2
\$\begingroup\$

I have a 5KV DC output from a Cockcroft-Walton multiplier. The current is around few microamps as per the simulation. To prevent discharge, the 200Mohm resistor holds the voltage of the Cockcroft-Walton multiplier stable. The 200Mohm resistor is rated for 22KV for the voltage tolerance.

The current sense resistor is rated for 600V.

Is it safe to connect the current sense resistor like this?

the actual circuit

The voltage across the resistor comes around 0.300V approximstely (calculated using Ohm's law.)

Is there a way to isolate the measuring setup from the high voltage?

They all share the same ground, and there is a seperate circuit that discharges the capacitors of the multiplier safely within itself without the current flowing to the common ground.

The ADC can measure differential voltages.

\$\endgroup\$
  • \$\begingroup\$ Precisely what circuit current are you attempting to measure? \$\endgroup\$ – Andy aka Jul 22 at 15:26
  • 3
    \$\begingroup\$ The only current being sensed is the current flowing through the 200M+10k voltage divider. It's not clear why you're referring to the 10k as a 'current sense' resistor. \$\endgroup\$ – brhans Jul 22 at 15:48
  • \$\begingroup\$ The 5KV comes from a another resistor. This is to test the resistance of resistor under High voltages. I am using the term "current sense" because it is a resistor especially designed to do so. \$\endgroup\$ – Deadpool Jul 22 at 16:47
3
\$\begingroup\$

You might want to slap a 1N4148 (or, better, two in series) across the 10K. That way if something were to spark or arc over it won't damage the instrumentation amplifier. A few K in series with the inputs wouldn't hurt.

There's not a lot of reason for the instrumentation amplifier, though the buffer helps because the input resistance of the ADC is not all that high (would load the input about 1.5%).

If you provide an isolated power supply you can add an I2C isolator and isolate the ADC from whatever is talking to it. You may have trouble finding parts (DC-DC and isolator) that are inexpensive and rated for 5kV continuous, so it depends on what the voltage across the isolation barrier will be.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks for the tip. I just figured the Input impedance of input pins of the Ins amp are high, it didnt need a resistor in series. I guess, I have to use this setup as it is. This thing is economically constrained. \$\endgroup\$ – Deadpool Jul 22 at 16:52
1
\$\begingroup\$

Galvanic Isolation

You need galvanic isolation.

Texas Instruments has a great series of videos on the subject of galvanic isolation which discusses inductive, capacitive and optical isolation techniques.

You also need to consider what happens if R1 – the component under test which has a good chance of failing – fails short-circuit:

$$ I = \frac{V}{R_2} = \frac{5000}{10000} = 0.5 \text{ A} \tag{1} $$

$$ P = I^2R_2 = 0.5^2 \cdot 10000 = 2500 \text{ W} \tag{2} $$

You could put a fuse or circuit breaker just after R1 to isolate the rest of the circuit when R1 fails short-circuit.

It would look something like this where the isolated 5 V supply could come from a 7805 voltage regulator supplied by a 9 V battery. The purpose of R3 is to equalise the potential difference between the two grounds to the left of the galvanic isolation barrier, while limiting current flow, so that the common mode input to IA doesn't drift out of tolerance. In your circuit, the equivalent R3 is 0 Ω because you are using a common ground:

Galvanic isolation for high voltage current sense resistor

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ This is a good idea!! Thank you for the suggestion! \$\endgroup\$ – Deadpool Jul 23 at 8:34
0
\$\begingroup\$

Use an optoisolator. Place the LED end in series with the HV current flow.

Place a 100,000 ohm resistor in series with the output of the optoisolator. Tie the top end of the resistor to +5v.

If the voltage at junction of opto and 100,000 ohm should change, you know the HV current changes. And perhaps you have a linear indicator of direction and how much changed.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.