Make before break circuit

Thread Starter

acheriti

Joined Apr 26, 2020
20
Hi,
I have a project where the load is operated from a battery. I want to be able to power it from a power supply if I'm in the lab for example. I want the circuit to automatically detect that the power supply is connected and thus disconnect the battery.
However I can't afford to lose the power so what I would most likely need is a circuit similar to a diode-or . The problem is that my power supply has a lower voltage than the battery and therefore I cannot use this configuration.
Does someone have a recommandation on how I could build such a make before break make circuit?
Ultimately I could use a chip like the LTC4417. This chip has a break before make circuit but the switching is made very fast. My power would remain always connected but I would have a current surge at the switching, which I am trying to avoid.

Thanks a lot for the help guys !
 

Thread Starter

acheriti

Joined Apr 26, 2020
20
Hi,
Firstly thank you for your reply. The load can go up to 150W max. The battery voltage is 50V and the power supply is 24V.
 

Ian0

Joined Aug 7, 2020
2,167
This has been discussed before (albeit for a lower voltage system), in a thread called "Least expensive circuit to cause one power supply to take precedence over another?".
There may be some interesting ideas there.


Mods Edit :
Please don't use large font that it looks like shouting.
 
Last edited by a moderator:

Ian0

Joined Aug 7, 2020
2,167
Simplest way is with a schottky diode in series with the power supply, then a 24V-coil relay with the coil connected across the power supply before the diode, which switches the battery off. There will be some power dissipation in the schottky diode, though.
 

AnalogKid

Joined Aug 1, 2013
9,244
First pass at a (((concept))) schematic. D1, D2, and K1 need to be changed to something appropriate for your power levels. I threw in C1 to smooth out the voltage change a little bit; adjust to taste.

ak
Dual-Power-Sw-1-c.gif
 

Ian0

Joined Aug 7, 2020
2,167
That's what I had in mind. Perhaps I should draw it out myself and not be so lazy.
I think it will manage without D1.
I think he mentioned a 150W load, in which case D2 needs to be rather more serious than a 1N4004. Perhaps a 10A schottky and a heatsink.
 

AnalogKid

Joined Aug 1, 2013
9,244
Another (((concept))), this time with a p-channel power MOSFET instead of the relay. As before, D1, D2, and Q1 should be selected for this application.

R1-C1 set a turn-off time of about 1/2 second because he said something about slowing the transition between voltages.

depending on the characteristics of the 24 V source and its switching/connection, D1 might not be necessary.

ak
Dual-Power-Sw-2-c.gif
 

Thread Starter

acheriti

Joined Apr 26, 2020
20
First pass at a (((concept))) schematic. D1, D2, and K1 need to be changed to something appropriate for your power levels. I threw in C1 to smooth out the voltage change a little bit; adjust to taste.

ak
View attachment 223605
Hi,
Thanks a lot for your answer. I got to admit I really like this topolgy . However I have a question for your : let's say my load is 150W and I switch from the 50V supply to the 24V supply. Theoretically the 24V source would have to supply 6.25A. If I put an amperemeter in series with the load, would I see the 6.25V directly or this current might experience some turbulences as the voltage at the load is changing rapidly ? Thank you very much.
 

AnalogKid

Joined Aug 1, 2013
9,244
For a constant load power and the relay circuit, the voltage will drop and the current will shoot up almost instantly when the relay opens. With the MOSFET circuit the same changes will occur over a brief interval, somewhere between 100 ms and 500 ms -ish. The MOSFET transitions will be monotonic, but the relay transitions might have disturbances caused by contact bounce.

ak
 

Ian0

Joined Aug 7, 2020
2,167
If the load is inductive (a motor perhaps) then the voltage fluctuations won't bother it, but the relay might arc badly on the off transitions.
If the load is capacitive then there might be a lot of inrush current as the capacitor charges from the low supply impedance of the battery.
If using the MOSFET circuit, don't underestimate how much power it will dissipate during the transition - an average of about 30 Watts. Although it only takes half a second it will need not necessarily a finned heatsink but some thermal inertia to absorb that heat.
 

MisterBill2

Joined Jan 23, 2018
8,714
Nobody has asked this yet, but what voltage does the actual load use? Is that 50 volts regulated to a lower voltage inside the system? One very simple arrangement will simply be to have a diode in series with the power supply 24 volt feed, and a simple switch in series with the battery. Then connect the supply and switch off the battery feed. And when leaving, switch on the battery feed and then disconnect the power supply. Semi-automatic, but very effective.
 

Thread Starter

acheriti

Joined Apr 26, 2020
20
Another (((concept))), this time with a p-channel power MOSFET instead of the relay. As before, D1, D2, and Q1 should be selected for this application.

R1-C1 set a turn-off time of about 1/2 second because he said something about slowing the transition between voltages.

depending on the characteristics of the 24 V source and its switching/connection, D1 might not be necessary.

ak
View attachment 223612
Another (((concept))), this time with a p-channel power MOSFET instead of the relay. As before, D1, D2, and Q1 should be selected for this application.

R1-C1 set a turn-off time of about 1/2 second because he said something about slowing the transition between voltages.

depending on the characteristics of the 24 V source and its switching/connection, D1 might not be necessary.

ak
View attachment 223612
Hi, Thanks a lot for your reply. I've been trying to simulate this circuit however even if my 24V is present I still see the 50V at the output. Am I missing something? Thank you for your time ! I simulate my circuit with a 1K load just to start my simulation.
 

Attachments

Ian0

Joined Aug 7, 2020
2,167
It might be circuit tolerances.
24V input, gives 23.4V across R6. With exactly 24V across D11 leaves 2.6V across D10. That 2.6V is across gate and source of the MOSFET, and it could turn on at 2V.
The voltage across D11 could be anywhere between 22.8V and 25.2V @ 50mA. At lower currents it will be even lower., and the current with R5=470k is of the order of microamps.
Try a lower power 27V zener (for which the breakdown voltage is rated at 5mA) and a lower value for R5.
 

AnalogKid

Joined Aug 1, 2013
9,244
Given that D10 has a resistor across it and D11 does not, it doesn't make sense that of the two diodes in series, it will be D11 that somehow maintains 24 V across it.

Leakage current: No semiconductor is ever completely "off". When the voltage across a zener diode is less than the zener voltage, there still will be a very small current throough the device. Back in the all-bipolar-transistor days this didn't matter, but FETs take picoamps of current to function. With R5 = 470 K, it takes less than 7 uA of current through it and D11 for Vgs to equal 3 V.

The 1N5252 in my schematic has 5 times less leakage current.

ak
 
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