3
\$\begingroup\$

Okay, a little background: I am designing a micro controller daughter board (uCDB) that is nominally powered at 3.3V by a voltage regulator on a mother board (MB). Also I will have a programmer board (PB) that I can plug directly into the uCDB. As an alternative to powering the uCDB from the MB, I could instead draw power from the PB at 3.5V. This would be very convenient for instances where I wanted to power and program the uCDM even when it was not plugged into the MB. What happens if I try to do this and the uCDB is already powered by the MB? Is it a problem?

An obvious fix would be to put in a jumper that selects what voltage source to use, but requires more thought from the user each time he/she hooks it up; "Okay, what state am I in so I don't burn things up?" Any thoughts on how to allow both voltage sources to be on without burning things up and without mechanical switches/jumpers. Perhaps it is okay to just connect the two lines together either directly or through a resistor or something? Maybe the jumper is the best solution? I attached an image that hopefully summarizes my problem.enter image description here

|improve this question
\$\endgroup\$
  • \$\begingroup\$ Well, you will be wasting power, but 200mV is not a lot. Still not a good idea. You could build it up where the act of plugging in the uCDB into the MB shorts a switch that sets the power source. \$\endgroup\$ – Branden Boucher Feb 12 '16 at 17:01
5
\$\begingroup\$

You have very close voltages for your supplies, but connecting them directly to each other could push/pull excessive power to/from your programmer (the MB will likely have a more powerful source, and not be the first component damaged).

One simple way to avoid problems, if your uC is like mine and has a wide supply voltage range (at least down to 2.5V for this purpose) would be to simply attach a switching diode in series with the +Vcc trace from each possible power source; then your uC would simply draw power from the highest-voltage connected power source.

Alternately, you can add a SPDT signal relay to your uC on the power trace. Attach the coil & NO pins to the MB power supply trace, and the NC pin to the programmer power trace. That way any time there's avail. power from the MB, the programmer power supply sees an open circuit, but when MB powers off, the uC won't back-feed power from the programmer to your MB (and probably overllad the programmer's supply).

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ I like your idea of the switching diode. I'm thinking it might work if I just had one diode in series with the 3.5V from the programmer. Programmer voltage would be around 2.9V and if the MB was powered up, then it would be the active supply. Am I correct here? \$\endgroup\$ – py_man Feb 12 '16 at 18:47
  • 1
    \$\begingroup\$ You are correct. With 1 diode the 3.3V from the MB will power the uCDB because it has the higher voltage. The problem is that when the MB is powered down the PB will still back feed to the MB and try to power it. You have to use 2 diodes to stop the back feed. \$\endgroup\$ – Steve G Feb 12 '16 at 19:13
  • \$\begingroup\$ @py_man I agree entirely with Steve G on back-feeding the MB. the only likely safe way to protect with only a single diode would be to place the diode on the MB supply. Then the programmer would be the 'default' power source for your uC any time it's connected, but the MB power supply could power any ti e the programmer isn't supplying power, and can't easily be back-fed. \$\endgroup\$ – Robherc KV5ROB Feb 12 '16 at 19:38
  • \$\begingroup\$ Here's a link to a 20V1A rated diode with ~0.35V forward-voltage drop to reduce the power loss from whichever supply you place it in series with (and it's only about $0.38/ea): digikey.com/product-detail/en/RB161M-20TR/RB161M-20CT-ND/926506 \$\endgroup\$ – Robherc KV5ROB Feb 12 '16 at 19:46
2
\$\begingroup\$

The 2 diode trick is the simplest and cheapest solution, but even using the low Vf schottky diodes that @Robherc found you will have 0.2V to 0.3V drop depending on current. If this is unacceptable then another solution is to use a Power Mux such as the TPS2115 from TI. You can configure this part to work in a number of ways but the simplest is use auto-switching where it selects the highest voltage of 2 inputs. Since it uses internal MOSFETs for switching you do not get the diode drop of other solutions. It will not back-feed, and it has a programmable current limit which can come in useful. No I'm not a TI salesman! The downside is that they are $2 each in 1-off quantities.

You can make a Power Mux out of discreet parts, but you will need 4 MOSFETs to do the same as a single IC.

Have a look at: http://www.ti.com/lit/ds/symlink/tps2115a.pdf

|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.