Transistor to switch power supply

[SystemsGuy]

Quick sanity check - please! ;-)


I have a circuit that's designed to be powered by 12vdc, and has a 5vdc converter to supply the digital side with 5vdc. It's also got an USB input to provide power to the 5v side.

Problem is that the damned thing ties the 5vd from the rectifier to the 5 vdc from the USB - which is in my mind a potentially "bad thing".

My though was to simply pop a PNP transistor onto the 5dvc USB feed, and tie the base to the 12vdc supply side so that when it's on, the transistor is off, so disconnecting power from the 5v side of the USB.

Am I missing something here, or will this work?

 

[(*steve*)]

You count isolate the power supplies using one (or more) schottky diodes, or do something like you suggest.

I'm not convinced that the solution is as simple as you suggest. You may find the voltage drop across the transistor to be more than the voltage drop across a schottky diode.

 

[CDRIVE]

If you're counting votes I second Steve's Schottky Diodes approach. It's a simpler solution than messing with a PNP.

 

[Electrobrains]

USB Supply automatic isolator-switch

How about this?
(Almost) no voltage drop and security that no feedback through the +12V branch would influence the internal voltage when using USB supply.
Everything for less than 0.5$ and super small!

 

[KrisBlueNZ]

Electrobrains, I suggest you show the parasitic diode in the P-channel FET to make it clear that your circuit will feed the internal 5V rail back onto the USB connector with one diode drop.

I don't think that will necessarily be a problem, but it would be helpful to show the diode, to make it clear that your circuit will do this.

 

[Electrobrains]

Electrobrains, I suggest you show the parasitic diode in the P-channel FET to make it clear that your circuit will feed the internal 5V rail back onto the USB connector with one diode drop.

I don't think that will necessarily be a problem, but it would be helpful to show the diode, to make it clear that your circuit will do this.

Hello Kris

I first thought of that possible problem, but didn't think it was worth mentioning, because the power flow is specified by the connectors.

But when you mention it here, a realize that it makes sense!

Problems could occur, if the external USB power supply (or whatever USB source is used) would be left connected, but not switched on!

Be aware of that the internal +5V (minus 0.7V) will appear on the USB side!
(compare: the original circuit does the same, even with the full 5V)

Thank you for the hint!
I have now changed the diagram

 

[Electrobrains]

Automatic and fool-proof USB Supply Isolator Switch

Ok, I have taken that problem seriously and made a fool-proof version!

It will cost you one more component, T2 ($0.16) and add 60 mOhm in the supply path.

This version additionally blocks the internal voltage from flowing back into the USB connector path, preventing any harm on that side - or harm to your own circuit if too much current would be lost that way.

I suppose this kind of circuit could be good in many other applications where external power supplies are combined with internal ones or batteries.

To save some little energy, I changed R1 to 47k. I suppose you could increase it even more. It must only be low enough to overcome the leakage current through Z1 in the case there would be reverse voltage appearing on the +12V pin, flowing back from the +5V spot.

The price of this circuit, still would be below 0.5$ and the thing could be soldered together on less space than a penny (well, here in Switzerland we would say 5 rappen coin =approx. 15x15mm)

 

[KrisBlueNZ]

Nice work Electrobrains.

I have the parasitic diode included in my schematic symbol. I've been caught out once before when I used one in the feedback path of an op-amp. It took me a while to figure out why the output was clipped on one side but not the other! Since then I made sure I could never forget it. It's shown in the schematic symbol on all the data sheets I've seen, too.

That's a useful circuit. It reminds me of the bridge rectifier made from MOSFETs, that has no forward voltage drop. They are a handy component to know.

 

[SystemsGuy]

Wow

Awesome electrobrains, much nicer than my quick thought! I'll have to go poke Farnel and get some bits ordered!

EDITED : Crap, naturally Farnell does not stock the NTGS4111PT1G - any suggestions as to a replacement MOSFET? Or a better source? (I'm in the UK - could do Mouser, but takes more time...)

Looks like NXP - BSH207,135 - MOSFET P-CH 12V 1.52A SOT457 might work?

 

[Electrobrains]

That's a useful circuit. It reminds me of the bridge rectifier made from MOSFETs, that has no forward voltage drop. They are a handy component to know.

Yes, but usually you put together the sources of N-FET's, but that doesn't work here, because you are blocking with a positive signal.

 

[Electrobrains]

Awesome electrobrains, much nicer than my quick thought! I'll have to go poke Farnel and get some bits ordered!

EDITED : Crap, naturally Farnell does not stock the NTGS4111PT1G - any suggestions as to a replacement MOSFET? Or a better source? (I'm in the UK - could do Mouser, but takes more time...)

Looks like NXP - BSH207,135 - MOSFET P-CH 12V 1.52A SOT457 might work?

If you want it small and cheap, try this one: TMS3455. If you want some more power and like to spend money, you can take this one: IRFTS9342TRPbF

TMS3455 is not as good as NTGS4111P, but I think even cheaper (at least in Swiss Francs). IRFTS9342TRPbF is very similar to NTGS4111P.

p.s. Don't use BSH207,135. It might even break down in your circuit (Vds=12V, Vgs=8V)

 

[SystemsGuy]

Thanks Electro - didn't notice the high Vds & Vgs. Being the big spender I am, I went for the IRFTS9342TRPbF! ;-)

If you want it small and cheap, try this one: TMS3455. If you want some more power and like to spend money, you can take this one: IRFTS9342TRPbF

TMS3455 is not as good as NTGS4111P, but I think even cheaper (at least in Swiss Francs). IRFTS9342TRPbF is very similar to NTGS4111P.

p.s. Don't use BSH207,135. It might even break down in your circuit (Vds=12V, Vgs=8V)