0
\$\begingroup\$

I need to implement a diode, which only operates, when enabled by external signal. Specifications:

  • When the enabling signal is off, there is no current irrespective of voltage (positive or negative, up to 40 V).
  • Enabled signal is on, after which the diode behaves normally.
  • I have searched a lot of alternatives: JFET, Mosfet, Bipolar transistor etc. but with no luck so far, except one having a two-way switch and diode in series (see attached picture). But I feel there must be simpler way. enter image description here Perhaps JFET could be used with suitable biasing and somehow allowing negative voltages in drain? Or, perhaps bipolar transistor enables this by some special arrangement?
\$\endgroup\$
11
  • 1
    \$\begingroup\$ what about an SCR ? \$\endgroup\$
    – Jasen Слава Україні
    Commented Dec 31, 2016 at 9:14
  • \$\begingroup\$ You don't need the diode in that circuit. Two back to back NMOS MOSFETs will do. But you need to get the gate drive voltages correct. \$\endgroup\$
    – Tom Carpenter
    Commented Dec 31, 2016 at 9:14
  • 1
    \$\begingroup\$ one diode + one mosfet will work, there's a diode blocking current in the other direction. \$\endgroup\$
    – Jasen Слава Україні
    Commented Dec 31, 2016 at 9:15
  • \$\begingroup\$ If timing is not a problem, get a relay. Or you could possibly replace the diode with a SCR, but you'll have the issue with its latching characteristic and it seems you do not want to remove the diode. \$\endgroup\$
    – Bradman175
    Commented Dec 31, 2016 at 9:15
  • \$\begingroup\$ Thanks for quick comments. I need speed, to the tune of microseconds, thus SCR and relay will not work. This idea of diode/single Mosfet might be workable, even though gate drive creates tough challenge. \$\endgroup\$
    – Jyrki
    Commented Dec 31, 2016 at 9:24

2 Answers 2

Reset to default
2
\$\begingroup\$

A p-channel MOSFET (source grounded) with a diode in series with the drain will almost perfectly perform your stated function. There is no need of a second MOSFET.

Your enable will be a negative voltage wrt ground (typically about -2 to -10 depending on the MOSFET) for 'on' and approximately 0V for 'off'. Choose the diode and MOSFET for your desired current, speed etc.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for the comment. Yes. This is the way to go: one Mosfet plus a diode. \$\endgroup\$
    – Jyrki
    Commented Dec 31, 2016 at 15:08
  • 1
    \$\begingroup\$ well, so please accept this answer, then, @Jyrki \$\endgroup\$
    – Marcus Müller
    Commented Nov 18, 2017 at 11:16
0
\$\begingroup\$

Since you basically want to construct a circuit equivalent to a diode and a switch in series, and want it to switch at high frequencies while not caring about leakage that much, you have various options. As Spehro Pefhany said, you can just have a mosfet and a diode in series and you'll be fine. However there are other ways to achieve what you want.

You can also use a transistor:

enter image description here

Or utilise a optocoupler, thus using a BJT instead of a diode.

enter image description here

HOWEVER THERE IS ONE HUGE PROBLEM WITH THE OPTOCOUPLER ONE. THE VEBO WOULD BE WAY OVER AND YOUR CIRCUIT COULD BLOW.

All you need to do now is go on digikey or element14 and search up the component that fits your needs as desired by using the filter option for frequency, current, voltage and etc.

\$\endgroup\$

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.