to 100mV/A division and upon connecting it to the scope, I change the probe setting ON SCOPE to be 0.100V/A.

I am getting this waveform :

enter image description here

Could you please tell me how to perform the conversion?

Or is the peak current (upon start-up) 7.8A as shown or is there some conversion. Can someone tell me how to perform the conversion?

I am thinking that there is a conversion because the inductor would get damaged if 7.8A flows because it is only 2.05A rated?

\$\endgroup\$
6
  • \$\begingroup\$ Link to the inductor please and also how is the inductor being driven? \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2019 at 9:03
  • \$\begingroup\$ product.tdk.com/info/en/catalog/datasheets/…. 68uH part. Driven by NCV3063 Boost IC \$\endgroup\$
    – user220456
    Commented Nov 12, 2019 at 9:37
  • \$\begingroup\$ And, is the input voltage to the converter at 10 volts when you did the above measurement? \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2019 at 9:39
  • \$\begingroup\$ If you set both the probe and the scope to 0.1V/A, then you should need no conversion: the scope will show you the current directly. To help understanding whether or not 7.8A is reasonable we need more information about the circuit. As far as inductor current ratings, they are typically for continuous current based on heat dissipation characteristics (there is also saturation current you should care about). I'm not surprised that the inductor can survive a 0.5ms long pulse of 7.8A peak as your screenshot shows. \$\endgroup\$
    – joribama
    Commented Nov 12, 2019 at 9:40
  • \$\begingroup\$ schematics would help us further assist you \$\endgroup\$
    – joribama
    Commented Nov 12, 2019 at 9:48

1 Answer 1

1
\$\begingroup\$

If you assumed that the current was really peaking at 7.8 amps AND, that the inductor is 68 uH (as per your comment) AND, that it is used in a boost converter as the energy storage inductor then, you could calculate what the current was (providing you know the power supply voltage).

But, assuming that the current is 7.8 amps and that it rises from zero amps in 50 us to 7.8 amps, you can back-calculate the supply. Firstly the rise time of the current: -

enter image description here

Based on your scope setting, I estimate the rise time to be about 50 us. This means that \$\frac{di}{dt}\$ = 7.8/(50E-6) = 156,000 amps per second.

Multiply this by inductance and you get a measure of what the input voltage was during the measurement you made.

So, \$V= L\frac{di}{dt}\$ = 10.6 volts.

If your input supply was around this value then, the current your probe measured is probably correct and you should rethink the inductor you have used because it will get hot and ruin the efficiency of your boost conversion in some cases.

\$\endgroup\$
14
  • \$\begingroup\$ I have added the schematic. Is the 7.8A acceptable for the inductor? Won't it get damaged? \$\endgroup\$
    – user220456
    Commented Nov 12, 2019 at 10:22
  • \$\begingroup\$ OK, so it is the energy storage component but, what was the incoming supply voltage to the circuit when you measured 7.8 amps? \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2019 at 10:59
  • \$\begingroup\$ It is 7.5V only \$\endgroup\$
    – user220456
    Commented Nov 12, 2019 at 11:01
  • 1
    \$\begingroup\$ Well, maybe the on-time I reckoned to be 50 us is more like 75 us. I think the 7.8 amps is not far off. Yes, that sort of current is not great for that inductor because the core will saturate and switching efficiency will be poor. However, if the inductor doesn't get too warm then it will likely survive. \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2019 at 11:08
  • 1
    \$\begingroup\$ I can only say that with 75 us rise time for current, 68 uH inductance and 7.5 volts applied, the current will peak around 7 amps. On that basis I would say that your measurements are likely to be accurate. \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2019 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.