In article <1fa742ba-1f65-4192-b9ad-fecf0b701d69@googlegroups.com>, langwadt@fonz.dk says...> > > both because the body diode has to be one way to function as a switch, and > > > the opposite way to function as polarity protection. The PNP looks like the > > > right solution. > > > > > > Thanks for any suggestions. > > > > Use a Pmos as polarity protection and you can soft start the > > gate. > > Jamie > > soft starting the gate doesn't help much when the body diode is conducting > > >Yes, I know, I miss understood the question, I thought he was looking for good protection and didn't see where he was also looking for switching. Anyway, I followed that up with another one after the fact. Thinking some more about this, somewhere in my history I remember using an protective switch IC, ready made.. But I guess he maybe looking for heavy loads too ? Jamie
PNP for soft switch and reverse battery protection
Started by ●February 9, 2017
Reply by ●February 10, 20172017-02-10
Reply by ●February 10, 20172017-02-10
On Fri, 10 Feb 2017 17:58:24 -0500, Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> wrote:>On 02/10/2017 05:29 PM, Winfield Hill wrote: >> Peabody wrote... >>> >>> It's the pesky PNP base that I'n worried about. >> >> An interesting fact is that many (most?) PNP >> BJTs actually have quite high Veb breakdown >> voltages. They may not say so on the specs, >> but when you measure them, you discover this >> is true. Consider, as an example, the LM339 >> comparator's PNP input transistors, Q2 and Q3. >> They can handle 36 volts of reverse biasing. > >Those are laterals, though, which is a bit of a special case--aiui they >don't have super high emitter doping. > >Cheers > >Phil HobbshCorrect! Discretes, on the other hand, DO have heavily doped emitters. ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Thinking outside the box... producing elegant solutions.
Reply by ●February 10, 20172017-02-10
On Fri, 10 Feb 2017 16:28:22 -0600, Peabody <waybackNO584SPAM44@yahoo.com> wrote:> >Jim Thompson says... > > > R2 causes the emitter-base junction voltage to be too > > small in the reversed input voltage mode to leak and > > pass large currents. > >Ok, I breadboarded the PNP circuit in both the normal >orientation of the transistor and with the transistor placed >backward. Since the base is the middle pin, I just turned >the transistor around 180 degrees. > >The two circuits are shown here: > >https://s28.postimg.org/8tdthq4il/Peabody_breadboard.jpg > >In the normal orientation, the 2N3906 functioned normally. >Opening the base via J1 turned off the transistor >completely, at least to the limit of my meters, at any >V+ up to 12V. > >In the reverse orientation, at low V+ levels, it also >functioned normally, but with a bit less output current. >This presumably results from the lower beta. > >However, as I ramped up the V+ power supply, C-E current >began to flow regardless of what was happening at the base - >even if the base was completely open. Using the best meter >I have for measuring very low currents, which is my old >analog meter on the 50 uA scale, I found that the needle >just began to move when V+ reached 8.03V, and current >continued to increase as I increased V+ further, reaching 75 >uA with V+ at about 11.9V. > >This current was minimized with the pullup resistor (your >R2, my R1) connected to the collector. "Off" current >increased if I connected it to the emitter where you had it. > >So this is either a characteristic of the 2N3906, or perhaps >I have two of them that just happened to fail in the same >way, or there's something about the resistor values I used >that's causing this result. > >But if none of those apply, and notwithstanding what Spice >may have to say, it's possible that this is a characteristic >of all PNP transistors, or possibly all bipolars. And if >that's the case, then what I wanted to do, which is use one >device for both switching and polarity protection, will work >fine at, say, 5V or less, but not if you're using a 9V >battery. > >I guess the good news is that for 3.3V or 5V, this actually >works, and you don't even have to turn the transistor around >because the reverse voltage is less than the Vebo maximum. >You can switch the power on and off with the PNP, and you >get free polarity protection to boot. But it appears this >doesn't work at higher voltages - unless of course I screwed >something up that can be corrected. > >If anyone wants to try confirming my test, I would be >interested in the results. Meanwhile, unless there's a fix >I haven't thought of, it looks like the backwards P-channel >MOSFET will be needed for polarity protection for the 9V battery.You have R1 the wrong way around in the emitter-toward-load mode. And it should be about the same value as the bias resistor. And lose the diode. ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Thinking outside the box... producing elegant solutions.
Reply by ●February 10, 20172017-02-10
Jim Thompson says... > You have R1 the wrong way around in the > emitter-toward-load mode. And it should be about the > same value as the bias resistor. > And lose the diode. With the transistor in the backwards orientation, and V+ at 12V: I made all three resistors 47K, and removed the diode. I still get C-E current with the base open. If I move R1 to the emitter side, it INCREASES the OFF current from 75 uA to 200 uA. So that makes things worse. I tested three NPN transistors the same way - connecting C or E to 12V through a 47K resistor with the other connected to ground, but with the base open, and got similar results. If I connect C and E in normal polarity, no current flows (because there is no base current), but when I reverse the polarity of the C and E connections, I get current - even though there is no base current. So this effect appears to hold for all my bipolar transistors so far. What I don't know is whether the voltage at which this current flow begins to happen is related to any datasheet parameter.
Reply by ●February 10, 20172017-02-10
On Fri, 10 Feb 2017 20:49:08 -0600, Peabody <waybackNO584SPAM44@yahoo.com> wrote:>Jim Thompson says... > > > You have R1 the wrong way around in the > > emitter-toward-load mode. And it should be about the > > same value as the bias resistor. > > > And lose the diode. > >With the transistor in the backwards orientation, and V+ at >12V: > >I made all three resistors 47K, and removed the diode. > >I still get C-E current with the base open. If I move R1 to >the emitter side, it INCREASES the OFF current from 75 uA to >200 uA. So that makes things worse. > >I tested three NPN transistors the same way - connecting C >or E to 12V through a 47K resistor with the other >connected to ground, but with the base open, and got similar >results. If I connect C and E in normal polarity, no >current flows (because there is no base current), but when I >reverse the polarity of the C and E connections, I get >current - even though there is no base current. > >So this effect appears to hold for all my bipolar >transistors so far. What I don't know is whether the >voltage at which this current flow begins to happen is >related to any datasheet parameter. >What is you load current? I think you're doing something wrong... don't open the friggin' base! ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Thinking outside the box... producing elegant solutions.
Reply by ●February 11, 20172017-02-11
Jim Thompson says... > https://s28.postimg.org/8tdthq4il/Peabody_breadboard.jpg > What is you load current? I think you're doing > something wrong... don't open the friggin' base! As I said, I changed all the resistors to 47K. So the base pullup resistor is 47K, the base resistor is 47K, and the load resistor is 47K. I removed the diode. And the power supply is 12V. When I insert the transistor backwards, the ON performance is essentially the same as when it's in normal orientation. The base current is about .25mA, and load current is also .25mA, both of which are right for 47K resistors. The problem comes when I try to turn the transistor OFF by disconnecting the base resistor from ground, leaving only the base pullup resistor connected. If the base pullup is connected to the collector (i.e. - directly to the 12V power supply), I still get 75 uA of current through the load. If instead I connect the pullup to the emitter as you suggest, the load current increases to 190 uA. The problem of course is that with the base pullup tied high, the transistor should be completely off, and there shouldn't be any load current at all. The OFF load current only happens if the transistor is in the opposite orientation from normal, and even then only if the power supply is above about 8V (for my 2N3906's at least). Even though this effect doesn't appear to involve the E-B voltage, I do wonder if the Vebo rating of a transistor affects the 8V threshhold. But I don't have a high-Vebo part to test. If anyone actually breadboards this (not Spice) and gets a different result, please let me know.
Reply by ●February 11, 20172017-02-11
On Sat, 11 Feb 2017 11:38:32 -0600, Peabody <waybackNO584SPAM44@yahoo.com> wrote:>Jim Thompson says... > > > https://s28.postimg.org/8tdthq4il/Peabody_breadboard.jpg > > > What is you load current? I think you're doing > > something wrong... don't open the friggin' base! > >As I said, I changed all the resistors to 47K. So the base >pullup resistor is 47K, the base resistor is 47K, and the >load resistor is 47K. I removed the diode. And the power >supply is 12V. > >When I insert the transistor backwards, the ON performance >is essentially the same as when it's in normal orientation. >The base current is about .25mA, and load current is also >.25mA, both of which are right for 47K resistors. > >The problem comes when I try to turn the transistor OFF by >disconnecting the base resistor from ground, leaving only >the base pullup resistor connected. If the base pullup is >connected to the collector (i.e. - directly to the 12V power >supply), I still get 75 uA of current through the load. If >instead I connect the pullup to the emitter as you suggest, >the load current increases to 190 uA. The problem of course >is that with the base pullup tied high, the transistor >should be completely off, and there shouldn't be any load >current at all. > >The OFF load current only happens if the transistor is in >the opposite orientation from normal, and even then only if >the power supply is above about 8V (for my 2N3906's at >least). > >Even though this effect doesn't appear to involve the E-B >voltage, I do wonder if the Vebo rating of a transistor >affects the 8V threshhold. But I don't have a high-Vebo >part to test. > >If anyone actually breadboards this (not Spice) and >gets a different result, please let me know. >What is "motor"? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Thinking outside the box... producing elegant solutions.
Reply by ●February 11, 20172017-02-11
On Sat, 11 Feb 2017 11:38:32 -0600, Peabody <waybackNO584SPAM44@yahoo.com> wrote:>Jim Thompson says... > > > https://s28.postimg.org/8tdthq4il/Peabody_breadboard.jpg > > > What is you load current? I think you're doing > > something wrong... don't open the friggin' base! > >As I said, I changed all the resistors to 47K. So the base >pullup resistor is 47K, the base resistor is 47K, and the >load resistor is 47K. I removed the diode. And the power >supply is 12V. > >When I insert the transistor backwards, the ON performance >is essentially the same as when it's in normal orientation. >The base current is about .25mA, and load current is also >.25mA, both of which are right for 47K resistors. > >The problem comes when I try to turn the transistor OFF by >disconnecting the base resistor from ground, leaving only >the base pullup resistor connected. If the base pullup is >connected to the collector (i.e. - directly to the 12V power >supply), I still get 75 uA of current through the load. If >instead I connect the pullup to the emitter as you suggest, >the load current increases to 190 uA. The problem of course >is that with the base pullup tied high, the transistor >should be completely off, and there shouldn't be any load >current at all. > >The OFF load current only happens if the transistor is in >the opposite orientation from normal, and even then only if >the power supply is above about 8V (for my 2N3906's at >least). > >Even though this effect doesn't appear to involve the E-B >voltage, I do wonder if the Vebo rating of a transistor >affects the 8V threshhold. But I don't have a high-Vebo >part to test. > >If anyone actually breadboards this (not Spice) and >gets a different result, please let me know. >Oooops! I misunderstood your original post. I only tried reverse voltage protection, not turn-off via the transistor. Do you really need to turn off via the transistor rather than just a switch at the battery? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Thinking outside the box... producing elegant solutions.
Reply by ●February 11, 20172017-02-11
On Thu, 09 Feb 2017 13:16:44 -0600, 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. > >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. > >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. 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. > >I guess if it's clear that, at 9V, 100K is gonna keep anything from losing >its smoke, I would just go with that. But I'm more than a little goosey >about that. The easiest alternative I can think of is to just insert a diode >between the base and resistor. That would prevent any reverse current >flowing into the PNP base up to the rating of the diode, which will be way >above 9V, while not really affecting how the transistor functions. In >particular, it would have no effect on the main E-C voltage drop in normal >operation. > >So what do you think? Am I right at least about the theoretical risk? If >so, does the 100K resistor cover me, or do I need the diode? Or is there a >better way? Well, I guess there's always a better way. As I said, I want to >combine a soft switch with polarity protection. A P-channel MOSFET can't do >both because the body diode has to be one way to function as a switch, and >the opposite way to function as polarity protection. The PNP looks like the >right solution. > >Thanks for any suggestions.<http://www.analog-innovations.com/SED/Peabody_Switch_v2.png> ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | STV, Queen Creek, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Thinking outside the box... producing elegant solutions.
Reply by ●February 11, 20172017-02-11
On 2/11/2017 1:07 PM, Jim Thompson wrote:> On Sat, 11 Feb 2017 11:38:32 -0600, Peabody > <waybackNO584SPAM44@yahoo.com> wrote: > >> Jim Thompson says... >> >>> https://s28.postimg.org/8tdthq4il/Peabody_breadboard.jpg >> >>> What is you load current? I think you're doing >>> something wrong... don't open the friggin' base! >> >> As I said, I changed all the resistors to 47K. So the base >> pullup resistor is 47K, the base resistor is 47K, and the >> load resistor is 47K. I removed the diode. And the power >> supply is 12V. >> >> When I insert the transistor backwards, the ON performance >> is essentially the same as when it's in normal orientation. >> The base current is about .25mA, and load current is also >> .25mA, both of which are right for 47K resistors. >> >> The problem comes when I try to turn the transistor OFF by >> disconnecting the base resistor from ground, leaving only >> the base pullup resistor connected. If the base pullup is >> connected to the collector (i.e. - directly to the 12V power >> supply), I still get 75 uA of current through the load. If >> instead I connect the pullup to the emitter as you suggest, >> the load current increases to 190 uA. The problem of course >> is that with the base pullup tied high, the transistor >> should be completely off, and there shouldn't be any load >> current at all. >> >> The OFF load current only happens if the transistor is in >> the opposite orientation from normal, and even then only if >> the power supply is above about 8V (for my 2N3906's at >> least). >> >> Even though this effect doesn't appear to involve the E-B >> voltage, I do wonder if the Vebo rating of a transistor >> affects the 8V threshhold. But I don't have a high-Vebo >> part to test. >> >> If anyone actually breadboards this (not Spice) and >> gets a different result, please let me know. >> > > What is "motor"?I believe that is "meter", the thing in series with the voltage source on the left. -- Rick C
