Tim Wescott <seemywebsite@myfooter.really> wrote in news:67KdnTauUsazgrPJnZ2dnUVZ5vOdnZ2d@giganews.com:> 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. >Yeah, I thought Ian's idea was clever. Al
Negative Voltage Load Switch
Started by ●October 1, 2014
Reply by ●October 3, 20142014-10-03
Reply by ●October 3, 20142014-10-03
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.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 by ●October 3, 20142014-10-03
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.The AO6602 has a P and N channel FET in one package, Vds of 30V. -- glen
Reply by ●October 3, 20142014-10-03
Al Clark <aclark@danvillesignal.com> wrote in news:XnsA3BB5F9981AE5aclarkdanvillesignal@69.16.179.21:> Tim Wescott <seemywebsite@myfooter.really> wrote in > news:67KdnTauUsazgrPJnZ2dnUVZ5vOdnZ2d@giganews.com: > >> 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. >> > > Yeah, I thought Ian's idea was clever. > > Al >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. -- 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
Ian Malcolm <See.My.Sig.for.email@totally.invalid> wrote in news:XnsA3BBDED20E9120xDEADBEEF@213.239.209.88:> Al Clark <aclark@danvillesignal.com> wrote in > news:XnsA3BB5F9981AE5aclarkdanvillesignal@69.16.179.21: > >> Tim Wescott <seemywebsite@myfooter.really> wrote in >> news:67KdnTauUsazgrPJnZ2dnUVZ5vOdnZ2d@giganews.com: >> >>> 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. >>> >> >> Yeah, I thought Ian's idea was clever. >> >> Al >> > > 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.I have a working circuit now with a pair of BJTs and a dual N/P channel FET. Al>
Reply by ●October 3, 20142014-10-03
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote in news:m0mkhm$vjk$1 @speranza.aioe.org:> AO6602Add a BC847BVN (NPN/PNP complimentry pair) using the PNP transistor in common base as a level shifter for the -ve switch (ground PNP base, strap emitters togater, control signal to NPN base), and we have a two package + 3 resistor solution. -- Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED) ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk [at]=@, [dash]=- & [dot]=. *Warning* HTML & >32K emails --> NUL
Reply by ●October 4, 20142014-10-04
Ian Malcolm <See.My.Sig.for.email@totally.invalid> wrote in news:XnsA3BCCCBEC6CE0xDEADBEEF@213.239.209.88:> glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote in > news:m0mkhm$vjk$1 @speranza.aioe.org: > >> AO6602 > > Add a BC847BVN (NPN/PNP complimentry pair) using the PNP > transistor in common base as a level shifter for the -ve > switch (ground PNP base, strap emitters togater, control > signal to NPN base), and we have a two package + 3 resistor > solution. >And now you have the same solution that I ended up with (with help from Tim). Place a resistor (R1) between the emitters, and bias each FET (R >> R1) from gate to source. I added .1u caps from gate to drain to control inrush currents during switching. Turn on threshold is 2 x Vbe or about 1.4V which will work well with 3.3V logic. I added a pulldown on the input to insure that the circuit stays off in tri-state or weak pullup conditions. It doesn't take very much current to turn on the FETs. I also used the SSOT-6 FET pair (AO6602) that was suggested earlier. These are very cheap. Fairchild & Diodes both make pin compatible substitutes. The BJTs can be just about any 2 cent transistors (3904/06 for example)or a dual pair like Ian suggests (5 cents). If you need more current the IRF7509PbF seems reasonable. Thanks everyone for ideas and contributing. Al
Reply by ●October 4, 20142014-10-04
Al Clark <aclark@danvillesignal.com> wrote in news:XnsA3BC68736AAFBaclarkdanvillesignal@69.16.179.20:> Ian Malcolm <See.My.Sig.for.email@totally.invalid> wrote in > news:XnsA3BCCCBEC6CE0xDEADBEEF@213.239.209.88: > >> glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote in >> news:m0mkhm$vjk$1 @speranza.aioe.org: >> >>> AO6602 >> >> 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 together, control >> signal to NPN base), and we have a two package + 3 resistor >> solution. >> > > 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. > > AlThe 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 [at]=@, [dash]=- & [dot]=. *Warning* HTML & >32K emails --> NUL
Reply by ●October 4, 20142014-10-04
Ian Malcolm <See.My.Sig.for.email@totally.invalid> wrote in news:XnsA3BCAA233F3CD0xDEADBEEF@78.46.70.116:> Al Clark <aclark@danvillesignal.com> wrote in > news:XnsA3BC68736AAFBaclarkdanvillesignal@69.16.179.20: > >> Ian Malcolm <See.My.Sig.for.email@totally.invalid> wrote >> in news:XnsA3BCCCBEC6CE0xDEADBEEF@213.239.209.88: >> >>> glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote in >>> news:m0mkhm$vjk$1 @speranza.aioe.org: >>> >>>> AO6602 >>> >>> 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 together, control >>> signal to NPN base), and we have a two package + 3 >>> resistor solution. >>> >> >> 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 > > 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. > > >SOA, not a problem in my case, but clearly could be an issue in other implementations. Thanks Ian Al
