0
\$\begingroup\$

I have a design that has a +24VDC supply, a 7805 voltage regulator converting it to 5 volts and then a microcontroller and some peripherals behind the regulator.

The design and the 24V supply connector isn't meant for hot swapping, but users still do it. If the 24V connects before the ground, it seems that the 7805 acts pretty much as a short and everything behind it is at 24V (minus some voltage drop). When the ground connects a moment later, the microcontroller gives up the ghost, since all bypass capacitors have now 24V over them.

What could be done to fix this? Unfortunately the connector cannot be changed. I could change the regulator, but no datasheet mentions anything related to this. 7805 has a Darlington pair as an output transistor, would a MOSFET based regulator, such as LP2950 fix this? Another solution that came to my mind is using a 5.6V Zener between the 5V and the ground.

Edit

Here are the simplified schematics:

schematic

simulate this circuit – Schematic created using CircuitLab

Simplified because there are a couple more ICs, but those are connected as the microcontroller, with a 100nF bypass capacitor. Those ICs typically die as well.

Microcontroller inputs have no external pull-down resistors to ground, but I2C and reset are pulled up to 5V.

Only connector on the board is the supply connector, so ground loops aren't possible.

I found this forum thread discussing about the same phenomenon, and to quote: "Just tested it. 12.00 in 11.52 out. --- So, I guess that loosing the ground will fry your circuit."

Improve this question
\$\endgroup\$
5
  • \$\begingroup\$ Need a schematic! If the bypass capacitors are from 5V to board ground, then the micro will have given up the ghost before the 7805 kicks in. If the bypass caps are from micro inputs to global ground, then I can see the micro dying when the board ground is suddenly lowered by 24 volts. Could you distill your no-doubt complicated case to the power input and the bypass caps in question, including exactly which ground they're referenced to? \$\endgroup\$ – TimWescott Feb 12 '19 at 16:03
  • \$\begingroup\$ Sure, I'll update the question with a schematic in a couple of hours \$\endgroup\$ – jerry_c Feb 12 '19 at 16:06
  • \$\begingroup\$ do not allow CMOS inputs or outputs to exceed the supply rail or else you get SCR shorting out the 7805. so its not the regulator alone. \$\endgroup\$ – Tony Stewart EE75 Feb 12 '19 at 16:10
  • \$\begingroup\$ Using a linear regulator to drop 19V sounds like you're asking for trouble--but I suppose you're not quite to that point yet. \$\endgroup\$ – Hearth Feb 12 '19 at 16:18
  • 1
    \$\begingroup\$ Need a bit more information as to what else is connected and any possible ground loops. If nothing else is connected and the 5V stuff is between 7805 GND and Vout then nothing bad should happen, so there is something else going on. \$\endgroup\$ – Spehro Pefhany Feb 12 '19 at 16:38
1
\$\begingroup\$

Hot swap connectors needs to guarantee GND contact first. In your case, you can't have it, and if you don't have an extra connected ground, a diode won't work.

The solution is to use a pass MOSFET on Vdd side, between connector and your regulator. The PFET gate can be driven by a sinking BJT (which only operates with Vdd AND ground). A more robust design should use a hot swap controller, such as STPW12 or ADM1177.

Improve this answer
\$\endgroup\$
1
  • \$\begingroup\$ I agree, but Coss may need pass-thru transients if a small cap for 10ns pulses to be blocked. \$\endgroup\$ – Tony Stewart EE75 Feb 12 '19 at 16:13
1
\$\begingroup\$

On connector insertion you will have a high inrush current. The hook-up lead inductance will resonant with the on board bypass capacitors up to 48V. The inrush transient event alone is enough for the 7805 to give up the ghost. It would be prudent to add a 30V TVS to the input of the 7805.

It would also be wise to limit the line transients to within the regulation bandwidth of the 7805. Often a slow silicon rectifier diode like a 1N400x combined with input bulk decoupling is enough to slow the incoming line transients down. Of course you can add a little extra series R.

Improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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