How to use the LTC6102 to measure capacitor charging current from 1uA to 1A at 1000V?


I'm working on a project where I need to measure a capacitor's charging current from a maximum of 1A all the way down to 1uA. The capacitor is being charged from a 1.8kW programmable power supply which can output a voltage up to 1kV. The tricky parts of this project are the high voltage at which the capacitor is being charged and also the fact that the current can go as low as a few microamps after several hours charging. This leakage current must be measured with 0.5uA resolution, if possible and an error not greater than 2uA.

Is this possible with the LTC6102? I have several doubts regarding how to supply power to the LTC6102 and also if its high dynamic range "feature" applies to this situation and what kind of output error should I expect in real world applications.

I've tried Texas Instruments' AMC1301 isolated amplifier but I'm getting lots of offset and nonlinearity errors and due to the small input range, if I use a large shunt resistor, for instance 100R to get good results at low currents (< 50uA) I need to protect the shunt with an IGBT to avoid damaging the AMC1301 as the voltage drop across the shunt will exceed 5V for currents higher than 50mA. The AMC1301's input range is only +/- 250mV, so about +/-2.5 mA for a 100R shunt. To measure up to 1A I need to use another shunt of 0R2 in series to read up to 1.25A. I'm measuring at the high-side, so the capacitor remains grounded.

I would really appreciate your help because I'm stuck with the issues of having to use the competitors' AMC1301 product. Ideally I would like to have only one shunt resistor for the entire 1uA to 1A range. Is this feasible?


Habil Silva

  • 0
    •  Analog Employees 
    on Jul 19, 2019 5:59 PM

    Hi Habil,

    I'll let someone else answer your LTC6102 question. 

    Another solution to consider might be a log amp such as AD8304. You could use a 1000:1 resistive current divider in the return leg of the capacitor, and run the i/1000 sense current into the AD8304, configured for split supplies and VSUM= 0, as shown in datasheet Figure 10. Of course the output voltage is now proportional to log(I), not linear, so your downstream circuitry would have to accommodate. This usually means a lower cost ADC, because less resolution is required.

  • 0
    •  Analog Employees 
    on Jul 19, 2019 11:36 PM

    Hi Habil,

    The LTC6102 would be hard pressed to provide this dynamic range unless a non-linear method was used (like a diode in parallel with the sense resistor, but then there would be tempco issues). I agree that the AD8304 solution might be the best solution.

  • hi ,

    I find you mention the AD8304, and point out the Figure 10 in datasheet, but AD8304 itself cannot support 1A currnet, and what 1000:1 resistive current divider do you suggest?