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I am making a linear PSU capable of O/P 600v @ 300mA. This is for vacuum tube experiments. I need to connect/disconnect the load from the PSU using a switch. I will not turn OFF the PSU itself. I cannot figure out how to switch 600V. My search did not produce any promising results. All I found close is 500V DC circuit breakers that are intended for solar PV application. Those are costly, but I am in doubt how long these will survive at 600V.

This is the preliminary design. I have made it to show the proof of concept. The 2nd mosfet on the right will be used to disconnect the load. Sequence will be first cutoff the mosfet and then open S1&S2 during disconnect. During connect first close S1&S2 and then turn ON mosfet. The 1st mosfet on the left is used as regulator.

enter image description here

The circuit was simulated at LTspice. I have given the NO LOAD and FULL LOAD voltages. S1,S2,S3 are relays that will be controlled by some OPTO COUPLER. S4 is the initial inrush current control switch. During the mains power ON the S4 will be kept OPEN. Load impedence is 4.5K and its a Tube RF amplifier. Current is approx 100mA

I have not designed the short circuit protection circuit yet.

All comments/critisisom are welcome

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  • \$\begingroup\$ Use a household light switch . 240Vac= 680 Vpp which is more than 600Vdc \$\endgroup\$ – Tony Stewart EE75 Jun 12 at 9:38
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    \$\begingroup\$ 680Vpp is only 340V pk which is a lot less than 600V DC. \$\endgroup\$ – user_1818839 Jun 12 at 12:06
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    \$\begingroup\$ Also, an AC rating for a switch which expects self-extinguishing arcs isn't anything close to like a DC rating for a switch which cannot expect self-extinguishing arcs. \$\endgroup\$ – jonk Jun 12 at 18:55
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    \$\begingroup\$ @TonyStewartEE75 I don't think that a household light switch is optimized to break 600VDC. Also, your logic 240VAC = 680Vpp > 600VDC is completely incorrect, as the peak-to-peak voltage is measured between a positive peak and a negative peak of the AC signal, which occur at 2 different points in time. Therefore, a household switch will never see 680V across it. Also, breaking DC currents is very different from breaking AC currents. AC has a zero crossing point due the which the current through the relay will go to 0 at some point of time, thus extinguishing the arc. This is not true for DC. \$\endgroup\$ – Prathik Prashanth Jun 13 at 9:48
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    \$\begingroup\$ @SamGibson I think I got in the unsanctioned habit of "short answers in comments" because often those that downvoted answers with real effort, give NO effort to communicate their lame discredit. Seems like that ought to be a rule too. The effort to give feedback is stronger in comments. TY for the link and for your respect. \$\endgroup\$ – Tony Stewart EE75 Jun 13 at 15:27

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