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
Negative Voltage Load Switch
Started by ●October 1, 2014
Reply by ●October 1, 20142014-10-01
On Wednesday, October 1, 2014 3:07:50 PM UTC-7, Al Clark wrote:> 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).A 8 pins relay, can switch 2A on + and - power rails. Only need a transistor to buffer and drive the coil. http://www.digikey.com/product-detail/en/EE2-5NU-L/399-11017-6-ND/4506558
Reply by ●October 1, 20142014-10-01
On Wed, 01 Oct 2014 22:07:50 GMT Al Clark <aclark@danvillesignal.com> wrote:> 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 > > AlA 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 Email address domain is currently out of order. See above to fix.
Reply by ●October 1, 20142014-10-01
On Wednesday, October 1, 2014 6:59:28 PM UTC-4, Rob Gaddi wrote:> 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.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 by ●October 1, 20142014-10-01
On Wed, 01 Oct 2014 22:07:50 +0000, Al Clark wrote:> 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).Two transistors and two resistors is "a lot" for the negative voltage part of it? -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●October 2, 20142014-10-02
On Wednesday, October 1, 2014 6:19:36 PM UTC-4, edward....@gmail.com wrote:> A 8 pins relay, can switch 2A on + and - power rails. > Only need a transistor to buffer and drive the coil.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 by ●October 2, 20142014-10-02
Tim Wescott <seemywebsite@myfooter.really> wrote in news:qI2dnQPhAuXwLbHJnZ2dnUU7-U2dnZ2d@giganews.com:> On Wed, 01 Oct 2014 22:07:50 +0000, Al Clark wrote: > >> 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). > > Two transistors and two resistors is "a lot" for the > negative voltage part of it? >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 by ●October 2, 20142014-10-02
Al Clark <aclark@danvillesignal.com> wrote in news:XnsA3BAABC56D77Baclarkdanvillesignal@69.16.179.21:> Tim Wescott <seemywebsite@myfooter.really> wrote in > news:qI2dnQPhAuXwLbHJnZ2dnUU7-U2dnZ2d@giganews.com: > >> On Wed, 01 Oct 2014 22:07:50 +0000, Al Clark wrote: >> >>> 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). >> >> Two transistors and two resistors is "a lot" for the >> negative voltage part of it? >> > > 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? >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. -- Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED) ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk [at]=@, [dash]=- & [dot]=. *Warning* HTML & >32K emails --> NUL
Reply by ●October 3, 20142014-10-03
On Thu, 02 Oct 2014 22:18:02 +0000, Ian Malcolm wrote:> Al Clark <aclark@danvillesignal.com> wrote in > news:XnsA3BAABC56D77Baclarkdanvillesignal@69.16.179.21: > >> Tim Wescott <seemywebsite@myfooter.really> wrote in >> news:qI2dnQPhAuXwLbHJnZ2dnUU7-U2dnZ2d@giganews.com: >> >>> On Wed, 01 Oct 2014 22:07:50 +0000, Al Clark wrote: >>> >>>> 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). >>> >>> Two transistors and two resistors is "a lot" for the negative voltage >>> part of it? >>> >>> >> 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? >> >> > 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.Aw, now that's too easy. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●October 3, 20142014-10-03
On Thu, 02 Oct 2014 21:53:04 +0000, Al Clark wrote:> Tim Wescott <seemywebsite@myfooter.really> wrote in > news:qI2dnQPhAuXwLbHJnZ2dnUU7-U2dnZ2d@giganews.com: > >> On Wed, 01 Oct 2014 22:07:50 +0000, Al Clark wrote: >> >>> 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). >> >> Two transistors and two resistors is "a lot" for the negative voltage >> part of it? >> >> > 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? > > AlThe 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 -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
