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For a design I am working on a wind energy control circuit. The circuit has a maximum voltage of around 450 Vdc. When powered off, I believe it is safe to discharge the capacitors through bleeder resistors. Simple calculations told me that a 100k equivalent bleeder resistor circuit draws around 2 W of power and this will still take around 11 minutes before an 4RC time is reached (98 percent voltage reduction). Obviously for efficiency and PCB space I do not wish to increase the discharge power to a very large extent if this is not necessary. I see examples of bleeder circuits that discharge the caps under 30 seconds. I had trouble finding general rules of thumb on what a "safe" discharge time is. Any ideas?

Another idea that came to mind, is a normally closed switch with a bleeding resistor circuit, that is open while the circuit is powered, and closed when the circuit has no power so that the bleeding circuit will be activated. I believe relays are not suited for switching high DC voltages so I am thinking about adding a depletion mode MOSFET. What do you think about this?

Thank you in advance for your response

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  • \$\begingroup\$ A normally-on transistor comes to mind so you can have the best of both worlds. \$\endgroup\$
    – winny
    Jul 8, 2021 at 13:36
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    \$\begingroup\$ "normally open switch" I think you mean normally closed. \$\endgroup\$
    – winny
    Jul 8, 2021 at 13:38
  • \$\begingroup\$ Ah yes, you are correct! thank you I edited that. \$\endgroup\$ Jul 8, 2021 at 13:41
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    \$\begingroup\$ I would definitely recommend a high voltage NC relay for reliability, Gigavac or similar. Also consider a voltage sensor/warning light and possibly a lockout. \$\endgroup\$ Jul 8, 2021 at 13:44
  • \$\begingroup\$ @JohnBirckhead At only 450 V, why just just a depletion transistor? \$\endgroup\$
    – winny
    Jul 8, 2021 at 14:22

2 Answers 2

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"I had trouble finding general rules of thumb on what a "safe" discharge time is. Any ideas?"

There are no general rules of thumbs, because the safe voltage and time really depends on the system at hand and what kind of troubles you are defending against.

This is a system engineering risk analysis problem, in general, not a purely electronics one.

If for "safe" you mean "safe for human operators to touch the circuit", you should guarantee that the voltage in any part of the circuit goes below a human-safe threshold in the time interval between the actuation of the power-down switch (or whatever other mechanism is used to power-down the thing) and the instant the circuit can actually be physically accessed by an operator.

So this may require some real-world simulation/assessment/guessing about how quickly an operator can open the chassis and access the circuit in the worst (i.e. quickest) scenario.

For high-safety environments, you may also want to design-in some kind of interlock between the power supply and the panel that allows accessing the circuit, so that no operator can even open the panel unless the circuit is safe to operate upon.

Defining a human-safe threshold voltage is a bit tricky. Usually 40V DC is considered safe for the risk of electrocution in normal conditions (i.e. the contact with the human body happens in a dry environment, with the operator's skin intact).

This is a relevant thread on this site.

This doesn't take into consideration other risks, e.g. sparks caused by mishandling the circuit that can induce fires in the system.

You may want to read this Wikipedia Article for general guidance, and then consult more thorough safety guidelines sources.

If you want to be sure, go for a <2V threshold. Or less than 1V if feasible by your bleeding circuit.

Once you choose the voltage threshold, it's a simple exponential decay with good approximation (if the bleeding circuit is complex and non-linear you may need to consider the switching time of your active parts).

Keep in mind that if the bleeding circuit is not a simple resistor, you may have some trouble determining the "time constant". Therefore apply generous tolerance margins. Anyway, once you determine you need to reach safe voltage in, say, 5 seconds, design your circuit to reach that voltage in half that time, or even less (don't taunt Murphy).

Note that if your caps need to be discharged because of some kind of emergency shutdown (e.g. water ingress detected in the plant), you may want to VERY CAREFULLY estimate how quickly the problem that caused the shutdown is going to "reach" the circuit and causes troubles.

You may also need to design-in some emergency discharge mechanism, perhaps using sacrificial components that will dissipate the stored energy and get "fried" SAFELY in the process (remember Fukushima: the emergency generators where themselves damaged, thus the energy of the core had nowhere SAFE to go).

BTW, don't think that discharging some stored electrical energy can be done in however small time you would need. Generating a very short, high-energy, electrical pulse can generate a lot of EMI (Electro-Magnetic Interference). This pulse could propagate as an EM wave or as a conducted electrical pulse and could wreak havoc in sensitive control circuitry nearby (e.g. an MCU control board).

If you don't take appropriate countermeasures (e.g. lengthening the pulse and/or smoothing its edges) when your bleeding circuit kicks-in, it could cause failures in other parts of the system, and you may be dealing with a chain-reaction of failures.

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  • \$\begingroup\$ Sorry for the late response. Very good explaination. I think I have some work to do on this! I will keep these comments in mind during the design. I will not make it totally fail safe for the lab test prototypes which only I will test. I will be extra cautious and use a simple 10 minutes bleeder circuit in mind and probably an external volt meter connection to check the existing voltage levels on the caps for now. For a later version I will improve safety by implementing these comments. Thank you all. \$\endgroup\$ Jul 16, 2021 at 9:45
  • \$\begingroup\$ @WillyBogard Yep. From your post it wasn't apparent whether you'r doing it for research purposes (e.g as a graduate project), for an advanced hobby project or you are developing a circuit that will be put into a well-engineered product from some company. If this latter is the case you may want to brainstorm the safety issues with all the engineering team. If this circuit is going to be placed in a wind turbine generator high above ground you may have all sorts of additional safety requirements. \$\endgroup\$ Jul 16, 2021 at 9:59
  • \$\begingroup\$ @WillyBogard Moreover, if you feel like, you could edit your post and disclose some more details of your project, so that people can put it in the right context. \$\endgroup\$ Jul 16, 2021 at 9:59
  • \$\begingroup\$ Well it is a university research project at the moment. The assignment is provided by a company with the idea to further develop it once it is something that is usable for them. \$\endgroup\$ Jul 16, 2021 at 10:11
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I propose you use an depletion MOSFET like this one:

schematic

simulate this circuit – Schematic created using CircuitLab

The downside is that it does not discharge your capacitor all the way to zero volt due to the gate threshold voltage of the depletion MOSFET, but if it's just for human safety, discharging down to ~5 V is good enough.

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  • \$\begingroup\$ Thank you for the response. I get the idea of your circuit. There are some errors in this circuit though. But thank you :) \$\endgroup\$ Jul 9, 2021 at 11:56
  • \$\begingroup\$ @WillyBogard Which errors? Let me know and I’ll correct accordingly. \$\endgroup\$
    – winny
    Jul 9, 2021 at 13:44
  • \$\begingroup\$ M2 is now doing nothing, R3 is attached to ground on its right side and when M2 switches on it is also grounded on the left side. \$\endgroup\$ Jul 9, 2021 at 16:29
  • \$\begingroup\$ @WillyBogard Dang it. Now I see it too. I’ll update it. \$\endgroup\$
    – winny
    Jul 9, 2021 at 20:28
  • \$\begingroup\$ I have not simulated this, but I feel that when M1 is conducting (suppose Vgs = 0V). There is no voltage drop across R2. All the current goes through Rpower and M1. All of the voltage will be across Rpower. In this way there will not be a voltage on the gate of M1. So M1 will remain conducting current. Switching M2 will not change that. I think Vgs will always be at 0 V. \$\endgroup\$ Jul 16, 2021 at 9:55

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