Diode failure monitor

Thread Starter

m121212

Joined Jul 24, 2011
96
Hi!

I'm working on a power electronics box where there will be 300VDC input power going to a load via a stud diode (maybe up to 200A rated!).

I need to have a circuit to monitor the diode for failures (open or short), and am doing some research to see what exists. There seems to be a lot out there already for brushless generators, but I believe most of those circuits are related to the AC nature of the application.

I'm considering something based on voltage and current monitoring.

If there is current flowing but the differential voltage across the diode is less than expected Vf, then you have a shorted diode.
If there is a large differential voltage but no current is flowing, then you have an open diode. (Not clear in this case what you do if the load is disconnected).

Curious if there is already a well tested circuit out there for this.
 

Thread Starter

m121212

Joined Jul 24, 2011
96
Yes, sure. It's a complex setup, but if I simplify it to the minimal relevant subcircuit it looks like this -

diomon.png

A power supply feeds a load via a blocking diode and a switch.

The idea is to add a circuit across A:B that can monitor the diode for failures.
 

joeyd999

Joined Jun 6, 2011
5,360
Yes, sure. It's a complex setup, but if I simplify it to the minimal relevant subcircuit it looks like this -

View attachment 313028

A power supply feeds a load via a blocking diode and a switch.

The idea is to add a circuit across A:B that can monitor the diode for failures.
What's the minimum forward voltage drop of D1 under non-failure conditions?

And what is the worst case reverse voltage seen by D1?
 
Last edited:

MisterBill2

Joined Jan 23, 2018
18,986
If the 300 volt supply is not present there would be zero voltage across the diode, the same as if the contactor is switched off. So it is also required to know the voltage on both sides of the diode, relative to the 300 volt supply negative side. If the diode os OK and the load current is flowing, there will be a specific, fairly low, voltage across the diode So that condition must be verified. If the switch is closed then the voltage on the diode is much less, then the diode is shorted . If the switch is closed and there are 300 volts across the diode, then it is open. So to be able to accurately know the condition of the diode you must also know the condition of the switch and the power supply. Otherwise you will only be able to verify when all is well with any certainty.
 
Last edited:

Thread Starter

m121212

Joined Jul 24, 2011
96
What's the minimum forward voltage drop of D1 under non-failure conditions?

And what is the worst case reverse voltage seen by D1?
Minimum forward drop may be around 0.7 or 0.8V at higher temps.
Worst case reverse may be several hundred volts if the load is inductive and there's a kickback. Diodes I can get with a high reverse blocking, say 1200V. This monitor circuit would probably have some large divider for protection.
 

Thread Starter

m121212

Joined Jul 24, 2011
96
If the 300 volt supply is not present there would be zero voltage across the diode, the same as if the contactor is switched off. So it is also required to know the voltage on both sides of the diode, relative to the 300 volt supply negative side. If the diode os OK and the load current is flowing, there will be a specific, fairly low, voltage across the diode So that condition must be verified. If the switch is closed then the voltage on the diode is much less, then the diode is shorted . If the switch is closed and there are 300 volts across the diode, then it is open. So to be able to accurately know the condition of the diode you must also know the condition of the switch and the power supply. Otherwise you will only be able to verify when all is well with any certainty.
I agree with that logic. The circuit needs A, B, and also C where C is the 300V return. I think you might not need to know the switch state, or even if the load is or isn't connected (it is really external to my box). If you sense the voltage between B and C is 300V it means the switch is open or the load is absent.
 

MisterBill2

Joined Jan 23, 2018
18,986
Protection against any inductive spike voltages can be simpler because the voltage sensing connections do not need to draw current, meaning that a fairly high resistance can be used, As for inductive spikes and the diode, if the switch opens then there would not be a voltage across the diode, but only across the switch. But as we have no hint of what the load is, all of that concern is based on unsupported speculation.
 

MrChips

Joined Oct 2, 2009
30,946
200A x 300V = 60kW

I don't think it is possible to do what you want to do.
You really need to give us the big picture.
 

MisterBill2

Joined Jan 23, 2018
18,986
200A x 300V = 60kW

I don't think it is possible to do what you want to do.
You really need to give us the big picture.
I described what is needed and what must be known in post #5. I did leave the details to the TS, who appears to be able to handle those details.
 

Thread Starter

m121212

Joined Jul 24, 2011
96
200A x 300V = 60kW

I don't think it is possible to do what you want to do.
You really need to give us the big picture.
The circuit shown in post #3 is certainly possible. This isn't an arduino circuit, this is power distribution. The bigger picture is that the load might not be a single load, but multiple loads in parallel, but that is not relevant to my question.

My question was whether anyone was familiar with circuits that would monitor a diode for failure.
 

LowQCab

Joined Nov 6, 2012
4,209
"" but that is not relevant to my question. ""
How do You "know" what is "relevant" ??????

A complete Schematic would do wonders towards getting more useful responses.
Please include expected Voltages, and Currents, and Part-Numbers.
.
.
.
 

Thread Starter

m121212

Joined Jul 24, 2011
96
"" but that is not relevant to my question. ""
How do You "know" what is "relevant" ??????

A complete Schematic would do wonders towards getting more useful responses.
Please include expected Voltages, and Currents, and Part-Numbers.
That is fair, i just don't have a schematic available. It's a hypothetical situation. It hasn't been designed yet. All I know is I will have a diode that is going to see some heat and could fail. If that is too general of a question and an entire system schematic and BOM is needed before consideration, then I am sorry I asked.

High level / notional I am thinking -
1) monitor the forward voltage during operation. If it goes outside an expected range, consider that a failure (fail open)
2) during non-operation (say contactor is open), try to reverse bias the diode. If current goes above something tiny, consider that a failure (fail short)
3) during non-operation (say contactor is open), try to forward bias the diode with something smaller than Vf and look for currents larger than expected (fail short).

Very hand-wavy I think it looks kind of like a DMM with some control logic for deciding when to do which check. But already in my mind this seems complex, so I am curious what others have done that is similar.
 

LowQCab

Joined Nov 6, 2012
4,209
Build it like a Battleship to start with.
Then make sure, as best You can, that the Circuit doesn't have any "Bad-Manners",
and then don't worry about it too much.

The whole reason for needing a Schematic is so that there can be a
sort of best-effort, "group-consensus" arrived at,
that the proposed Circuit is not likely to have any bad-manners
that are likely to put any Components at risk of smoke in the first place.

And, sometimes, the overall approach to the Circuit-Design needs some re-thinking
in order to be closer to the best solution to the overall problem that needs to be solved.

What is the overall Goal, and what is the best way to get there,
are both unknowns at this point.
All factors have to be on-the-table, and subject to change.
.
.
.
 

MisterBill2

Joined Jan 23, 2018
18,986
I can easily imagine a servomotor driver for an industrial machine that would 30 amps at 150 volts. It would not take many of those drives to draw 100 amps at a peak. And when creating systems for manufacturing, reliability and performance are much higher on the list than shaving a bit on cost by providing "just barely enough" capability.
What machines use that much power?? A programmable metal folding machine for one, and most robots, for another. Or another would be a machine with a bunch of hydraulic servo-valves and all DC control circuits. That would be a lower voltage, but still lots of current.
 

Thread Starter

m121212

Joined Jul 24, 2011
96
I can easily imagine a servomotor driver for an industrial machine that would 30 amps at 150 volts. It would not take many of those drives to draw 100 amps at a peak. And when creating systems for manufacturing, reliability and performance are much higher on the list than shaving a bit on cost by providing "just barely enough" capability.
What machines use that much power?? A programmable metal folding machine for one, and most robots, for another. Or another would be a machine with a bunch of hydraulic servo-valves and all DC control circuits. That would be a lower voltage, but still lots of current.
1. Lots of loads have high power ratings, for example https://stealthev.com/product/tesla-front-drive-unit/

Each drive unit houses an AC induction propulsion 3-Phase/4-Pole motor that can spin up to 16,000 RPM in large drive units with 335 – 475kW output, and 18,000 RPM in small rear and front Drive Units with 220kW output. Drive Unit housing includes the Motor, Inverter, and the Differential.
That's an AC load, and in my case the loads are not AC. But still valid. A good DC example is the DC fast charger for a Tesla battery, which caps at 250kW.

2. The load could be multiple loads in parallel. The loads could be momentary or non-continuous.

3. 200A is a max rating, not a use case. My wall wart says 12V 5A, but that doesn't mean i'm drawing 60W when something is plugged in. If it did draw 60W it could handle that.
 
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