Better choice of voltage attenuation/signal conditioning

Thread Starter

mcculled

Joined Dec 14, 2020
3
I'm looking to measure an analog voltage signal in the range 0 - 100 V. The system has a 15 kV power supply and is typically running about 500 Amps.

I want to drop the voltage by a factor of 10 or 20. In the past I've used an optical isolator system like this to attenuate the voltage as well as do some signal conditioning.
I'm generally happy with the results, but product requires a module with a large footprint (their smallest module supports 8 channels and I only have the one).

My question is whether a simple voltage attenuator will work, and where to look to find one designed to handle 100 V signals. Would something as simple as these 20:1 attenuators that mechanics use work? I need 0 - 10 kHz bandwidth, but the more the merrier.

Papabravo

Joined Feb 24, 2006
16,094
I'm looking to measure an analog voltage signal in the range 0 - 100 V. The system has a 15 kV power supply and is typically running about 500 Amps.

I want to drop the voltage by a factor of 10 or 20. In the past I've used an optical isolator system like this to attenuate the voltage as well as do some signal conditioning.
I'm generally happy with the results, but product requires a module with a large footprint (their smallest module supports 8 channels and I only have the one).

My question is whether a simple voltage attenuator will work, and where to look to find one designed to handle 100 V signals. Would something as simple as these 20:1 attenuators that mechanics use work? I need 0 - 10 kHz bandwidth, but the more the merrier.
Is it going to be used to measure a fuel injector or primary ignition? If not, then I wouldn't use it without taking a look inside to see what it is doing.
If something goes wrong the manufacturer has absolved himself of liability when you use the product in an application for which it was not intended. You are dealing with dangerous voltage and current levels. A judgement against you or your company could reach into the multi-million dollar range if somebody dies as a result of your negligence.

How accurate does the measurement have to be and what does it represent?

Thread Starter

mcculled

Joined Dec 14, 2020
3
Is it going to be used to measure a fuel injector or primary ignition? If not, then I wouldn't use it without taking a look inside to see what it is doing.
If something goes wrong the manufacturer has absolved himself of liability when you use the product in an application for which it was not intended. You are dealing with dangerous voltage and current levels. A judgement against you or your company could reach into the multi-million dollar range if somebody dies as a result of your negligence.

How accurate does the measurement have to be and what does it represent?
The voltage through the system is what I'm trying to monitor. iIt's not a converted signal. Certainly nobody would die if it exploded or caught. It's very far away from any people, and I would test it out without immediately blasting full power through it...

That being said, short of designing the attenuation circuit myself I'm looking for ideas of other ways to attenuate the voltage signal. Accuracy of 1% of the measurement would be enough.

Papabravo

Joined Feb 24, 2006
16,094
I would start with two "high-value" resistors with a smaller one in between them. For example take two 5.11 K resistors and put a 511 Ω resistor between them. The total resistance of the three resistors is 10,731.0 So 100 volts across the three resistors would give a current of 9.32 mA. Take that 9.32 mA across the 511 Ω resistor and you have a differential voltage of 4.76 volts. Now at the other end of the scale, take 1 volt across 10,731.0 and you get a current of 93 μA. That 93 μA across 511 Ω will produce a differential voltage of 47.6 mV

That should get you in the ballpark. A differential amplifier with a gain of ≈ 1.05 will normalize that range to 50 mV to 5 Volts suitable for a microprocessor A/D converter.

The power dissipation in the 5.11K resistors will be 444 mW, so don't use surface mount 1/10 watt resistors for this application.
$(9.32 ma)^2\cdot5.11K\;\approx\;444mw$
1 Watt resistors should be used. do the similar calculation for the 511 Ω so you can size it appropriately as well.

I should point out that this solution is not unique. You can use larger resistors to pass less current through the attenuator and lower the power dissipation. I've tried to give you the flavor of how to design such a thing, but only you can understand your requirements and limitations. You could also incorporate a switch so the resistors are dissipating power all the time, but only when you need to make a measurement.

Thread Starter

mcculled

Joined Dec 14, 2020
3
I would start with two "high-value" resistors with a smaller one in between them. For example take two 5.11 K resistors and put a 511 Ω resistor between them. The total resistance of the three resistors is 10,731.0 So 100 volts across the three resistors would give a current of 9.32 mA. Take that 9.32 mA across the 511 Ω resistor and you have a differential voltage of 4.76 volts. Now at the other end of the scale, take 1 volt across 10,731.0 and you get a current of 93 μA. That 93 μA across 511 Ω will produce a differential voltage of 47.6 mV

That should get you in the ballpark. A differential amplifier with a gain of ≈ 1.05 will normalize that range to 50 mV to 5 Volts suitable for a microprocessor A/D converter.

The power dissipation in the 5.11K resistors will be 444 mW, so don't use surface mount 1/10 watt resistors for this application.
$(9.32 ma)^2\cdot5.11K\;\approx\;444mw$
1 Watt resistors should be used. do the similar calculation for the 511 Ω so you can size it appropriately as well.

I should point out that this solution is not unique. You can use larger resistors to pass less current through the attenuator and lower the power dissipation. I've tried to give you the flavor of how to design such a thing, but only you can understand your requirements and limitations. You could also incorporate a switch so the resistors are dissipating power all the time, but only when you need to make a measurement.
Thanks for taking the time explain how to calculate the current draw.

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