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If you havn't 400V NPN transistor for using in 325V net, is it acceptable use 1kV diode in series to NPN transistor (150V) collector? This is low current part of circut so additional losses is not important.

VT2 is used to increase Hfe of VT1 but has low Vceo. In case VO3.2 is closed VT2 is open and collector-emitter voltage less than 1V because almost all voltage will be on load. In other case closed VT2 will be burn by high voltage. So my question is will additional diode before VT2 collector protect VT2 from high voltage?

the part of schema

Second image is result of check transistor after tests. enter image description here

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    \$\begingroup\$ A small picture showing the voltages of the various terminal voltages will help understand your question because I am not able to visualize how the diode will reduce the collector voltage. \$\endgroup\$
    – sai
    Sep 27 at 12:26
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    \$\begingroup\$ 1kV diode in series to NPN transistor (150V) collector a forward-biased diode will not drop 100s of volts across itself so the 150V-rated transistor will die under 325V. A reverse-biased diode will not allow current draw so your transistor will not be in conduction. So, in any case, it's not acceptable. As sai stated above, put a schematic so people here can give you a better answer. \$\endgroup\$
    – Rohat Kılıç
    Sep 27 at 12:28
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    \$\begingroup\$ The only thing suitable is another transistor, in a cascode arrangement. \$\endgroup\$
    – Hearth
    Sep 27 at 12:50

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If you properly cascode several transistors (yours will need 3) it can be done: -

enter image description here

Image from here: https://en.m.wikipedia.org/wiki/File:Cascode-voltage-ladder.png

See also this document that offers an LTspice model. I can't vouch for it because I use a slightly different circuit for my 13 kV capacitor discharger (which unfortunately I can't disclose).

is it acceptable use 1kV diode in series to NPN transistor (150V) collector?

It's unclear how you think this could work. I can't imagine any way that it will.

So my (modified) question is will additional diode before VT2 collector protect VT2 from high voltage?

It's the same answer: It's unclear how you think this could work. I can't imagine any way that it will.

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No.

When the transistor is off, there will be minimal voltage drop across the diode (as no current flow), and all the voltage will appear across the transistor.

Unless it is rated for that voltage it will likely fail.

There are a wide range of NPN transistors that are suitable. For instance the ST STR1550.

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I have assembled simple test circuit and it works fine. I used load resistor, led for visualization, HER108 (1kV) diode and 2N5551 transistor (Vceo = 160V) in 230V AC net. How it works (as I thought). When transistor is open almost all voltage is on resistor. When transistor is close current flow is almost 0. Transistor as well as diode has parasitic capacitance so we have capacitor voltage divider, C (2N5551 ) = 36pF, C (HER108) = 50pF. Transistor peak voltage = 230x1,41x36/(36+50) = 135V that is less 160V

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  • \$\begingroup\$ This is incorrect, as can be verified by measurements. Indeed, for diode capacitance to matter, it needs to be charged to an even higher voltage, making things even worse. Most likely, the transistor simply breaks down around 300V, and you haven't measured it, or it hasn't failed in this service yet. \$\endgroup\$
    – Tim Williams
    Sep 29 at 14:57
  • \$\begingroup\$ Please don't leave comments as answers. That's not how you communicate here at Stack Exchange. Use the Add a comment link instead. Thank you for your cooperation and have a good day. \$\endgroup\$
    – MiNiMe
    Sep 29 at 15:19
  • \$\begingroup\$ Sorry for my comment. \$\endgroup\$
    – Y Sh
    Sep 29 at 15:26
  • \$\begingroup\$ Measurements isn't acceptable in this situation due low capacity values. Voltmeter will be a part of circuit and has influence on it. I will add image of checking this transistor in first post. \$\endgroup\$
    – Y Sh
    Sep 29 at 15:39
  • \$\begingroup\$ Measure with a multimeter anyway. The transistor is probably in avalanche. You really can't rely on capacitive dividers, because imbalance in leakage currents will affect things dramatically. \$\endgroup\$
    – Hearth
    Sep 29 at 15:43

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