In a npn transistor with 5 volts VCE and current limited to 10ma by a resistor, it needs to drive its base with a DC voltage ranging from 3 to 500 volts DC, using a base resistor of 10 K ohms the voltage in the base never exceeds 900mv, is my drive correct? In this case, how do you set the maximum current you can receive at the base?

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    \$\begingroup\$ That would need a 25 watt base resistor and I'm sure you'll agree that it isn't what you really want. \$\endgroup\$
    – Andy aka
    Jan 15, 2021 at 12:33
  • \$\begingroup\$ Since BJT driving by current, you need current stabilizer with wide range of voltage input. Circuit may be complicated. Use some automatically switching voltage divider, like used in autorange multimeters. \$\endgroup\$
    – user263983
    Jan 15, 2021 at 13:18
  • \$\begingroup\$ You can use 500k 0.5W voltage divider with 1mA to 5Vinput with a 5k load to drive it. Using 1W 500k \$\endgroup\$ Jan 15, 2021 at 14:06

1 Answer 1


If you want the device ON with an input voltage between 3 and 500V, you are going to need more circuitry, because at the high voltage, the base current will be too much.

I suggest putting a comparator at your input so you have a constant voltage to turn on the transistor.

  • \$\begingroup\$ I was thinking of using a 500k resistor at the base, and changing the transistor for a Darlington of gain close to 2000 hfe, that way it would trigger with 5 or 500 v. \$\endgroup\$ Jan 17, 2021 at 13:57
  • \$\begingroup\$ Don't confuse hfe with base current required to keep the device in saturation. That gain is typically in the range of 10 to 20. The hfe you mention is typically specified with Vce at 5V or more. electronics-notes.com/articles/electronic_components/transistor/… \$\endgroup\$ Jan 18, 2021 at 1:46

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