0
\$\begingroup\$

[![enter image description here][1]][1]I am trying to build a simple circuit to power a load with two different power supplies (5 V and 0.7 V). enter image description here

When one is active the other one is not and it's disconnected from the load. I know that this would be straightforward with a manual switch but I need this application to be automatic, maybe using a MOSFET as a switch controlled by a uC but it's not clear to me how I can achieve that. I wouldn't use a relay since I have constraints on cost and size of the footprint. Thanks. Claudio

\$\endgroup\$
13
  • \$\begingroup\$ What are your constraints on ohmic resistance and speed of operation? \$\endgroup\$
    – Andy aka
    Commented Jan 24, 2017 at 14:01
  • \$\begingroup\$ Speed of operation is not that important in my case and few mohm as resistance should be fine. \$\endgroup\$
    – Csh90
    Commented Jan 24, 2017 at 14:14
  • \$\begingroup\$ Your R1 confuses me. Is it there for a specific reason? \$\endgroup\$
    – Marcus Müller
    Commented Jan 24, 2017 at 14:15
  • \$\begingroup\$ @MarcusMüller Yes I am powering a oxygen sensor. Before start this O2 sensor (RLOAD) which is connected with a shielded connector to my PCB, i need to polarize it with 700mV for 30 minutes and then switch to measurement mode (5V) power. This circuit was provided by the manufacturer of the sensor but they did not include the part for the polarization so I am trying to figure it out myself \$\endgroup\$
    – Csh90
    Commented Jan 24, 2017 at 14:18
  • 1
    \$\begingroup\$ Here is another answer on this community about switching between two supplies to power a Raspberry Pi. Seems very similar to the solution you might be looking for. \$\endgroup\$
    – electrophile
    Commented Mar 31, 2017 at 3:51

2 Answers 2

Reset to default
1
\$\begingroup\$

First of all: add a capacitor in parallel to your load. Every switching operation will inevitably lead to some short disruption, and a sufficiently sized capacitor will allow your device to continue working through the switchover.

Then, as M D said, you could just have a microcontroller on the load side of things that monitors both power sources and initiates switching when necessary. How that switching is done, and what "necessary" means in this context, is up to you to define.

Furthermore, this is often solved by using dedicated circuitry, since it's a very common problem and needs a fast and reliable solution in many cases.

So, companies like Linear Technologies have ICs that do that – the LTC4412 has an especially nice datasheet with examples of how to use it to control a MOSFET to do your switching.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for suggesting this. In the past I had used the LTC4353 which is more expensive in my part of the world and is sometime not available entirely. The LTC4412 seems just right. \$\endgroup\$
    – electrophile
    Commented Mar 31, 2017 at 3:45
0
\$\begingroup\$

Use a load or analog switch (pending power level) to handle the switch component, and select power supplies with a low-power shutdown (enable pin) to turn them off when unused. Pending component selection and your application , you might be able to accomplish this with a single GPO from the MCU.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks but it's still not 100% clear. Is one nMosfet enough? Or I need two? The thing is that I cannot find a way to connect the MOSFET so that either one line (5V) or the other (700mV) is connected. \$\endgroup\$
    – Csh90
    Commented Jan 24, 2017 at 14:13
  • \$\begingroup\$ You may be able to do this with a single pass device pending the individual supplies' ability to handle voltage applied to their Vout when disabled, and your ability to control sequencing and timing of the voltage slew. You should also be cognitive of the body diode conduction if doing this with FETs. \$\endgroup\$
    – M D
    Commented Jan 24, 2017 at 14:21

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.