Hey guys,

I have equipment that has a DC bus of around 800Vdc. How can I make a circuit to read the voltage value and when there is no voltage on the bus, activate a relay.

I thought about measuring the voltage on the dc bus with an op amp and then using a voltage comparator to activate the relay.

What would be your suggestions? Any examples?

 

what kind of equipment? servo drives etc that operate at those voltages normally have relay output for this.
if not, you could use voltage divider to bring the signal down, feed it to a comparator and on its output have a relay.

since this is decent size voltage, i would use more than one resistors connected in series for the top side of the voltage divider, and have them spaced nicely. also i would add zener diode across low side resistor - just in case.

for example 4x 220k for R1 and a 4k7 for R2, which should give you roughly 213:1 ratio so output of divider would be 3-4V.

 

I failed to mention the purpose, which is for security. The relay contact would prevent the door from opening if there was still power on the dc bus.

 

No need for a comparator if the voltage on the bus is zero.
Operate a relay directly off the 800 volt bus.

 

The relay should only be activated if the voltage is 0V or very close to 0V

 

Would it matter if the relay is normally on using spdt contacts?
How much variance is on the 800 volt bus?

 

The bus is under voltage or has a voltage of 0V, because when it is turned off until complete discharge occurs, the relay has to activate to allow access to it

 

OK but that can be accomplished using only a relay powered from the 800 volt bus

 

The relay will indicate that there is no voltage on the bus allowing access to it without risk of electric shock

 

I think what @sghioto is getting at is that if you choose your relay and set up thoughtfully, you can come up with a simple design in which when the relay has 800 V going to it the relay prevents the door from opening, and only when the relay has no power can the door be opened.

 

Hi,
you can use the relay with 220VDC coil from Finder :

55.32.9.220.0040 FINDER - Relay: electromagnetic | DPDT; Ucoil: 220VDC; Icontacts max: 20A; 55.32.9.220.004 | TME - Electronic components

and add the three resistors in series , 43 kOhm ,1W each.

Current is I=P/U=1/220=0.0045A. Voltage to be dropped is 800-220=580V. Additional resistor is R=U/I= 580/0.0045=129 kOhm. Resistor's power is P=I^2*R=0.0045^2*129 000= 2.6W.
It is better to use several resistors in series, so if you use three resistors with power rating of 1W , R=129/3=43 kOhm

 

My recommendation would be to use a 12 volt DC relay. Depending on how stable the 800 volt bus is either several series resistors or possibly in addition a voltage regulator.

 

My recommendation would be to use a 12 volt DC relay. Depending on how stable the 800 volt bus is either several series resistors or possibly in addition a voltage regulator.

The mentioned relay can operate at 176-242VDC. So, if DC bus voltage lies in range of 640-880VDC, everything is OK

 

you can use the relay with 220VDC coil from Finder :

the mentioned relay can operate at 176-242VDC. So, if DC bus voltage lies in range of 640-880VDC, everything is OK

How does that meet the TS's requirement below?

The relay will indicate that there is no voltage on the bus allowing access to it without risk of electric shock

The relay should only be activated if the voltage is 0V or very close to 0V

 

Hi,
you can use the relay with 220VDC coil from Finder :

55.32.9.220.0040 FINDER - Relay: electromagnetic | DPDT; Ucoil: 220VDC; Icontacts max: 20A; 55.32.9.220.004 | TME - Electronic components

and add the three resistors in series , 43 kOhm ,1W each.

Current is I=P/U=1/220=0.0045A. Voltage to be dropped is 800-220=580V. Additional resistor is R=U/I= 580/0.0045=129 kOhm. Resistor's power is P=I^2*R=0.0045^2*129 000= 2.6W.
It is better to use several resistors in series, so if you use three resistors with power rating of 1W , R=129/3=43 kOhm

 

That's why I suggested a 12 volt relay.
They typically drop out at appx 4 volts.

 

That's why I suggested a 12 volt relay.
They typically drop out at appx 4 volts.

But with a series resistor to drop the voltage from 800V, the relay will drop out at a much higher bus voltage than 12V.
I would think the bus voltage should be below 50V before the relay opens.

 

I have equipment that has a DC bus of around 800Vdc.

How much current can be drawn from this bus for the test?

I thought about measuring the voltage on the dc bus with an op amp and then using a voltage comparator to activate the relay.

Do you have lower voltages available to power the circuit?

 

How does that meet the TS's requirement below?

If TS need to activate relay at very low voltage, the only solution is to use the Voltage Relay. It's hard to find a relay for such a high voltage- 800VDC, so it is possible to use a lower voltage relay with voltage divider. Or to use the voltage comparator LM 311, LM 393, etc. Voltage divider needed

 

Here's my approach to keeping the relay pulled in until the voltage drops below 50V if the 800V supply can tolerate a 6mA load:
It uses a high-sensitivity, 48V relay (e.g. Omron G5V-2 48 VDC) to minimize the current draw.
The circuit has a high-voltage MOSFET in a constant-current circuit to maintain the relay coil current (and thus voltage) until the 800V drops below 50V.
(It's simpler for this application to regulate current rather than voltage).

The MOSFET will have to dissipate about 5W.
Edit: Since that's a lot of wasted power, a circuit using a resistive divider and a comparator would likely be better.