Negative Voltage Load Switch

I have an application where I want to turn on a bipolar +/-15 volt supply from 3.3V. The supply already exists, I am disconnecting a part of a circuit to save power during an idle state.

Integrated load switches comprised of P channel and N channel enhancement mode FETs are readily available for positive supplies.

I also know how to achieve a negative voltage version with FETs using discrete components.

The parts are not expensive, but there are quite a few of them.

Does anyone know of an inexpensive integrated solution? Ideally this would be an 8 pin part with a logic level enable and both positive and negative supply switching. In my case, I don't need to switch more than 100mA. Alternatively, I might consider just a negative voltage version (combined with a standard load switch).

Thanks

Al

Reply to
Al Clark
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from 3.3V. The supply already exists, I am disconnecting a part of a circu it to save power during an idle state.

mode FETs are readily available for positive supplies.

crete components.

d be an 8 pin part with a logic level enable and both positive and negative supply switching. In my case, I don't need to switch more than 100mA. Alte rnatively, I might consider just a negative voltage version (combined with a standard load switch).

A 8 pins relay, can switch 2A on + and - power rails. Only need a transist or to buffer and drive the coil.

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Reply to
edward.ming.lee

A pair of 4 pin SSRs? I know we're paying about $0.70 a pop for CPC1008Ns, which I think would cover your problem.

Alternatively, if you're generating the +/-15 rather than taking it, could you shut down the supply generation itself, either with a shutdown pin or by cutting the power to it?

--
Rob Gaddi, Highland Technology -- www.highlandtechnology.com 
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Reply to
Rob Gaddi

Another vote but I've used CPC1018 (slightly more expensive but lower turn-on current, not that CPC1008 take so much).

Hope that helps, Best Regards, Dave

Reply to
Dave Nadler

Two transistors and two resistors is "a lot" for the negative voltage part of it?

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Tim Wescott 
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Reply to
Tim Wescott

And if you use a relay, don't forget the diode!!

Seen the results of that a few times ;-)

Hope that helps, Best Regards, Dave

Reply to
Dave Nadler

Tim Wescott wrote in news: snipped-for-privacy@giganews.com:

My circuit took three fets for the negative supply switching (some can be BJTs. This was so that on was active high.

Otherwise my positive supply is the wrong polarity.

So my overall circuit uses 3 cheap FETs and a Dual N/P channel lower Rds type (IRF7209 for example)

What do you use for 3.3V control with two transistors?

Al

Reply to
Al Clark

Al Clark wrote in news:XnsA3BAABC56D77Baclarkdanvillesignal@69.16.179.21:

Can you stand to drive the -ve rail switch off the +ve rail coming up?

Hang an 18V Zener off the +ve rail to the gate of a N channel MOSFET switching the -ve rail. Add a gate-source pulldown resistor to keep it hard off when the +ve rail is off.

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Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)  
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Reply to
Ian Malcolm

The upper resistor limits the current going into the emitter of the PNP, and thus the PNP collector current. The lower resistor helps to turn off the NPN -- you could leave it off if you wanted to be really sleazy.

You do need to size the upper resistor to flow enough current to really turn the lower transistor on: I suppose that if you don't have a pin that can source several this may not be the way to go.

LATE BREAKING NEWS: I got the schematic all done, then realized that if the bottom transistor is an N-channel FET, the only problem with lightly driving the upper transistor is the turn-off speed of the supply. If you can stand it working at a mosey, you can size the upper resistor for 1mA or so of current (or whatever your pin will drive), then size the lower resistor to drop 10-15V at 1mA. Turn-off won't be snappy, but that may not matter to you.

____ o--|____|-. | >| |----. /| | | /// | .------o switched -15V | |/ o-----------| | |>. - | | | | | | | - | | | '--------------o | | o -15V

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Tim Wescott 
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Reply to
Tim Wescott

Aw, now that's too easy.

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Tim Wescott 
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Reply to
Tim Wescott

Tim Wescott wrote in news: snipped-for-privacy@giganews.com:

Yeah, I thought Ian's idea was clever.

Al

Reply to
Al Clark

If it has P and N channel FETs, isn't it supposed to work both ways?

The only one I know about, the AO9926B, is described as "uni-directional or bi-directional load switch".

I believe the AO9926B is designed for 3.3V or 5V supplies, and can switch up to 7A or so. It is N channel, so needs a positive voltage (Vgs) to turn on. Two fit in a tiny SOIC-8 package.

There might be some others in the family that would work for you.

(I found out about this one when working on a circuit that used it. I looked up the data sheet to see what it was.)

-- glen

Reply to
glen herrmannsfeldt

The AO6602 has a P and N channel FET in one package, Vds of 30V.

-- glen

Reply to
glen herrmannsfeldt

Al Clark wrote in news:XnsA3BB5F9981AE5aclarkdanvillesignal@69.16.179.21:

It will have problems if there is too much capacitance on the +ve rail as the rail must come up from 3V to approx 10V fast enough to get the MOSFET through its linear region before it gets too hot.

An alternative to directly switch both rails is P and N channel discrete MOSFETS, each source to the corrisponding supply with a gate-source resistor to turn them off, and an optocoupler shorting the gates togger to turn them on. It will need a limiting resisistor for the opto's LED so thats three active parts and three resistors total. The MOSFETs each get 15V of gate drive - if you want to reduce that, add an extra resistor in series with the opto's collector.

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Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)  
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Reply to
Ian Malcolm

Ian Malcolm wrote in news:XnsA3BBDED20E9120xDEADBEEF@213.239.209.88:

I have a working circuit now with a pair of BJTs and a dual N/P channel FET.

Al

>
Reply to
Al Clark

glen herrmannsfeldt wrote in news:m0mkhm$vjk$1 @speranza.aioe.org:

Add a BC847BVN (NPN/PNP complimentry pair) using the PNP transistor in common base as a level shifter for the -ve switch (ground PNP base, strap emitters togater, control signal to NPN base), and we have a two package +

3 resistor solution.
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Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)  
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk  
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Reply to
Ian Malcolm

Ian Malcolm wrote in news:XnsA3BCCCBEC6CE0xDEADBEEF@213.239.209.88:

And now you have the same solution that I ended up with (with help from Tim). Place a resistor (R1) between the emitters, and bias each FET (R >> R1) from gate to source. I added .1u caps from gate to drain to control inrush currents during switching. Turn on threshold is 2 x Vbe or about 1.4V which will work well with 3.3V logic. I added a pulldown on the input to insure that the circuit stays off in tri-state or weak pullup conditions. It doesn't take very much current to turn on the FETs.

I also used the SSOT-6 FET pair (AO6602) that was suggested earlier. These are very cheap. Fairchild & Diodes both make pin compatible substitutes. The BJTs can be just about any 2 cent transistors (3904/06 for example)or a dual pair like Ian suggests (5 cents).

If you need more current the IRF7509PbF seems reasonable.

Thanks everyone for ideas and contributing.

Al

Reply to
Al Clark

Al Clark wrote in news:XnsA3BC68736AAFBaclarkdanvillesignal@69.16.179.20:

The resistor between the emitters controls both collector currents which I suppose you need if you are slugging the switching time by adding Miller caps on both MOSFETs.

I *ASSUME* you've verified the MOSFETs remain within their S.O.A while switching.

--
Ian Malcolm.   London, ENGLAND.  (NEWSGROUP REPLY PREFERRED)  
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk  
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Reply to
Ian Malcolm

Ian Malcolm wrote in news:XnsA3BCAA233F3CD0xDEADBEEF@78.46.70.116:

SOA, not a problem in my case, but clearly could be an issue in other implementations.

Thanks Ian

Al

Reply to
Al Clark

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