If we take your question literally then the answer would be 16 ohms BUT as others have told you your question does not make sense. This answer is interpreting your question as you have the base emitter junction of 100 PNP transistors connected in parallel and you are passing a current of 2.25 amps through these junctions in parallel and you measure a voltage of 36 volts across the junctions. The polarity cannot be forward biasing the junction as that would give a voltage of about 0.7 volts. If the junction was reversed biased then I would expect the voltage to be about 5 volts (This is a typical base emitter junction reverse breakdown voltage.) This is the reason members are trying hard to find out what your question REALLY IS.

Les.

PNP: Transistor Circuit 1 Circuit 100

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Emitter +36V DC



Base 16 OHM



Collector GND



Burn-in Board Connection: 5 sets of 3 connectors – 1 set of connectors to 20 PNP.

All 5 rows of 20 connected in parallel = 100 devices. A mete will read 16 Ohms on

Device 1 Base to Collector or 16 ohms from Device 1 Base to Device 100 Collector.

The base-to-collector resistance would be quite high. The base-to-collector capacitance would be approximately 100x that of a single transistor biased that way, which is often in the datasheet. The base-to-collector resistance is independent of current.

Did you really me impedance? Or emitter resistance re?

 

Thanks for providing clarification about your question. Your calculation is not quite right as you have not taken into account the voltage drop of about 0.7 volts (Assuming a silicon device.) across the base emitter junction of the transistors. So you have 35.3 volts across the 16 ohm resistor which will give a current of 2.20625 amps. The actual current taken from the supply will be much higher than the current through the 16 ohm resistor. As the collector current will be the base current times the hfe value for the transistor. If you look at the data sheet for your transistors you will find that that there is a big spread of hfe values for the transistor. It can be anything between the minimum and maximum value quoted on the datasheet. For example for a BC557 the hfe can be anything between 125 and 800. So with a base current of 22.0625 mA the collector current could be between 2.76 amps and 17.65 amps. (So the total current from the power supply would be between 276 amps and 1765 amps. This is a simplified explanation as the transistors would probably be operating in the saturated mode in that circuit.

Les,

 

Thanks for providing clarification about your question. Your calculation is not quite right as you have not taken into account the voltage drop of about 0.7 volts (Assuming a silicon device.) across the base emitter junction of the transistors. So you have 35.3 volts across the 16 ohm resistor which will give a current of 2.20625 amps. The actual current taken from the supply will be much higher than the current through the 16 ohm resistor. As the collector current will be the base current times the hfe value for the transistor. If you look at the data sheet for your transistors you will find that that there is a big spread of hfe values for the transistor. It can be anything between the minimum and maximum value quoted on the datasheet. For example for a BC557 the hfe can be anything between 125 and 800. So with a base current of 22.0625 mA the collector current could be between 2.76 amps and 17.65 amps. (So the total current from the power supply would be between 276 amps and 1765 amps. This is a simplified explanation as the transistors would probably be operating in the saturated mode in that circuit.

Les,

The average gain among the 100 pnp is 160 so thank you and you answered some questions for the problems we were seeing.

 

Can you now see why everyone wanted to know how your circuit was wired ? I can't see what your circuit achieves other than generating a lot of heat. (About 12.3 Kw total or 123 watts for each transistor.) If you tell us what you are trying to achieve we can probably help with a solution.

Les.

 

Can you now see why everyone wanted to know how your circuit was wired ? I can't see what your circuit achieves other than generating a lot of heat. (About 12.3 Kw total or 123 watts for each transistor.) If you tell us what you are trying to achieve we can probably help with a solution.

Les.

SSOP Testing - Steady State Operation Testing at 80% power which in this case is 800mW per device and there are 100 devices. We are

 

What is the Base to Collector resistance be of 100 pnp transistors in parallel with total power 36V DC at 2.25A

Wrong question- you need to be asking about Thermal Junction Temperature- look at your datasheet for the transistor in question. Perhaps, if you told us which transistor your interested in, we could help you decipher the datasheet.

 

I think a much better method would be to have control over the emitter to collector voltage and control the emitter current.
I would suggest applying your 36 volts between the base and collector (- to collector + to base.) You would then need to set the emitter current to 0.8/36 = 0.022 amps = 22 mA. so if you had a second supply of say 5 volts with the negative to the base and the positive to the emitter via a resistor you could set the emitter current which will be almost the same as the collector current.
As the base emitter voltage will be about 0.7 volts the voltage across the resistor will be 4.3 volts. So to get 22 mA emitter current the resistor will need to be 4.3 volts/0.022 mA = 195.5 ohms. You would need a resistor on the emitter of each transistor. In practice it would be easier to use standard value resistors of say 180 or 220 ohms and adjust the "5 volts supply" to give 22 mA through the chosen value of resistor. This method would mean that the dissipation did not vary much for the full spread of hfe values of the transistor.
Edit, I have made a slight error in the calculation. The collector emitter voltage will be 0.7 volts more than the 36 volts of the first power supply. You could re do the calculations to take that into account OR set the 36 volts down to 35.3 volts.

Les.

 

Wierd question:

#1. Why 100 transistors for 2.2Amps?
#2. Usually there would be emitter resistors for each transitor in parallel.
#3. resistance of B-E really doesn't matter. You have a base current necessary for each base, and a Vbe drop (hopefully the same or close), and an emitter resistor.
#4 You would want to match Hfe of those 100 transistors

The problem is impractical!

 

Wierd question:

#1. Why 100 transistors for 2.2Amps?
#2. Usually there would be emitter resistors for each transitor in parallel.
#3. resistance of B-E really doesn't matter. You have a base current necessary for each base, and a Vbe drop (hopefully the same or close), and an emitter resistor.
#4 You would want to match Hfe of those 100 transistors

The problem is impractical!

It is what it is - if you can't wrap your head around the given parameters just be silent.

 

You should exercise some humility

Really - ? I think that your position that your text description in post #5 is so perfectly complete and unambiguous that a schematic is not needed - in an Electrical Engineering forum (!!!) - is the height of arrogance.

Paraphrasing Rear Admiral Joshua Painter,

"Engineers don't take a dump, son, without a schematic."

From your writing, it appears that English is not your native language. Where are you located?

ak

 

Your descriptions still are not clear, but it looks like everyone is inferring that there are 100 transistors and only one single 16 ohm resistor. If so, this circuit does not have stable and predictable performance, and looks more like a classroom exercise than a mature design. It is a form of "dangle-biasing", the worst possible way to bias a transistor for linear operation.

The circuit is critically dependent on the gain of the transistors, which can vary greatly with manufacturer, production lots, temperature, etc., and from part to part within production lots. A better approach is to derive the power dissipation from some form of relatively stable voltage reference.

Also, a far more reliable approach is to use one or two small power transistors to control the current through 100 resistors.

All of which raises a question: If the 36 V source voltage is at least semi-stable, why not use nothing but resistors? There is nothing in your posts about adjustability or needing to turn the load on and off electronically. 100 resistors, 1.6 K, 2 W each, would give you 1% accuracy and basically zero drift, with over 100% thermal margin.

ak

 

the problems we were seeing.

What problems are you seeing?

With the board you have now, what is the wattage rating of the 16 ohm resistor? Also, how hot does it get?

ak

 

First pass *guess* at the load board schematic. The 4403 is a placeholder from my design library, because the actual part number is a secret. This is not a good design.

The only current through R1 is the total base current of all transistors. If they average a gain of 160, and the current in R1 is over 2 A, then the total current for the circuit will be over 320 A. With the parts shown, LT Spice puts it at 230 A; the lower value reflects the lower gain of a 2N4403. To get that, I put a 0.001 ohm resistor in series with the +36 V to measure the total current, and that dropped the voltage to the emitters about 0.25 V. In Spice-land, all transistors are perfectly exactly equal, so the base currents are distributed perfectly, as is the total power dissipation. In the real world, the power sharing among the transistors will be horrible, in the milliseconds before they burst into flames.

ak

 

This is the method I suggested in post #27.

Each transistor together with it's 195 Ω resistor forms a constant current circuit of 22 mA. The voltage between the emitter and collector will be 36.7 volts so the power dissipation is well defined. (807.4 mW) This is ignoring the small dissipation due to the base current flowing trough the base emitter junction. (About 0.01 mW) The 807.4 mW value is due to the error I made (Not taking vbe into account.) in my calculation in post #27. The calculation to get 800 mW dissipation will need to be re done to correct this error and enable the use of a standard value of emitter resistor.

Les.

 

Maybe it's a model rocket ignitor?

 

That requires an additional multi-amp power supply. You can get there without it by creating a 30-31 V -ish regulated node to sink the combined base currents, less than 25 mA. A 1/2 W zener can do it.

ak

 

Maybe it's a model rocket ignitor?

At 200 A, it can ignite a *real* rocket.

ak

 

Hi ak, I agree that that a zener could be used to supply the regulated voltage to the base. I thought that showing two power supplies would make it easier for the TS to understand as he seems to have almost no understanding of electronics. I am now starting to think this is some school homework or project as a company testing the transistors they manufactured would have more knowledgeable people.

Les.

 

My read is that he has a load board designed by someone else, and has to explain either how it works or why it doesn't work as expected. A load board is a common item in system development and production testing. We had dozens of them in various flavors - VME, VXI, cPCI, VXS, etc., plus custom ones for oddball projects. As someone above pointed out, all they do is get hot.

ak