Need Help With Transistor Biasing Math - Page 1
Forum section: electronics / beginners

author: Jwillis
author_rank: Super Contributor
timestamp: March 02, 2018, 12:52:21 am
content: I think I know now what my problem is in balancing parallel transistors.I'm getting confused about what values of resistor I need at the emitter and base.Now clearly a 0.1 ohm resister at the emitter is not doing the job of limiting load hogging.I'm burning transistors to much and it sucks .So I need to add a base resistor as well.Since I'm using a well know transistor (the 2n3055) theirs more information on how to calculate the resistors.My problem is my math is so rusty that I'm not sure if I'm getting this right or even if I'm using the right values from the data sheet.http://www.onsemi.com/pub/Collateral/2N3055-D.PDFI would like to try to get this working with the components I have ,then make improvements as better ones become available.Just need to get the Math right!!These are the parameters I'm working with.I have 4 amps from the transformer at 40 volts. I would like 4 2N3055's to share the 4 amps (1 amp each).According to the Active Region Safe Operating Area I should have around 2.5 amps at 40 volts. So that leaves me a little head room .Around 1.5 amps. These are mounted on a fairly descent heat sink.Now to calculate the base resistor I need the hFE ( I've been using 5)The Vce is 1.5 (that's the voltage drop across collector and emitter?)Ic will been 4 amps.Vb is 40 volts.It says Vbe is 1.5 or should it be 0.6 -0.7Now the 2n3055's are being driven by a TIP100. the reason is I need to have that gain (hFE 1000) to limit the draw from a regulator which needs to be around 0.5mAdc.This is explained in the data sheet of the regulator.https://www.mikrocontroller.net/attachment/160010/mc1466l_tiff.pdfNominal Design Procedure and Design Considerations : note 10So that would give me 4amps at the base of the 2N3055's ( probably have that wrong as well)But I figured I'll use the maximum possible assuming the TIP 100 is is full saturation.I know its not good to assume.Rb=(V-Vbe)/Ib=(40-1.5)/4=154ohmsNow since the transistors are parallel do I need to multiply that value by 4 from each transistor?I tossed in the schematic as well because when the transistor blows (only one because of the diodes from collector to emitter saved my back side) the voltage adjust potentiometer also blows.At least I'm not cratering the chip.I can get the darn thing to a little over 2 amps but then every thing goes south. I don't want to wreck another pot because that was one of my favorite one's

author: Paul Price
author_rank: Super Contributor
timestamp: March 02, 2018, 01:27:19 am
content: Make sure you mount the TIP100 on a fairly large heatsink!See the attached spec sheet clip for base resistor values.100 to 150 ohms B-E resistors should be ok to prevent thermal runaway. What you really need is a series high wattage resistor in series with the collector of the TIP100, having a value about 10 to 50 ohms. The TIP is failing in your >2A tests..right?I don't understand with the circuit shown how you are not always going to be burning up the v-adj. pot. As the voltage is being adj. down from maximum or any set voltage, the power supply output capacitor is being discharged into the V-adj. pot. Very bad if you turn the pot quickly, very bad if you turn the pot slowly over longterm use, especially bad for the pot near lowest voltage out..bad design!The v-adj. pot must always obey the ratio relationship shown in the second attachment, meaning it cannot be allowed to be adj. to 0-ohms. Q2 unloads the pot to some extent but must dissipate the full power output stored in the output storage capacitor. It requires a high-beta transistor on a large heatsink with up to 200W short-term dissipation.What the typical circuits explicitly show is that this chip should be set and left set to some fixed output V, and not to be used in an adj. output voltage bench power supply!

author: Jwillis
author_rank: Super Contributor
timestamp: March 02, 2018, 02:12:41 am
content: Quote from: Paul Price on March 02, 2018, 01:27:19 amTIP100 is on a heatsink and its Ok.Its the 2N3055's that are failing.The protection diodes between the collectors and emitters prevent them all from failing.The failed transistor will just pass the raw voltage and current from the transformer.So that's what it seems to be happening anyway.Because the voltage and current at output jacks matches the measured voltage and current after the rectifier and filter caps after failure.I doubt that the TIP100 even has 1 amp going through it because its always cold to the touch.The Q1 PNP with the (R10)5K resistor is the part that prevents the output cap from discharging through the V-pot Its described as such in the data sheet for the regulator.The Transistor Q1 has since been changed to handle a higher current than the B647AC. That all works fine until the 2N3055's fail.

author: Paul Price
author_rank: Super Contributor
timestamp: March 02, 2018, 02:25:05 am
content: Exceeding the max base current of the 3055's through the TIP100 drive will blow them up. That's why you need a resistor in series with the TIP100 collector.The protection diodes across the 3055's do nothing to protect them from overload except if the output voltage is negative or >than the 3055 collector voltage, and they will short out if this condition persists for more than an instant from some external connected voltage source.Transistor Q1 does a very poor job of protecting the V-pot if the V-pot is at all turned quickly or near zero resistance..poor circuit design.The v-adj. pot must always obey theratio relationshipshown in the second attachment, meaning it cannot be allowed to be adj. to 0-ohms. Your circuit obeys this ratio only at full output voltage. At all other lower output voltages the output capacitor discharges itself through the B-E of the transistor directly into the V-adj. pot. since the high-valued emitter resistor doesn't permit the PNP transistor to conduct any current > than ( Vout/PNP collector resistor) away from the V-adj. pot. The rest goes directly into the pot, this is a very bad design.

author: Andy Watson
author_rank: Super Contributor
timestamp: March 02, 2018, 02:51:00 am
content: Where did you find this circuit? It seems to be a modified version of the original Van Roon schematic - which presumably worked.Note this:Quote from: Paul Price on March 02, 2018, 01:27:19 amThat is base to emitter of the 2N3055s. I would also consider adding a few k ohms between the base and emitter of the TIP100 - for the same reasons. You could also double or triple the emitter resistors in the 2N3055s without much loss.Other thoughts:Are you using the more robust versions of the 2N3055s that are hardened against thermal runway? (Used to be the ones with the "H" suffix.) Are the 2N3055s genuine?Put a smallish sacrificial resistor in series with the V-pot whilst you are debugging to save the pot."Precision" power supply ? You want us to ignore the burden of the ammeter in series with the output!

author: Jwillis
author_rank: Super Contributor
timestamp: March 02, 2018, 03:13:45 am
content: I was simply going by what the data sheet says to do about the V-pot.But what I don't understand is if the voltage is brought to zero then the current drops to zero as well so there would be no charge in the output capacitor at zero volts.It works fine at 2 amps .Only when the tolerance of one of the 2N3055's is exceeded and the 2N3055 fails (around 2.5 amps) is when the V-pot fails as well.My problem is the load hogging at the 3055's.And after calculation of the TIP100 because theirs only .5mAdc applied to the base 'SoIe=(beta+1)Ib=(200+1)*0.5mAdc=100.5mAdc being applied from the TIP100 to the base of 2N3055's which are rated for a Maximum of 7 amps at the base.I'm not trying to be a jerk or anything but it seems the problem is at the 2N3055's load hogging .So i need to balance them so they each pass 1 amp.

author: Jwillis
author_rank: Super Contributor
timestamp: March 02, 2018, 03:28:14 am
content: Darn sorry I posted before I could reply Andy.Yes it is a modified version of Tony Van Roons.And yes it does work but I changed a couple things like adding the Q1 to save the V-pot(as described in the Datasheet for the MC1466L.Because I was having trouble with that with the original design.I added the extra transistors to boost to 4 amps instead of the original 2 amps.The amp meter as described in the datasheet for it does not require a shut so thats why its hook up that way.Yes I had indee tested it on a bench supply and a load to check accuracy.So I would connect the the 100 - 150 from base to emitter for each transistor.What is the math for that so I can apply it to other transistors.Also bump up the resistance of the emitter resistors to say 0.4How do you calculate that?

author: Paul Price
author_rank: Super Contributor
timestamp: March 02, 2018, 03:39:23 am
content: The problem is that the PNP transistor protects the pot only when it's collector resistance is low enough to absorb the output capacitors change in charge voltage,the pot is damaged during changing output voltage settingsand is especially stressed in your design as the voltage drops close to zero output. The PNP collector has to be 1/10 of the V-adj resistance setting.The power stored in the output capacitor is C*V*V/2, that's quite a lot for the pot to dissipate. The PNP transistor doesn't help at all if it's collector resistor is high valued and so the V-adj pot cannot handle the current it must take from the output cap with even a small change in output voltage.The collector resistor of the PNP transistor must be 1/10 the value of the V-adj pot for the PNP transistor to protect the pot at all See the attached clip from the data sheet.The current driving the TIP100 is required to be <.5mA to keep the circuit in regulation, but this current can be greatly exceeded by an out of regulation condition and so the base current of the 3055's can be high enough to destroy them. A base overdrive condition is created by the output capacitor when the voltage output is being adjusted up. The value of Beta for the TI-100 is typical min. value but is likely to be much greater than 100.The emitter resistors will balance their current sharing only if there is a base to Vout resistor to bias the 3055 base voltage to the bottom of the .1 ohm resistors, or else the voltage drop in the .1 ohm emitter resistors does nothing to balance 3055 current sharing.What is usually used to protect pass transistors is a NPN transistor whose emitter connects to the output voltage(after the .1 ohm), collector to the bases of the 3055's and the base to 100 ohm resistors each connected to the top of the .1 ohm resistors connected to each 3055 emitter. This approach still would require the 3055 emitter resistors to be around .33 ohm each to work well. Also, a series current-limiting resistor at the TI100 collector is needed or the NPN protection transistor just described could be damaged by the TI100 supplying the base-drive current during overload conditions, as in changing output voltage or output short circuit.

author: floobydust
author_rank: Super Contributor
timestamp: March 02, 2018, 03:41:56 am
content: I suggest backing up and reviewing the design, I see a few problems.First, it can eat Vadj potentiometers and Q1. Any output voltage over 0.6V will flow through Q1 E-B to the pot, with nothing to limit current there.I have to figure out why that transistor is used in the first place, I'll have time tomorrow. When LM317's short, they take out the adjust pot. I learned to add a small fuse to a multiturn pot, they are expensive.For the 2N3055 pass-transistors, the emitter current-sharing resistors you want as large as possible to give the most equal sharing. You also want the resistors as small as possible to be compact and not waste energy as heat. In practice 0.1R-0.47R is used. We can plow through the math, I don't see a problem using 0.1R's, yet you say things are failing. Current-sharing is not the problem.I think Q2 needs some protective current-limiting resistor as there's nothing to limit base-drive current to the 2N3055's or current from the MC1466's output pin. Around 33-100 ohms.D8 needs ~0.7V to conduct and with 0.05R, that is 14A before it does anything, so it might be helpful.There's a few other protective diodes needed. I can contribute more tomorrow.

author: Paul Price
author_rank: Super Contributor
timestamp: March 02, 2018, 04:22:48 am
content: The problem destorying the pot is that the PNP transistor protects the pot only when it's collector resistance is low enough to absorb the output capacitors change in charge voltage,the pot is damaged during changing output voltage settingsand is especially stressed in your design as the voltage drops close to zero output.The PNP collector resistor has to be 1/10 of the V-adj resistance setting to protect the pot.The emitter resistors will balance their current sharing only if there is a base to Vout resistor to bias the 3055 base voltage to the bottom of the .1 ohm resistors, or else the voltage drop in the .1 ohm emitter resistors does nothing to balance 3055 current sharing.What is usually used to protect pass transistors is a NPN transistor whose emitter connects to the output voltage(after the .1 ohm), collector to the bases of the 3055's and the base to 100 ohm resistors each connected to the top of the .1 ohm resistors connected to each 3055 emitter. This approach still would require the 3055 emitter resistors to be around .33 ohm each to work well. The idea here is that the voltage developed across any of the 3055 .1-ohm resistors turns on the NPN protection transistor which clamps the 3055's B-E voltage to zero.Also, a series current-limiting resistor at the TI100 collector is needed or the NPN protection transistor just described could be damaged by the TI100 supplying the base-drive current during overload conditions, as in changing output voltage or output short circuit.It is a very bad idea to allow any significant current to flow out of the slider pin of a pot, it always causes damage.Also, as a pot is connected as a rheostat in your circuit it is likely that there will be created or has been created dirty or bad spots that cause the pot to be at its an instantaneous max resistance value for maybe just a few milliseconds during rotation at which time the output voltage will attempt to jump up to maximum voltage output and the the 3055's base current might exceed many amperes, enough to destroy one or more of the 3055's.

author: Jwillis
author_rank: Super Contributor
timestamp: March 02, 2018, 04:53:18 am
content: My bad!! that Q1 has been changed but I didn't say what it was changed to in the first post.Its a D45H11.And I did use the formula for selecting the 5k resistor for R10 which is 1/10 that of the 50k resistor pot.I presumed that is what they meant in the given formula .So I should calculate R10 to be what resistance if P1 is not 50k?Ok ya the min beta for the TIP100 is 200.And the MC1466L will indeed supply up to 1.5 amps.My bad i based my calculations on Imax/beta1*Beta2 should be less than or equal to 0.5mAdc so I worked it as 4Adc/200*5=2000mAdc ..woops thats 2 amps .I see now. See told you my math is bad.I had set the voltage to 30 volts and left it there to run load test.I only adjusted the amperage up.When it reached around 2.5 amps that when I heard the V-adj pop.The voltage ran up to max 40Volt and the amps went up to 4 amps instantly.I shut down and checked all components.The pot was blown and only one of the 3055's was blown.All others tested ok .all other componants tested ok.This is all good information your giving me I really appreciate it.

author: Paul Price
author_rank: Super Contributor
timestamp: March 02, 2018, 04:58:16 am
content: The problem here is that the PNP's collector resistance needs to be 1/10th of the changing V-Adj resistance, that means any fixed value is not anywhere near the required value for most or nearly all the different resistance settings of the V-adj pot. If you attempt to make the collector resistance too low the PNP transistor will self-destruct. This power supply design is made to be work properly at and only over a small output voltage range...bad design. The pot will be destroyed if you attempt to set the output voltage close to zero V.Remember that the current driving the TI100 transistor is only <.5mA when the supply is regulating but can greatly exceed this current when an out of regulation condition exists.

author: Jwillis
author_rank: Super Contributor
timestamp: March 02, 2018, 06:22:12 am
content: I'll fix the output transistors first.I'll set the voltage with a fixed resistor to test then I'll tackle the V-pot issue.So I would be looking at something like this?

author: Paul Price
author_rank: Super Contributor
timestamp: March 02, 2018, 11:25:07 am
content: Something like this, add protection diodes.R1=.22 for your P/S

author: David Hess
author_rank: Super Contributor
timestamp: March 02, 2018, 01:02:08 pm
content: It is possible to calculate the needed values for the base-emitter shunt resistors and low valued emitter ballasting resistors to prevent thermal runaway and enforce good current sharing but I usually start with a pair of conservative rules of thumb:1. Select the value of the base-emitter shunt resistor to double the drive current. The drive current is the Ic/hfe and the Vbe is about 0.8 volts so this is easy enough. If the driver cannot handle doubling of its output current, then it lacked enough margin to start with.2. Select the emitter ballast resistor such that at the maximum collector current, the voltage across it is equal to the Vbe of about 0.8 volts. This doubles the voltage drop from the base to the output. If this is unacceptable, then matching the transistors for close Vbe at the maximum collector current will allow for a lower value of emitter ballast resistor but there is little point going below a voltage drop of about 1/2 Vbe unless you are designing a class-AB amplifier where transconductance needs to be maximally flat at crossover.3. In dynamic applications, there is some advantage to placing the base-emitter shunt resistor between the base and the other side of the emitter ballast resistor. If rule 2 above is followed, then its resistance is doubled for the same current and the negative bias supplied under dynamic conditions results in better high frequency performance. Audio amplifiers should take advantage of this or even connect to the base-emitter shunt resistor to the complementary side for even more negative bias.4. In high current applications, it may be a good idea to measure the nominal value of Vbe at the maximum design current before doing any of the above for more consistency.