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I'm trying to measure the output voltage of a constant current LED driver that ranges from 140VDC to 170VDC. I'm using an STM32 MCU which is powered by a separate external power source to measure the voltage. The most common circuit is to use a simple potential divider to step down the voltage from my research. However, this means the ground will have to be shared between the two systems which I would like to avoid. I have multiple LED drivers and MCU pairs. All the LED drivers have their own power source whereas all the MCU shares the same power source.

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1. How can I achieve this at low cost?

2. Can I simply connect the LED- to my MCU ground?

Max
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    https://circuitcellar.com/research-design-hub/high-voltage-differential-probe/ Not isolated per se, because isolating analog signals is a pain. But plenty of scopes use this method. Note the multiple 1M resistors in the divider to prevent dangerous single point failures if a resistor fails short. – DKNguyen Jul 12 '21 at 02:53
  • HCNR200 HP linear optocoupler. https://www.digikey.com/short/2t5n8bzn – Russell McMahon Jul 15 '21 at 05:23

4 Answers4

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I can give you some solutions but without a schematic etc of your circuit this is only a guess. You could use a VCO (Voltage Controlled Oscillator) and use an optocoupler to connect to your computer. The frequency would be dependent on the voltage, and then the micro could measure this frequency and determine the voltage. My prefered method is to use linear optocouplers, they may do what you want and can be extremely accurate. Application Note #50 Designing Linear Amplifiers Using the IL300 Optocoupler by Vishey gives you many tested options along with explanations as to how it works. This will allow you to keep the grounds separate.

Gil
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  • I like the linear optocoupler option but based on the application note it requires an amplifier. The amplifier requires a low-voltage power that must be separate from my MCU. How do I obtain this power? – Max Jul 14 '21 at 09:50
  • Possible solutions: They make buck converters (DC/DC) that will allow you to power with the 170VDC. My suggestion since you have AC is use a small power supply, which you can get for less then $10.00 to your door from your favorite china supplier. They typically automatically switch from US to European voltages on the input. The ones I use output 1A. – Gil Jul 14 '21 at 15:25
  • Unfortunately, a $10.00 power supply kind of exceeds my budget as I have to replicate this almost 200 times. I was thinking if it's possible for me to put a DC/DC isolated converter module at the power input of each MCU? Essentially creating a virtual ground at each set. – Max Jul 14 '21 at 16:08
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The approach taken by mains powered bench multimeters and differential probes is to scale each of the inputs by a large factor (such as 100 or 1000) using a resistive voltage divider. The grounds of the two system do not need to be connected, but there is still going to be some issues:

  1. There still will be a limit to the maximum voltage difference between the two grounds. Bench multimeters, for instance, still require any input to be within like 1000V of earth ground, and they use protection devices like MOVs to handle transients above that limit.

  2. The inputs may be negative relative to your MCU's ground so you'll need a negative rail to measure them.

I think a more practical solution is to use an idea like @Gil suggested -- use a little power from the LED driver to power a measuring circuit which then transmits the data optically to your main MCU. For a power supply, have a look at Figure 2.2 of the application note:

Atmel AVR465: Single-Phase Power/Energy Meter with Tamper Detection

It's for drawing a small amount of DC power from mains, but if you remove C1 you can use it with a high-voltage DC source.

ErikR
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Other answers provide good options for using an optocoupler or a differential amplifier methods.

There is an easier method. Effort and risk wise that is.

You can use an Isolated Amplifier. TI sells this one, but also Silabs and Analog Devices have some.

amc1311 isolated amplifier circuit example

Benefit:

  • Low work, no complicated sensitive amplifiers.
  • Low risk of noise through ground loops.

Drawback:

  • Price, you also need an isolated power supply.
  • You may still need an attenuator and ADC frontend, since it's still an ADC.
  • Delay and bandwidth are limited.
  • Differential output requires two "simultaneous" sampled ADC channels.
  • Board space. It's an isolator, it requires significant clearances.

Isolated often isn't the answer to all your problems, but in this case it will allow you to develop a measurement circuit for a high side voltage measurements without in-depth analog knowledge about differential amplifiers.

Jeroen3
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  • I think the main issue is isolators are expensive and difficult to find. They seem to never be manufactured for very long before going obsolete. Still, one expensive part might be cheaper than several less expensive parts. – DKNguyen Jul 15 '21 at 01:24
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There is possibly another way, you will have to play with the transfer characteristics of an optocoupler. You can place the 170V to the LED via a resistor start with something in the 200K range. Then use the NPN output of the optocoupler as a pseudo current mirror. I do not expect this to be super accurate but with some calibration it may work for you. Use 1/4 watt resistors. If there is any chance of this being reversed bias you will need to add a diode in series to protect the optocoupler from that.

Gil
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  • @max. As above - and HP make an optocoupler designed for this purpose. It has a single LED input and two matched output transistors. One transistor is used in a feedback loop and the other on s may he'd cct in the output. – Russell McMahon Jul 15 '21 at 05:19