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I am planning to use the Maxim Current Sense Amplifier(MAX9634HEUK+T)with a gain of 100V/1V implemented as HIGH-SIDE to measure the system current in both normal working mode and sleep mode.

I am using a Shunt Resistor of around 1R.

In normal operation the device consumes around 200mA MAX operating at 3.3V. So I am calculating a Vsense=0.2A*1R= 0.2V or 200mV

However, in sleep mode, the device draws a maximum of 200uA that gives a Vsense=0.0002A*1R=0.2mV

Q. What size of shunt resistor should I choose to accurately measure current as low as 200uA and as high as 200mA?

Q. The MAX9634HEUK+T has a max offset voltage value of 250uV meaning I won't be able to accurately measure the device current when it operates in sleep mode.

Q. The MAX9634HEUK+T comes with various gain options. What gain version should I select along with the shunt resistor to accurately and efficiently measure the current in both normal and sleep mode?

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  • \$\begingroup\$ How accurate is "accurate"? Can your load circuit cope with a 200 mV drop in its supply rail under full load conditions due to the shunt? Your second question (Q) is not a question. \$\endgroup\$
    – Andy aka
    Feb 19, 2022 at 14:10
  • \$\begingroup\$ Try this \$\endgroup\$
    – Andy aka
    Feb 19, 2022 at 14:45

2 Answers 2

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The way I see it, immediate things to consider are:

  1. Greater shunt resistor. So that for MAX current you get 3.3V output. Thus it has to be 3.3V/0.2V = 16.5 times your 1R, 16.5R. You may want to go to some more common value under 16.5, such as 15R. Downside: 15R * 0.2A = 3V burden voltage. So it's a no go. How much burden voltage lost on shunt can you afford? 0.2A * 1Ohm is already 0.2V lost.

  2. You can pre-amplify voltage difference across the shunt before feeding it into the current sense amplifier. With another differential amp. If you use 1R that gives 200mV vsense, you can amplify it by 15, therefore 200uV will become 3mV, which with 250uA offset of your MAX IC may be acceptable depending on device tolerance specs you aim for. Beware, this chip has its own offsets.

  3. Another current sense amp or maybe differential amp. There are chips with 25uV offset voltage which are still within affordable price range. If they're in stock, that is.

EDIT:

I would personally start with reducing your shunt to lose less voltage on it and first try 3.

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  • \$\begingroup\$ Thank you for the valuable feedback. Here is what I have come up with so far. I went with the Texas Instrument's INNA190 that has an offset voltage of ±15 μV (max). To reduce the burden voltage, I decided to go with a 0.15R Shunt resistor. The 0.15R shunt gives a voltage drop of around 15uV when sleep current is around 100uA and gives a voltage drop of around 30mV when normal operating current is around 200mA. The INA190A3 has a gain of around 100V/V so that gives me an output voltage of around 100V*30mV= 3V at max current and 100V*15uV=1.5mV at minimum sleep current. Would that work? \$\endgroup\$
    – Adnan
    Feb 24, 2022 at 7:13
  • \$\begingroup\$ @Adnan sounds reasonable, I don't see why not. Give it a go. \$\endgroup\$
    – Ilya
    Feb 24, 2022 at 7:59
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You can "enable" a shunt with a pMOS. Use two different shunts in parallel (each or at least one with a MOSFET) and you get gain selection. Take care that the MOSFET is fully saturated so it has as little effect on the measurement as possible.

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