# Circuit to identify the voltage on the dc bus

#### massanetabr

Joined Jun 3, 2019
44
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?

#### panic mode

Joined Oct 10, 2011
2,524
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.

#### massanetabr

Joined Jun 3, 2019
44
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.

#### sghioto

Joined Dec 31, 2017
5,099
No need for a comparator if the voltage on the bus is zero.
Operate a relay directly off the 800 volt bus.

#### massanetabr

Joined Jun 3, 2019
44
The relay should only be activated if the voltage is 0V or very close to 0V

#### sghioto

Joined Dec 31, 2017
5,099
Would it matter if the relay is normally on using spdt contacts?
How much variance is on the 800 volt bus?

#### massanetabr

Joined Jun 3, 2019
44
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

#### sghioto

Joined Dec 31, 2017
5,099
OK but that can be accomplished using only a relay powered from the 800 volt bus

#### massanetabr

Joined Jun 3, 2019
44
The relay will indicate that there is no voltage on the bus allowing access to it without risk of electric shock

#### WBahn

Joined Mar 31, 2012
29,508
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.

#### Pyrex

Joined Feb 16, 2022
214
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

#### sghioto

Joined Dec 31, 2017
5,099
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.

#### Pyrex

Joined Feb 16, 2022
214
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

#### crutschow

Joined Mar 14, 2008
33,355
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

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#### JWHassler

Joined Sep 25, 2013
306
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

#### sghioto

Joined Dec 31, 2017
5,099
That's why I suggested a 12 volt relay.
They typically drop out at appx 4 volts.

#### crutschow

Joined Mar 14, 2008
33,355
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.

#### crutschow

Joined Mar 14, 2008
33,355
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?

#### Pyrex

Joined Feb 16, 2022
214
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

#### crutschow

Joined Mar 14, 2008
33,355
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.

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