6
\$\begingroup\$

I'm working on a device and I'm encountering some problems. Basically the device pushes varying currents through a variable load using a current source and an H-bridge. For safety purposes, the load needs to be grounded on both sides in case the power drops so that built up voltages over the load are able to dissipate.

Thus far I've solved this with the circuit down below, using two P-channel JFETs. If Vsupply drops below voltages on either side of the load, the P-JFETs will turn on and current can flow to ground. However, recently the requirements for the voltage supply have changed from 15V to 40V. I cannot find any P-JFET yet that complies with these voltage requirements.

enter image description here

I did consider using PMOSFETs instead of the P-JFET, however they do have the disadvantage that Vthreshold of voltage remains over the load. Also the choice for a JFET is not so easy, since many JFETs start to leak current if voltage over the load (Vload) starts to equal the Vsupply. The LSJ74A was one of the best options I could find, but it doesn't comply with the 40V requirements.

In short: I am looking for a solution that grounds both sides of the load if Vsupply drops to zero. Additionally, not leaking any current (or at least as little as possible), if Vsupply is at 40V. All while complying with a voltage-rating of 40V or more.

Any help is welcome, thanks already in advance!

\$\endgroup\$
1

2 Answers 2

4
\$\begingroup\$

Any help is welcome, thanks already in advance!

I don't think you need to use a JFET here. You can use diodes to ensure that the load voltage is clamped to nearly 0 volts when the power fails: -

enter image description here

Basically, as power fails (falls towards 0 volts), the diodes become forward biased and "drag" any residual voltage on the load to the falling Vsupply.

There will be a diode drop but, this will fairly quickly leech-away any residual charge. If the supply is good, the diodes won't conduct.

\$\endgroup\$
5
  • \$\begingroup\$ Note that if the load is e.g. a big capacitor, this will not completely discharge it. \$\endgroup\$
    – jpa
    Aug 18, 2023 at 17:51
  • \$\begingroup\$ @jpa it's a recognized method used far and wide in electronic circuits. Diodes leak and it may take a bit longer over the last few hundred millivolts. \$\endgroup\$
    – Andy aka
    Aug 18, 2023 at 18:27
  • \$\begingroup\$ Yes, it will ensure there is no positive voltage remaining compared to ground. But if you put a floating capacitor (with no other ground connections) in place of the load, it only sees two opposing diodes and will not discharge. This is unlike what a relay or the original JFET circuit would do. Some extra diodes to clamp any negative voltages to ground might help. \$\endgroup\$
    – jpa
    Aug 18, 2023 at 18:31
  • \$\begingroup\$ I think you'll find that the H-bridge lower MOSFETs will clamp and discharge negative excursions. Of course, the H bridge hasn't been disclosed so that's just an assumption that the body diodes will be there and conduct. If not, add two more diodes to restrict and discharge negative excursions @jpa (oh you just said that hehe) \$\endgroup\$
    – Andy aka
    Aug 18, 2023 at 18:49
  • 1
    \$\begingroup\$ Hmm yeah, that's probably true about the H-bridge. It probably even has the diodes proposed in your answer built-in. \$\endgroup\$
    – jpa
    Aug 19, 2023 at 5:54
4
\$\begingroup\$

An cheap alternative is to replace the transistor with 2 normally closed contacts of a relay. The coil will be powered from Vsupply. Some 48 Vdc coil can work with 40 V, check minimum operating voltage of relay. Another option is a 24 Vdc coil, but you need a auxiliary power supply.

\$\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.