On 2017-02-09, Peabody <waybackNO584SPAM44@yahoo.com> wrote:> I'm working on a small circuit that will be powered by a 9V battery. I would > like to combine the functions of a soft power switch and reverse battery > protection in a single device. I think the answer may be to use a PNP > transistor on the high side, with a base resistor to ground - actually to > ground through a physical momentary switch, or indirectly to ground via an > NPN transistor controlled by a microcontroller output pin. But for purposes > of this post, just think of a base resistor to ground. I only need a few mA > of current, and the base resistor needed will be 100K.ok> My question concerns reverse polarity protection. I believe I understand > correctly that current will not flow backwards from collector to emitter of > a PNP so long as the transistor's voltage rating is not exceeded. So the > "main" power flow will be protected.nah. backwards its just like a worse PNP. VCE max may be 5V, Hfe takes a hit by a factor of 3 to 30> But the absolute maximum base-emitter reverse voltage for this transistor > (BC560C) is 5V. So if 9V were to be applied directly to the base, with the > emitter at ground, I think the transistor would be blown. The question is > whether the 100K resistor limits current enough so that the transistor is not > damaged.it does does not protect. current will flow from the base to the collector, turning the reversed PNP on.> I just don't have any experience with this situation, and don't > knpw what actually works. There is also a separate question as to whether > the limited reverse current flowing through the resistor would damage the > microcontroller (max 3.6V Vcc) since all of that current would actually flow > through it via its protection diodes.through 100K the microcontroller should be safe, but the PNP collector will be trying to pull current out of the voltage regulator. pnp -9 --+----. .------[VREG]-- | \ / | | ------- | | | ---+--- -->Z----| 5V | 100K | | | \| NPN |-+-100K-+-uC /| | / +--|<-+ ... uC protection diode. | | | | ----+-----------------+-- so you've got current flowing from the microcintroller, through the base resistor and the NPN B-C junction this turns the reverse NPN on as an emitter follower since it's feeding the PNP B-E junction you get maybe 5V drop there (avalanche breakdown, potentially damaging - shown as a zener diode) and the rest in the two diodes and 2 resistors. you get about 25uA in the main 100K and maybe 3 uA from the microcontroller, A diode in series with the PNP base will fix that. So the PNP is biased off but current may flow from the voltage regulator through the collector to the base then through the emitter. if it does you could destroy the PNP. -- This email has not been checked by half-arsed antivirus software
PNP for soft switch and reverse battery protection
Started by ●February 9, 2017
Reply by ●February 14, 20172017-02-14
Reply by ●February 14, 20172017-02-14
On 2/12/2017 12:26 AM, Peabody wrote:> Clifford Heath says... > > >> It's a long thread, maybe this was stated. Is there a > >> reason why a Schottky diode can't be used for reverse > >> polarity protection? > > > Sometimes you simply can't spare 400mV, such as in > > single-cell applications. > > > A MOSFET that will pass (say) 10amps is a lot smaller > > and cheaper than a Schottky that will. > > I need very little current, so a Schottky would be cheaper. > But it's mainly the voltage drop. Anything that produces a > voltage drop just shortens the life of the battery - > assuming the circuit will only operate properly down to a > certain supply voltage.If there is very little current, the voltage drop on the Schottky is less, possibly as low as 200 mV.> Also, Schottkys aren't known for having all that great > reverse leakage values, which might make them problematical > for protecting against reverse polarity.Your circuit uses less current than the reverse leakage current of a Schottky diode? Wow! That's low.> From everything I've read, the P-channel mosfet, installed > backwards, is the school-approved solution for reverse > polarity protection. They are more expensive, but for my > needs, the TP2104 should work fine, and it's 60 cents at > Digikey, versus maybe 40 cents for a Schottky.How many are you making? -- Rick C
Reply by ●February 14, 20172017-02-14
On Monday, February 13, 2017 at 10:31:04 PM UTC-8, Jasen Betts wrote:> On 2017-02-09, Peabody <waybackNO584SPAM44@yahoo.com> wrote: > > I'm working on a small circuit that will be powered by a 9V battery. I would > > like to combine the functions of a soft power switch and reverse battery > > protection in a single device. I think the answer may be to use a PNP > > transistor on the high side, with a base resistor to ground> > ... current will not flow backwards from collector to emitter of > > a PNP so long as the transistor's voltage rating is not exceeded.> > But the absolute maximum base-emitter reverse voltage for this transistor > > (BC560C) is 5V.So, use a PNP with higher base-emitter reverse voltage. This one is good to 25V: <http://www.mouser.com/search/ProductDetail.aspx?R=0virtualkey0virtualkeyCMPT404A-TR>
Reply by ●February 15, 20172017-02-15
Jasen Betts says... > you get about 25uA in the main 100K and maybe 3 uA from > the microcontroller, A diode in series with the PNP base > will fix that. Yes it will, but - > So the PNP is biased off but current may flow from the > voltage regulator through the collector to the base then > through the emitter. if it does you could destroy the > PNP. Detroy the PNP and all the 9V unregulated circuitry outside the regulator and controller. I was hoping that with the base protected by the diode, or even open, no current would flow from collector to emitter, thus protecting everything from reverse battery connection. And that does work at 5V or 6V, but not at 9V. In my tests, current began to flow at 8V, so it wouldn't protect against a backwards 9V battery. I've seen a lot of descriptions of the Zener effect when you backward bias the B-E junction of a PNP, but what I'm seeing is the same Zener effect from C to E, although at a higher voltage, no matter what's happening with the base - it can be open, or tied high, and C-E current still flows. After a couple hours of Googling and reading, I've yet to find any discussion of this phenomenon or why that happens. It's a shame the reverse protection doesn't work at 9V (or actually 9.6V with a new battery), because then the single PNP could both switch the power on and protect against reversal, which is what I wanted to do.
Reply by ●February 15, 20172017-02-15
whit3rd says... > So, use a PNP with higher base-emitter reverse voltage. > This one is good to 25V: > http://www.mouser.com/search/ProductDetail.aspx? R=0virtualkey0virtualkeyCMP T404A-TR Yes, someone else also recommended that one. It's not clear to me that the Vebo rating necessarily matters in this case. I mean, you certainly would think it would matter, but I'll have to order one at some point and see what happens. At this point, though, I've been using through-hole parts from Digikey, so the surace mount item from Mouser is a bit off the trail. But I do need to test it. But you know, the world is not full of PNPs with that kind of Vebo rating, or even something as modest as 10V, which would be enough for me.
Reply by ●February 15, 20172017-02-15
On Tue, 14 Feb 2017 22:16:22 -0600, Peabody <waybackNO584SPAM44@yahoo.com> wrote:>Jasen Betts says... > > > you get about 25uA in the main 100K and maybe 3 uA from > > the microcontroller, A diode in series with the PNP base > > will fix that. > >Yes it will, but - > > > So the PNP is biased off but current may flow from the > > voltage regulator through the collector to the base then > > through the emitter. if it does you could destroy the > > PNP. > >Detroy the PNP and all the 9V unregulated circuitry outside >the regulator and controller. > >I was hoping that with the base protected by the diode, or >even open, no current would flow from collector to emitter, >thus protecting everything from reverse battery connection. >And that does work at 5V or 6V, but not at 9V. In my tests, >current began to flow at 8V, so it wouldn't protect against >a backwards 9V battery. > >I've seen a lot of descriptions of the Zener effect when you >backward bias the B-E junction of a PNP, but what I'm seeing >is the same Zener effect from C to E, although at a higher >voltage, no matter what's happening with the base - it can >be open, or tied high, and C-E current still flows. After a >couple hours of Googling and reading, I've yet to find any >discussion of this phenomenon or why that happens.The c-b junction is forward biased, in series with the b-e zener. So c-e zeners about 0.6 volts higher than b-e. Some transistors magically become 6.2 volt "reference zeners", super low TC zener diodes, used backwards c-e with base open. But longterm, zenering a b-e junction damages the part.> >It's a shame the reverse protection doesn't work at 9V (or >actually 9.6V with a new battery), because then the single >PNP could both switch the power on and protect against >reversal, which is what I wanted to do. >Why not use a series schottky diode, instead of complicated circuits? -- John Larkin Highland Technology, Inc lunatic fringe electronics
Reply by ●February 15, 20172017-02-15
John Larkin says... > Why not use a series schottky diode, instead of > complicated circuits? I just wanted to avoid even that voltage drop. Any drop just means the battery has to be replaced sooner. Also, a P-channel mosfet requires no additional parts, has essentially no voltage drop, and in case it even matters, only costs a little more that a Schottky diode. One other possible problem is that Schottky diodes tend to have larger reverse leakage currents than regular diodes, and I'm just not sure how much reverse current my circuit can withstand. This may be a non-problem. I just don't know.
Reply by ●February 15, 20172017-02-15
On 2017-02-15, Peabody <waybackNO584SPAM44@yahoo.com> wrote:> John Larkin says... > > > Why not use a series schottky diode, instead of > > complicated circuits? > > I just wanted to avoid even that voltage drop. Any drop > just means the battery has to be replaced sooner. Also, a > P-channel mosfet requires no additional parts, has > essentially no voltage drop, and in case it even matters, > only costs a little more that a Schottky diode.if preserving the battery is important, why are you using "9V", which is one of the most expensive forms? or is that 6 AA cells in series? -- This email has not been checked by half-arsed antivirus software
Reply by ●February 15, 20172017-02-15
Jasen Betts says... > if preserving the battery is important, why are you > using "9V", which is one of the most expensive forms? or > is that 6 AA cells in series? Part of the circuit needs unregulated 9-ish volts, and the 9V battery is just easier to deal with for size and weight than multiple AAs.
Reply by ●February 15, 20172017-02-15
