Attenuation of ±100V to ±5V

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raj1391982

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Hello sir,
I am working on a project, making Data Acquisition System in which the signal from sensor is having strength of ±100V with current in mA. Along with this I have to provide isolation of 1.1KV, filter and drive up to 10 meter through cable….
Options that I have thought of are:
1. Attenuation: attenuate using ultra precession resistors network.
• Will this work?
• But what should be criteria in selecting resistors? And specifications?

Please suggest me something to kick-start my project on to the first step….
Thanx…..
 
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100 volts to 5 volts
You need two precession resistors. One will have 95 volts across it and the other with 5 volts. (95k and 1k) (I know; you cannot get a 95k resistor.) This divider will use 1mA and the top resistor will burn about 0.1 watts. If the 1mA is too much, then increase the resistors. (5X)

1.1kv isolation
I need to know, is there power available where you want to measure the voltage. This greatly effects how the isolation is done.

Ultra precession
What is your idea if precession? 5%, 1% 0.01% This changes things!
 
the 100V signal (+/-) will change at which freqiency? If it is high freqncy, you may have stepdown tansformer. perhaps any electrical cable can transport that voltage for 10m. At the max you may go for a twin core shielded cable. return wire will be the secon wire and the sheld will guard from ouside indution for power lines etc.,
after you get it to the test location, you can scale down as suggested by Ronsimpson. you can go for a 20:1 ratio you can also do by multiturn pot of say 1meg ohm.this will help reduce the load on the signal. with all that you have to be careful that any moment the lower side resistor becomes break or open circuit-- the interfacing electronics will JUST FAIL and Distruct few ICs. So think of another way with an isolation type arrangement.
 
"Along with this I have to provide isolation of 1.1KV, filter and drive up to 10 meter through cable…."

You need to solve this before dealing with the attenuation factor which will be easier. As posted already there is a need to know the frequency response needed for the input signal, not just it's voltage range.

Lefty
 
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