PNP Saturation Base Current Question... - Page 1
Forum section: electronics / beginners

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 27, 2015, 11:56:48 pm
content: i want to make a high pnp switch to a load 0.5-2A, but to ensure proper saturation operation to avoid heat waste (as in linear mode operation), i need to set R1, R2 value properly. R1, R2 also will make a voltage divider to avoid Vbe greater than max rating ie around -5V is the idea correct?so i made a circuit using 2SB1443 PNP. my heart content says i want Vbe to be around -5V max, Ib max is 10mA, so i made some unknown calculation resulting R1 = 500ohm, R2 = 1.5Kohm approximately. thinking there will be much current going from 15V to GND, i upped the value to R1 = 1K, R2 = 3K. then i made the circuit and measured. i've changed R1 R2 to different value closed to the calc value but the measured Vbe is still the same, Vbe measured to be around -0.3V only. its out of my expectation, although the pnp switch works, but i certainly missing something and i cannot guarantee that the PNP is in saturation mode, and i cannot guarantee it will dissipate less heat (efficient) for higher load. please advice, thanks.ps: circuit and pnp datasheet are provided below for reference.note: load symbol shown is incorrect. i quick drew it in tina spice and i dont see any proper symbol to use. please assume its just a normal load like resistor etc... actually currently i put 5V dc-dc converter there driving some other load.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 28, 2015, 12:27:22 am
content: Hi,With a PNP transistor the B-E junction would be reverse biased if the Base is more Positive than the emitter.In this configuration you would need to have a base current about 1/50 th of the collector current. So for 2A of collector current you would need 40 mA of base drive. So with 15V, that would be(15 -0.7V) x 0.04A = 0.57W so use a 2W resistor. The base-emitter resistor is used to improve switching performance.Remember to a diode from ground to collector, the anode should be connected to ground, to protect the transistor from an inductive load.You could replace the PNP with a P Ch.MOSFET and significantly reduce the dissipation in the resistor.Regards,Jay_Diddy_B

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 28, 2015, 12:41:37 am
content: Quote from: Jay_Diddy_B on March 28, 2015, 12:27:22 amyes but i need a path from base to ground to activate the pnp, thats the problem since gnd is -15V compared to +ve (emmiter), shorting base to gnd resulting in glowing red pnp yeah i killed one earlier with spectacular glowing red, and no load connected yet at collector pin. with your 2ohm as R1, i need 6ohm at R2 to avoid excessive than -5V at base, but that is 8ohm @15V = 2A of waste energy at base/battery side, conflict! i know p-mosfet performs better but 1) i need to master this pnp biasing matter and 2) i dont have medium power pmosfet around, the only in my stock is 600V 5-10A size which i can use for another properly rated use.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 28, 2015, 01:42:23 am
content: Hi,Like this:Jay_Diddy_B

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 28, 2015, 02:59:35 am
content: yes thank you thats the common setup and not a problem when Vce or Vsupply is within 5V range. what confuses me is that Vrail (15V) is more than Vbe max in datasheet hence i went into dilemma i was afraid without R going to 15V to divide the voltage, the GND will pull the Vbe to -15V damaging another PNP. just as in "no load" condition, ie Ib will be too small hence pulling Vb node closer to 0V hence making Vbe nearly -15V again, no?

author: w2aew
author_rank: Super Contributor
timestamp: March 28, 2015, 03:08:42 am
content: Quote from: Mechatrommer on March 28, 2015, 02:59:35 amNo. The base-emitter is basically a diode, so the voltage across it is only going to be .7v give or take, the rest of the voltage is dropped across the resistor that connects it to ground. The resistor value determines the base current (Vsupply - Vbe)/R.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 28, 2015, 05:14:46 am
content: Quote from: w2aew on March 28, 2015, 03:08:42 amconfusion deleted, problem solved. one piece of knowledge gained. thanks w2aew and jay! since i'm at it.... is pmosfet works like that too? the source to gate is equivalent to a diode too? i heard for mosfet we need to provide R between gate to source (equivalent to base to emitter in transistor) to discharge capacitance, i thought transistor works the same way somehow it got base to emitter capacitance, thats why i always thought and resort to a voltage divider which sometime confuses me...

author: Zero999
author_rank: Super Contributor
timestamp: March 28, 2015, 10:02:01 am
content: No, a P-MOSFET's gate acts like a capacitor, same as an N-MOSFET and the maximum gate voltage rating shouldn't be exceeded.The VEBrating for a BJT is specified when it's operating in reverse biased mode.It depends on the transistor. A forced beta of 50 may not be enough to fully saturate the transistor. Typically the saturation voltage is specified when IB=1/10ICbut you may be able to get away with much less than this, depending on the application.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 28, 2015, 12:18:48 pm
content: Quote from: Hero999 on March 28, 2015, 10:02:01 amok, now i know, for centuries i'm reading transistor's datasheet, i thought it forward bias since no mentioning "reverse biase" condition in datasheet. and my lack understanding of what a transistor's base is.Quote from: Hero999 on March 28, 2015, 10:02:01 amso i guess the stated 82 min hfe and 270 max is for linear mode? most transistor's datasheet are confusing then because there is no mentioning "linear mode" in "electrical characteristic" data/table. and also no mentioning about max rating in "reverse biased" operation. i guess we need a teacher and practical experience burning some, if we want to learn a transistor. anyway thanks hero, another knowledge gained, things become much more clear right now.

author: Psi
author_rank: Super Contributor
timestamp: March 28, 2015, 12:24:19 pm
content: Floating base ?Quote from: Jay_Diddy_B on March 28, 2015, 01:42:23 am

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 28, 2015, 12:40:25 pm
content: i never heard issues with floating BJT base, though repeatedly adviced to be avoided for mosfet. from my practical experience tells that too, my circuit never accidentally activated due to floating BJT base, but floating MOSFET gate problem is pretty real almost everytime i made a circuit in such a floating way. but i'm not sure, maybe other have better experience.

author: SeanB
author_rank: Super Contributor
timestamp: March 28, 2015, 12:50:49 pm
content: Current operated BJT and voltage operated MOSFET. It is very easy to have leakage current charge the very high impedance mosfet up to conduction ( or to keep it there if it is on and there is no discharge path) while the BJT needs either enough collector voltage so that leakage turns it on or really high gain ( darlington without integral discharge resistors) so that even low leakage is enough to do the same. If you get it hot enough ( over 150C) then the leakage is going to be enough to turn it on.Reverse biased operation is rarely specified, as most of the time transistors are not operated in there, it is more a stand off voltage set that it will survive. The only time you really have it used is in power devices, where the base is reverse biased during turn off to turn off the device faster, as then the charge in the base region is actively removed.Turn on voltage might be a peak of 2V at 3A base current, and turn off is a peak reverse voltage of around the same magnitude, provided by a driver transformer.

author: Zero999
author_rank: Super Contributor
timestamp: March 28, 2015, 01:56:49 pm
content: Quote from: Mechatrommer on March 28, 2015, 12:18:48 pmIt's quite common to add a base-emitter resistor to speed up the off time, otherwise it can take a long time for the BJT to turn off due to the base-emitter capacitance.The Hfe is often specified at a certain VCEwhich is sometimes quite high.Fortunately Hfe increases with temperature, so if it heats up a little, the voltage drop will be less for a given base current.If in doubt, post the datasheet for the transistor you're using.

author: flynwill
author_rank: Regular Contributor
timestamp: March 28, 2015, 03:28:31 pm
content: I'd be very tempted to add a transistor, or use a darlington in place of Q1 to avoid that 1/2 watt wasted in the drive resistor.I would also add the base-emitter resistor -- probably ~220 or so in this circuit, to speed up that off transition. You don't say how often you are switching the load, but the slow turnoff can contribute significantly to the heat load on the drive transistor.

author: SeanB
author_rank: Super Contributor
timestamp: March 28, 2015, 03:38:53 pm
content: Quote from: flynwill on March 28, 2015, 03:28:31 pmHalf watt in the resistor as opposed to 3W in the non saturated darlington device. Darlington transistors do not saturate, they always will have at least 1V drop across them, irrespective of base current. If you are driving it at low base current it will be closer to 1V5.

author: TimFox
author_rank: Super Contributor
timestamp: March 28, 2015, 04:17:48 pm
content: Adding a base-emitter resistor is often a good idea to avoid problems with the collector-base leakage current at high temperatures in the Off state. Otherwise, the leakage current flows into the base and can result in unwanted collector-emitter current, possibly leading to thermal runaway.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 28, 2015, 07:58:42 pm
content: Quote from: flynwill on March 28, 2015, 03:28:31 pmi can make direct mechanical switch to the load (or relay), but since the load is dc-dc converter, it has a nasty spike during startup (or i think due to big Vinput capacitor on the dc-dc module). thinking mechanical switch will not last long, i resolved to this pnp switch. now the mechanical (manual) switch will be at the pnp 's base, exactly like i drew it in the 1st post. since this is a switch on->use the device and ->switch off by hand when done, like normal appliances, so switching loss is assumed negligible. this is not automated switching like PWM etc.

author: flynwill
author_rank: Regular Contributor
timestamp: March 29, 2015, 04:08:20 pm
content: Quote from: SeanB on March 28, 2015, 03:38:53 pmYep, you're right I didn't think that one through.Quote from: Mechatrommer on March 28, 2015, 07:58:42 pmAh. In that case, unless you are really pressed for space, I would stick with a mechanical switch with the appropriate current ratings. With a good snap-action switch the inrush rarely causes any issues. It's the arc when the switch opens caused by inductive loads that kills switches.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 29, 2015, 11:02:44 pm
content: Quote from: SeanB on March 28, 2015, 03:38:53 pmisnt any transistor will have voltage drop even in saturation? let me pick any random transistor datasheet in my keeping... BD245 power transsitor. it speced c-e saturation voltage max = 1V @ 3A Ic, and 4V drop @ 10A Ic. i wonder why transistor are not speced like mosfet's Rds-on, bjt must also have Rce-on saturation right? but i havent seen such thing speced in datasheet for bjtQuote from: flynwill on March 29, 2015, 04:08:20 pmexactly this is the problem in my mind. at switch open and close, the arc (spike).

author: flynwill
author_rank: Regular Contributor
timestamp: March 29, 2015, 11:51:37 pm
content: Quote from: Mechatrommer on March 29, 2015, 11:02:44 pmYes that is true, but typical saturation Vce for a BJT is in the 0.3-0.5 volt range. In the case of a darlington, the first transistor can go into saturation, but it can't force the second into saturation so you get typical saturation Vce of 1-1.2 volts ie 2 or 3 times greater all else being equal. I'm not sure what worst-case conditions the manufacturer for that device is assuming with that 4V spec, if you look at the graph on the next page you'll see that the typical Vce for those conditions is 0.8 V which is still rather high. I did a very quick search (it's been a while since I've done any serious transistor shopping) a turned up a more typical device:http://www.mouser.com/ds/2/308/ENA1799-D-462253.pdfQuote from: Mechatrommer on March 29, 2015, 11:02:44 pmBecause BJTs almost never behave in a way that is well modeled by a resistance. They are best modeled as a current source.QuoteYes but a large capacitive load (which indeed is probably what your DC-DC converter looks like) won't cause an arc when the switch opens, the capacitor will hold up the voltage on it's side so there isn't any significant potential across the switch as it opens. There will be a large current when the contacts close, so contacts that close slowly (like a slide switch) can have issues under such circumstances. But I think if you get a switch rated for at least 10A, that is snap-action (like a typical toggle switch) from a decent manufacturer you'll be just fine.If the switch has to be some distance from the DC-DC converter and it's power source (is this going into your car?) then I can see the advantage to using the transistor. You can run relatively light wire to and from the switch.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 30, 2015, 12:17:49 am
content: Quote from: flynwill on March 29, 2015, 11:51:37 pmits for my rc helicopter. the power is from 4S lipo, the whole circuit incl dc-dc need to be as small as possible in heli compartment, the said 5V dcdc load, will supply mcu and radio receiver power, probably with some other LEDs. and then i need to wire the switch some distance back to the heli body where hand can access. yes small wiring is much preferable. i believe power loss in transistor is negligible compared to power that will be used by main motors.

author: flynwill
author_rank: Regular Contributor
timestamp: March 30, 2015, 03:46:50 am
content: Ah! and obviously you can't afford the weight of heat sinks (or 10A rated switches). I would spend some serious time looking at available MOSFETS. I suspect that there are lots of devices available that will easily beat the saturation voltage of a BJT with your 2A load, and you won't be wasting that 1/2 watt with the base series resistor either. Your switch can probably be the tiniest lightest thing you can find.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 30, 2015, 07:13:57 am
content: Quote from: flynwill on March 30, 2015, 03:46:50 amcant argue anymore, i agree 100% the load is not a 2A flat maybe like 0.5-1A tops, or 0.3-0.5A typical thats why the compromise. i need to measure amperage later to be sure, if heat loss is significant, there is no better suggestion than yoursthanks for the suggestion.

author: Zero999
author_rank: Super Contributor
timestamp: March 30, 2015, 12:54:08 pm
content: The BD245 doesn't have much gain and has a high saturation voltage.You can get much better transistors. Try the ZTX1149A.http://www.diodes.com/datasheets/ZTX1149A.pdf

author: Ton
author_rank: Regular Contributor
timestamp: March 30, 2015, 01:28:57 pm
content: If you want to go down the mosfet route then On semi have a nice application note explaining how to do it with P-mosfet even including a current limit for the turn on scenario. (capacitive loading )http://www.onsemi.com/pub_link/Collateral/AND9093-D.PDFas a P-fet you could use something like this one in a super sot6 housinghttps://www.fairchildsemi.com/datasheets/FD/FDC5614P.pdf, continous current up to 3A, at 2A the loss will be 2A*2A*0.135ohm=0.54Wtemperature raise if no cooling (at 2A) 0.54W*156degC/W=84degC plus ambient, so a bit toastyat 1A it will humm along happily at 21degC above ambient as long as it is soldered to a bit of PCB.not bad for a 3x3mm componentBR Ton