0
\$\begingroup\$

To switch 1kV DC at 5A, I usually use a mechanical contactor (<- Wikipedia) that switches both highside and lowside of a resistive load.

Any suggestions on a semiconductor (or non-mechanical part) to do the job? For me, especially the high side switch makes it difficult.

Both highside and lowside have to be switched to meet the guidelines.

Schematic

\$\endgroup\$
7
  • 1
    \$\begingroup\$ Keywords: IGBT, GTO, SiC MOSFET. \$\endgroup\$ Commented Apr 15, 2016 at 6:47
  • \$\begingroup\$ IGBT would work. Solid state relays. Maybe GaN MOSFETs. \$\endgroup\$
    – user76844
    Commented Apr 15, 2016 at 6:48
  • \$\begingroup\$ How to drive the high side gate? \$\endgroup\$
    – m00wn
    Commented Apr 15, 2016 at 7:14
  • \$\begingroup\$ Use isolated gate driver and isolated piwer DC/DC supply. I think, there are even combo of both. \$\endgroup\$
    – user76844
    Commented Apr 15, 2016 at 7:20
  • 1
    \$\begingroup\$ What guidelines? For example, you are unlikely to meet any safety guidelines with a solid-state isolation device. \$\endgroup\$
    – Transistor
    Commented Apr 15, 2016 at 8:03

1 Answer 1

2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Combined relay / switch + SSR.

Keep the relay or switch for safe isolation and lockout. A high-voltage DC SSR can be installed on either or both legs of the circuit. You need to meet:

  • Blocking voltage (> 1k V and maybe > 2 kV for safety) for your 1 kV supply.
  • Isolation voltage (> 2 kV) to prevent insulation breakdown between the high voltage and low voltage circuits.
  • Switching current (> 5 A) and allow for any surge due to cold resistance of the load, etc.).

A quick search found Sensitron devices, for example.

Your control system can now turn on RLY1 and leave it on in normal operation while using the SSR for normal control. This gives you the advantages of low cycles and wear and tear on the relay while maintaining the ability to isolate safely.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.