Hello all,
So to be up front, I'm NOT an EE! I know just enough to be dangerous.

I'm working on a project where I want to take 480VAC, step it down to be able to use a device that can measure 0-10vdc to sense either on or off. ...without using a transformer (don't want the bulk of the transformer)...

I've attempted a small full wave rectifier (KBP307) and a voltage divider circuit (1/4watt 2M & 27K resistors) that got me down to around 7.5vdc output, and appeared to be working... (famous last words!) I had it setup and all contained in some clear heat shrink and when I came back from the bathroom it was smoked quite nicely. Upon further research I found that a full bridge AC to DC will give you Vout =~Vin*3*Sqrt(2)/PI() so my 480vac ended up being ~650vdc (I didn't actually measure that output, but this is what the interwebs is telling me the theoretical output should be). There wasn't enough left to see what actually went first, but I'm thinking that the rectifier went as it has a max rating of 600V @ 3A. The calculated current across the should have only been about 0.2mA so it shouldn't have been an overcurrent issue.

I'm thinking a higher voltage rated rectifier? ...or am I just approaching this incorrectly?

Thoughts?

Thank you,
chuggins143

 

Vin 480vac
Vout 10V (at what current!)

without using a transformer

The negative side of the output voltage will be neutral or 480v without a transformer. In some cases that is OK but in most it is not!

 

I'm thinking a higher voltage rated rectifier? ...or am I just approaching this incorrectly?

This circuit will work forever, because reverse voltage on diodes never exceed 10 V.
To prevent electrocuting, resistors 1 Meg are connected in series with every HV wires:

 

First, consider that 480 volts AC is a dangerous voltage to work with. Not only are any shocks more likely to be fatal but in addition, at least in my experience, the available fault currents are much higher. !20 volts@20anps will damage your tool, 480 volts at 60 amps will evaporate that tool out of your hand. (Mild ARC FLASH damage).
So the bulk of a transformer can make it a lot safer.
The circuit in post #3 could work, but the choice of diodes is not what seems prudent. I suggest 1N4007 diodes rated for 1000 peak reverse volts. That is a safe amount above the peak voltage by an adequate margin. The fact is that with zero current the voltage drop across a resistor becomes very small. THAT is why picking diodes at the higher voltage is a better choice.
My guess is that this is to monitor the power to verify if it is on or off. Usually 480 volts is three phase, but here it may not be.

Here is a serious caution: Power presence monitoring systems are life critical systems because they monitor dangerous voltages. So the system must be very reliable.
The warning on the one system I connected cautioned: Warning, Hazard of DEATH in 4 inch high red letters. Amazingly, it has not been bothered with unaware folks poking around inside.

 

We really cannot suggest anything without knowing the current needed.

 

Good morning all,

Okay, so what I'm actually doing is sensing if an thermal cutout has tripped on an electric heater... measuring across said cutout... under normal operation there is no voltage differential yielding an output of zero, but if the thermal trips then there will be a potential across said device. I'm looking at 480vac as that will be our worse case scenario... the potential across the devices actually varies from 24vac up to 480vac, but typically on the lower end. Basically just looking to see if there is potential there or not so my DAQ system will catch it.

As far as current goes, as little as possible.... just enough to generate a voltage signal. I'm sensing the voltage with National Instruments hardware (LabVIEW driven system) so I just need to know if the thermal has tripped or not.

Trying to keep this as compact, simple, and robust as possible as I have to give it to our lab technicians who are actually doing the hookup and testing. I wrap everything in heat shrink tubing so no one can even think of touching "hot" bits. We are all aware of the hazards of working with 480, 277, 240-208, and 120. I will work up one of the units in #3 and let it sit powered up for several days before I even think of giving it to the lab to use just to make sure nothing gets to hot or has other issues.

Thank you all for the advice, words of caution and wisdom!
chuggins143

 

Is this a typical NA 480v Star connected supply, if so, the N will be at earth GND usually.

 

Is this a typical NA 480v Star connected supply, if so, the N will be at earth GND usually.

Yes it is 480v NA service. ...some of the heater circuits are leg to leg so that's the reason for the 480V differential...

 

A neon optocoupler across your contacts tells if they are open. When I did need such and could not find in the market, used superglue to attach a Ne-2 bulb to a half optocoupler and all wrapped in black shrink tubing. Will edit if can find the coupler link later.

Edited : at surplussalesdotcom:


Much better option than the robbery above :

 

Why not to use an optocoupler? PC817 , for instance. Your circuit will be galvanically isolated from the mains, that's the point. A resistor, a diode bridge and PC817-it's all you need

 

Never even heard of the optocoupler. Will these also work down to 24vac?

 

Never even heard of the optocoupler. Will these also work down to 24vac?

https://www.jameco.com/Jameco/workshop/Howitworks/what-is-an-optocoupler-and-how-it-works.html

 

Yes, you definitely don't want to try to detect that voltage without isolation from the mains.

Below is the LTspice sim of a circuit using a relatively high-gain Darlington output opto isolator driving an emitter-follower to minimize the current/power draw from the mains, which should do what you need over the range of 24Vac (green traces) to 480Vac (yellow traces):
The output is ≈4.3V when the AC voltage is detected with a 5V supply powering the output.

Anyone touching the circuit to the right of the resistors should not receive a lethal dose of current, even if the circuit to the right is not connected.

Note the bridge rectifiers don't see the high voltage, only the resistors do.
The bridge just has the 1.25V drop of the opto input LED across it.

 

Why proposing increasing Vp-p x 1.41 the already high mains with a bridge rectifier just to feed a LED ?? A single diode does it; and probable the opto LED alone can, properly buffered. Smoothing can be done at the isolated side.

 

I am working on a project where I need to detect 24VAC and 120VAC through isolation.

For the 24V I am using a AC isolator. That simplifies the circuit. C13 is used to the current to 3.6mA. R90 limits the current if the input is switched on at the peak AC voltage. Using C13 and not a resistor saves watts, in high voltage applications. On the computer side (right) there must be a pull up resistor to VCC. C9 and the pull up resistor makes a filter. This is needed because there are two short times when there is no LED current during the power line cycle.
For high voltage: I used a solid state relay in this case. It detects 120VAC and turns on a low voltage, low current relay.
C31 again is the current limiting device with out burning power. For 480V use a smaller capacitor value and 1kv rating. Many people use "X" or "Y" caps but it is not required. The "power line" caps are really good but with series resistors to limit current there is no high current so I don't see the need for those caps.
Many small resistor are not rated for more than 150 or 200V. The 1meg resistors need to add up to the power line peak voltage. Also R63, 65 will see the power line voltage at power up. They should also add up to the power line voltage.
I used dual diodes to make a bridge.
C32: If R64 =0 then C32 must be very large to keep current flowing during "zero crossing". With R64=1k then the voltage across C32 can have many volts of ripple and we still have LED current.
I probably would make R63,65 10k for HV applications.

This is how I am doing it. Isolation is your friend.
RonS.

 

This is how I am doing it.

You likely want a voltage rating of at least 1kV for the capacitor, for some margin.
If that capacitor shorts, you will have some fireworks.

 

I vote for the Vactec neon isolator. If they don't make one rated for 480 Vac. add an external resistance. Something in the 470K - 2 M range should work. The neon bulb operating current is way less than that of a typical LED-driven optoisolator, reducing the external resistor wattage rating.

ak

 

If that capacitor shorts, you will have some fireworks.

From doing that kind of testing, the surface mount resistors act as a fuse. I believe in adding series resistors with the capacitor.
For through hole resistors, metal film opens up much better than carbon comp resistors.
For 480V, I would use several capacitors in series with resistors across each.

 

Yes, you definitely don't want to try to detect that voltage without isolation from the mains.

Below is the LTspice sim of a circuit using a relatively high-gain Darlington output opto isolator driving an emitter-follower to minimize the current/power draw from the mains, which should do what you need over the range of 24Vac (green traces) to 480Vac (yellow traces):

Note the bridge rectifiers don't see the high voltage, only the resistors do.
The bridge just has the 1.25V drop of the opto input LED across it.

Okay, I'm about to show my lack of electronics knowledge here... (removed the pics to conserve space)... Won't the input side of the rectifier portion of the circuit still have the full peak to peak voltage coming into it? ...+650V, where the 1N4004 "Repetitive peak reverse voltage" is 400V (according to the data sheet I pulled down from Mouser)... Wont this damage the diodes?? ...or am I incorrect in my thinking here?

If I remember my basic electronics from college correctly the current though the two 250k resistors in series should be around 1mA? ...which is roughly 1/2watt...

Would increasing the input resistance from the mains to say two 1M resistors only have the effect of increasing the time that the 24v signal takes to move up to the full output? (Still trying to figure out why one needs to use two resistors in series rather than just a single resistor, but that's now a research project for me...) Only asking because I have 1M 1/4w resistors and the KPB307 rectifiers in hand today... I'd still have to order the 4N32 opto isolators, but I should have the rest of the parts here... Thought is that the 2M total resistance would pull the power consumption down to about 1/8 watt and allow me to use what I have on hand... or is my logic off here?

Thank you again for all of the knowledge and experience! It is greatly appreciated.

chuggins143

 

Won't the input side of the rectifier portion of the circuit still have the full peak to peak voltage coming into it? ...+650V,

Starting at the LED of U1. There is 1.5V across the LED. The diodes add more to about 3V. The voltage drop is across R1 and R2.