Unexpected Transistor Behavior - Page 1
Forum section: electronics / beginners

author: MarkKn
author_rank: Contributor
timestamp: October 25, 2022, 04:55:17 pm
content: Being a diligent student of matters electronic, I decided to take some Ib versus Ic on a transistor and calculate its beta and find the various regions. I got a motorola MJ10042 on ebay which is described as a power darlington transistor (NPN) designed for high voltage switching.I hooked it up, and rounding off the voltages a bit, at 8ma base current the collector current was in the microamps. I increase it and when it gets to around 8.9 v volts the transistor switches on--going up a tenth of a ma at a time, there was no linear region at that resolution, it just switched on. Then as I dialed the base voltage back down, the transistor stayed on until I got back down to around 8 ma.Given my limited knowledge of bjt, I was surprised that once it was 'switched on', it was kind of sticky and the shut off base current was almost 1ma lower than the turn on voltage--its kind of sticky. I can understand in general that a switching transistor does not have much of a linear region. Essentially its base to collector current relationship changed once it was full on. I will do some more reading to get the proper terminology for the transistor regions of operation.Its also possible that this transistor is damaged--it is surplus from somewhere.Does anyone have any thoughts or observations. Thanks in advance.

author: Vovk_Z
author_rank: Super Contributor
timestamp: October 25, 2022, 06:19:36 pm
content: You better begin with the usual BJT but not Darlington and get it from a reputable supplier. Otherwise, it is an equation with too many unknowns.

author: MarkKn
author_rank: Contributor
timestamp: October 25, 2022, 06:34:11 pm
content: Yes, I can see that is a good way to look at it. Thanks for the insight. It was interesting to see its behavior, but not a good place to take measurements and calculate hfe.

author: srb1954
author_rank: Super Contributor
timestamp: October 25, 2022, 07:22:21 pm
content: Quote from: MarkKn on October 25, 2022, 04:55:17 pmA switching BJT is designed and characterised to switch rapidly between the on and off states but otherwise it doesn't differ markedly from any other BJT in terms of the shape of its on-off behaviour. What is different with the MJ10042 is that it has some very low value internal base-emitter resistors, which are there to ensure that the transistor switches off quickly when operating at high voltages and currents.The internal resistors have the effect that, as you increase the base current from zero, most of the base current initially flows through the base-emitter resistor and the first transistor doesn't start conducting appreciably until there is sufficient voltage (>0.5V) developed across its base emitter resistor to start turning the first transistor on. The effect of the input current being shunted away by the base-emitter resistors is to make the effective hFEof the transistor appear very poor at low currents. If you look at the hFEgraphs in the data sheet you will see this effect. In fact they don't even provide any information on the transistor at collector currents operating below 0.5A as no-one would be interested in using such a big and expensive transistor down in that region.

author: MrAl
author_rank: Super Contributor
timestamp: October 25, 2022, 08:03:55 pm
content: Quote from: MarkKn on October 25, 2022, 04:55:17 pmWhat you need to do if you havent already is install your own base resistor Rb. Even 100 Ohms is better than nothing. Then you can measure the current into the base with Ohm's Law and where Vs is your applied voltage:I=(Vs-Vbe)/Rb (Vbe can be measured or estimated to be around 0.7 volts)Usually you dont apply a voltage to the base you apply a current unless the emitter has a resistor to ground. That's because the base emitter voltage is very short ranged so when you increase the voltage by even a little the collector current goes up by a lot. It would be hard to adjust it just right. If you use a base resistor though (in series with your power input supply voltage) you will see a much smoother transition. That's provided it is mainly just two transistors in that package and just some resistors.Very often an emitter resistor is used also and that makes it even more linear. This resistor value is usually low like 1 Ohm to 100 Ohms, you can start with 10 Ohms.Typically the apparent Beta is much less than the normal Beta when the transistor gets near saturation. Saturation causes the collector base 'diode' to conduct and that steals current from the base, making the Beta look lower. This could occur at around 0.7v collector to emitter voltage, and will get worse with even lower CE voltage. If you happen to see a CE voltage of maybe 0.2 volts, the Beta will be very low unless you have a special bipolar that is made as a higher grade than the typical. The Beta will always decrease in that condition though.If you draw a picture of your circuit (schematic) it will be easier to tell what is not working right.

author: strawberry
author_rank: Super Contributor
timestamp: October 25, 2022, 08:15:23 pm
content: colector current might be too low for 20A BJTand BE internal resistor is ~50+8 ohms (22mA@1.3Vbe) first BJT barely turned on and second transistor too low current gain to turn it on with 10mA

author: MarkKn
author_rank: Contributor
timestamp: October 26, 2022, 06:32:15 pm
content: Thanks, all, that is some very interesting information I will dig into. When I saw the resistors and diode on the data sheet for this device, I didn't really think about how they would impact the behavior of the device. The spec sez it can switch 850 volts and 37 amps, if properly heat sinked--and if it is really up to spec.My 32 volt power supply does not appear to be able to get the device into its normal operating region, but I was able to switch it on and that was interesting.

author: Picuino
author_rank: Super Contributor
timestamp: October 26, 2022, 07:11:20 pm
content: Datasheet

author: MarkKn
author_rank: Contributor
timestamp: November 28, 2022, 03:47:52 pm
content: Just a follow-up. I performed some measurements on an MJ15003, a straight-up NPN transistor, not a darlington. I got some results that more closely resembled a the cutoff, active, and saturation regions. I only put 15 volts on it through a 1k resistor. I noted what I consider unexpected resuts but within the realm of reason. The device was an open circuit with .388 volts/.83ua B to E. It appeared so be saturated at .591 volts/463 ua, with Vce of .18 volts. In the active region I noted a beta of 52, which is within spec. FYI--I used a 10k resistor in series with the base and the power supply to bias the base.Now the voltage was quite low as the device is rated for 140 volts Vce, so maybe it would be closer to spec with its normal environment. I was expecting the transistor to be cut off until .7 volts, but by then it was in saturation. Whatever the case, I was glad to see that some actual measurements matched the theoretical ones to some degree. Thanks all for the suggestions up to this point.

author: fourfathom
author_rank: Super Contributor
timestamp: November 28, 2022, 04:21:26 pm
content: Quote from: MarkKn on November 28, 2022, 03:47:52 pmThat typical 0.6V to 0.7V BE threshold is very loose, and depends greatly on the transistor and the temperature. What you measured is quite typical. In small-signal, room-temperature work 0.6V is probably a better general approximation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 28, 2022, 04:34:39 pm
content: "0.7V" is just a ballpark figure. Power transistors have massive huge junctions, shifting your tiny signal currents into very low current densities and therefore low Vbe. Up around typical levels, you'll see, well, typical voltages, but also higher than "usual" because of the additional resistance -- the base and emitter layers are more or less normal thickness, but they span a wider die, meaning relatively more ESR in these electrodes. They do deal with this by interleaving connections, and probably using thicker metallization, but the distance from metal to furthest points of the base layer are still longer on most types. Which is also easily seen by the much lower fT (which in part manifests due to base resistance into junction capacitance). So you can have Vbe(on) in the 1.2V range.Tim

author: Picuino
author_rank: Super Contributor
timestamp: November 28, 2022, 04:52:41 pm
content: You need 0.65V and 13mA on the base pin for the first transistor to start working.Then you will need about 1.3V at the base and about 14mA for the second transistor to start working.From here, any increase in base current or any voltage rise at the base will translate into a large current increase at the collector.You should analyze the circuit to understand it better.