need help with math  - Page 1
Forum section: electronics / beginners

author: Jwillis
author_rank: Super Contributor
timestamp: December 06, 2017, 08:58:58 am
content: I've havn't done any algebra for over 30 years and I,m really rusty.I need to calculate the base resistor and emitter bias resistor and I keep getting odd results.I've attached the data sheet of the transistor that needs biasing.Also I'm not 100% sure if I'm using the correct information from data sheet.The hfe I'm using is the lowest presented ,which is to say 25. The Vbe is 2 volts .The Vin is 28 Volts.Transformer current is 26 ampsI calculate the Ib (base current) as Ib=Ic(collector current) /hfeSo Ib =20amps/25 Ib=0.8now I calculate the Base resistorSo Rb = Vin-Vbe/3xIb =28-2/3x.8=26/2.4 Rb=10.8 omhsthen I calculate emitter resistor for 10 ampsRe=(Vin-Vbe)/Ie+hfe/Rb=(26/10)+(25/10.Re=4.9But this just burns up the transistorUsually I find that the emitter bias resister should be under 1 ohm.So I recalculate with a 0.1 resistor for 10 ampsRb=hfe[((Vin-Vbe)/Ie)-Re] Rb=25[26/10-0.1] Rb=62.5Am I diong the math wrong or did I read the data sheet wrong I dont know.Every thing was working fine up until now.Now its all going south .

author: bd139
author_rank: Super Contributor
timestamp: December 06, 2017, 09:16:34 am
content: Are you using this for switching or in the linear region? What is your application? Can you post the circuit diagram too?

author: rstofer
author_rank: Super Contributor
timestamp: December 06, 2017, 01:48:45 pm
content: A schematic would help. For example, are you building a switch or an amplifier?Have you calculated the power dissipation for each resistor?P = E^2 / R or P = I^2 * R10 Amps through a 5 Ohm resistor would be20500 Watts.With an emitter resistor, the required base voltage is moving around with emitter current.

author: StillTrying
author_rank: Super Contributor
timestamp: December 06, 2017, 03:11:21 pm
content: "A schematic would help."+10"10 Amps through a 5 Ohm resistor would be 20 Watts."+480 (Watts)

author: rstofer
author_rank: Super Contributor
timestamp: December 06, 2017, 04:10:19 pm
content: Quote from: StillTrying on December 06, 2017, 03:11:21 pmMy bad... It was really early and I hadn't had my first dose of caffeine. I corrected my earlier post but the short answer is 10 Amps through a 5 Ohms resistor is 500 Watts. That's a BUNCH!

author: Connoiseur
author_rank: Regular Contributor
timestamp: December 06, 2017, 04:11:13 pm
content: Quote from: StillTrying on December 06, 2017, 03:11:21 pmHow?Unless I've been living in some parallel universe it should be 10x10x5=500W

author: rstofer
author_rank: Super Contributor
timestamp: December 06, 2017, 04:12:35 pm
content: Quote from: Connoiseur on December 06, 2017, 04:11:13 pmYup! You got in just after I fixed it. As I said, too early, no caffeine.

author: IanB
author_rank: Super Contributor
timestamp: December 06, 2017, 04:22:59 pm
content: Quote from: Connoiseur on December 06, 2017, 04:11:13 pmYes, but 20 W + 480 W = 500 W (i.e. add on another 480 W to get the correct answer)

author: Jwillis
author_rank: Super Contributor
timestamp: December 06, 2017, 08:43:51 pm
content: Sorry folks I forgot the schematic . Its a power supply experiment .I've tried the 2N3055 and get the same out come .I keep burning out the base.None of the resistors get close to warm on load.I'm trying to bias the transistors to limit hogging and thermal run away . But also want to have a fair degree of current over head on each transistor to keep heat down.But I'm also limited with the components I have.Takes months for stuff to get here.

author: Connoiseur
author_rank: Regular Contributor
timestamp: December 07, 2017, 02:48:29 am
content: Quote from: Jwillis on December 06, 2017, 08:43:51 pmWhat I think is, you are operating your BJT's out of safe operating area (SOA). Right now, I don't have time to go through all of that, maybe I'll get back in the evening (according to my time zone). Just check the datasheet for combinations of voltage and currents that lie inside SOA and keep your load line within it, no matter what.

author: radiogeek381
author_rank: Regular Contributor
timestamp: December 07, 2017, 04:18:04 am
content: A few quick points:0. I'm not sure where your original equations came from. They don't smell right. First, Vbe = 2V is way way beyond normal operation for a single BJT. Most folks use 0.6 to 0.7 when they're doing back-of-the-envelope calculations. In real life, 2V across the base-emitter terminals is an indication that not all is well.1. As Connoiseur indicated, a key concept that is behind some of the troubles with this design is the "SAFE OPERATING AREA" In general, for a power transistor, this relates tempearature, power dissipation, and likelihood of smoke. Specifically, for a BJT it is a graph giving a bounding polygon where the transistor is in a "safe" operating condition if the voltage (Vce) and current (Ic) are within the polygon. Note that the safe area is normally specified assuming some duty cycle, and a case temperature. Knowing the case temperature requires a little math on its own.2. What you've got in the drawing is an emitter follower. Unless there's some other criteria that you're considering, this is not often the best choice for a regulated power supply. See the pretty good discussion here:https://www.eevblog.com/forum/beginners/variable-regulator-pass-transistor/But let's look at what's going to happen with the circuit in your drawing.Assume for the moment that the input voltage to the LM138 is 40V. (Just for the sake of discussion). And now let's say that we want a 20V output at about 4 A. Working back from the output, we've got four transistors in parallel, each with its own emitter resistor.Those resistors need to be small, or they'll have to be very high power widgets. At 4A, with four parallel pass devices, you'll see 1W per ohm in each of the Re . Push that current up to 10A and we're talking 6.25 W/ohm -- A few watts here, and a few watts there, pretty soon you're talking real power. Your original post suggested Re = 5 ohms.If we go to the 5ohms that your initial figures suggested, this is what happens:If we set the LM138 feedback when we have 0 load, the voltage at the 138 output will be something like 21 V. Let's take the load up to 4A. Then we'll be dropping 1A * 5 ohms or 5V through each of the series devices. That means your original 20V open circuit voltage dropped to 15V under load.If the load draws 20A then you're realling in the soup, since each Re is passing 5A through 5ohms and the 25V drop across Re puts the emitter above the base. That's clearly nonsense, so what will really happen is Vout will drop to some voltage where the load current, emitter drop, Vbe, and Vb all conspire to please Kirchoff. But Vout won't be anywhere near 20V.So, 5ohms for the emitter resistor is clearly not right. Normal values should put you on the order of 0.1 ohms or so...Emitter followers don't make for very good regulation unless you sense the actual output. But there are interesting problems when you introduce that kind of feedback, since it is possible to design a combination DC supply and oscillator all in the same widget.-------A second consideration is the power dissipation. I'm assuming that you want this power supply to cover some range of voltages, perhaps 5 V to 20V? Let's now consider what happens in the transistors when you draw 4A at 5V. Vce will now be 35V (remember the input voltage was something like 40V). And each transistor is passing 1A. So each transistor has to dissipate 35 Watts of thermal energy somewhere. The data sheet shows a contour for "1s" duration. Frankly, I don't do power electronics much (I design loads, not supplies..) but I suspect that continuous duty won't be quite as generous as the 1s contour.The big deal here is that you're going to want to remove 35W of heat. The thermal resistance from the junction to the case is about 0.7 C/W. So, the junction is going to be about 20 C above the case temperature. If you get a heat sink of about 46mm x 46mm with some air blowing across it, you'll see a thermal resistance of about 1.5 C/W. Assuming a perfect bond between the case and the heat sink, the junction temperature is going to be about 70C above ambient. If you don't have a fan on the heat sink, the resistance rises to more than 6 C/W... Then the junction will be about 200 C above ambient. WHOOOSH! The smoke gets out. (And we were not even close to 10A load current...)Heat sinks are important and easy to get wrong.Take a look at the other thread, it had some good thinking points.

author: Jwillis
author_rank: Super Contributor
timestamp: December 07, 2017, 08:11:19 pm
content: Quote from: radiogeek381 on December 07, 2017, 04:18:04 amI got the 2V Vbe off of the data sheet so obviously I read that wrong. So I should calculate for .6V to .7VQuote from: radiogeek381 on December 07, 2017, 04:18:04 amIt seems that I'm within SOA except that according to the transformer data which is supposed to deliver 26 amps .And I think my DMM is lying to me because I'm getting 29.4 Vdc after rectifier then 44.2 Vdc when filter caps are connected.but still well below the 140v Vce and Vcb I think that might be the way the DMM is measuring .Because I get 28 Vdc after the LM138 (on its own sink)which works as expected without any temperature increase.Anyway the circuit worked with the 0.1 emitter resistor and no base resistor but I didn't load test in that configuration.But when I put in the 10 ohm base resistor in I can't even load an 1 amp at 2 volts before the transistor burns up.That's on a decent heat sink.This is why I thought my math has to be messed up.