DC/DC Converter Instant Off

Question

The output voltage of my DC/DC converter and my power supply both slowly descend. I believe this is due to capacitors being close to the output of each device.

I want to get an instantaneous, or as close as possible, off signal from the power supply so that I can connect my backup supply at that moment.

I currently signal the base of a p-channel mosfet with the output of the power supply. Because the voltage slowly descends the mosfet is not signaled until much after the supply is disconnected which leads to a break in power.

It looks something like this, Where the 5V supply is the main off of the DC/DC converter or Power Supply, the 4.7V is the backup supply, and the 100k resistor is the load. (The load draws much more than a 100k resistor.)

I have tried adding a voltage divider at the base of the mosfet to lower the voltage and a cap at the output to smooth out the main signal. The problem with this is depending on the load the output capacitor has to be very large.

I have also tried using two NPN transformers to pull the mosfet directly to GND. They are signaled sooner than the mosfet but not fast enough.

Is there any way to quickly drain the power supply output caps without adding a constant power drain on the system or signal the mosfet as soon as the power supply starts to decrease its power?

Answer 1

I have the feeling, unless I misunderstand or just miss some extra requirement, that you are over-engineering this.

You have a supply that is higher than your backup-supply by more than 0.1V.

Just the diodes will give you the best and quickest response: Leave out the P-MOSFET.

So long as you have a back-up supply that is a little lower than the main supply, you need to do nothing other than have those two diodes. Once the 5V drops below 4.7V the 4.7V back-up will take over instantaneously. Well, yes, there is a transition period, but it's in the order of millivolts for most diode types of the same type and same batch, if not less. I'm too fuzzy right now to do the full maths.

If your back-up is higher than the main supply (this is an exceptionally weird situation, to be honest), you will need a switching element, the easiest way to get that is to use a comparator. They can give a very nice, sharp and hard response on a fixed voltage.

If you have a 2.4V reference and divide the main supply by 2 through a resistive divider, you can compare those with a comparator directly and it will be able to switch a transistor. If the comparator has an open-collector output you can use it directly for your higher-voltage back-up supply, if it doesn't you'll likely need a support-NPN-transistor to translate its output upward to the P-MOSFET of the higher back-up voltage.

But again: Your back-up is below the main voltage, as is very usual, so just the diodes are more than enough.