[SOLVED] 2N1203 - little package but powerful ?

Hey, i found this transistor and it says "I am P120V,3,5A,34W,Ge".
My question is how much of this 34W i can get from it without getting it warm ?

8.5W ? This seems unrealistic and strange. Look this transistor is so tiny..



Bonus question -> how much power i can get with 90W transistor and a big heatsink. 45W ? or 75% ?
another bonus one -> how much power i can get from BD139 with some heatsink ?
another bonus one -> how much power i can get from 625mW transistor without heatsink ?

This is so confusing.

There is some mistake with your picture. The 2N1203 is a bulky stud type old germanium power transistor.

The printing on the thing says 1203 it DOES NOT say 2N1203. Maybe it was made in 1991 or maybe it was the 91st one ever made.

OK, thanks. About the other questions, bonus ones ?

The power ratings of a transistor have little to do with how much power you can "get out of it". You assumption is wrong that the devices are designed to produce power such as in the case of an audio or RF amplifier, please remember that the majority of transistor applications are for signal or power switching, oscillators or low level amplification.

The voltage ratings are to tell the designer the maximum they can allow across the pins without damage occuring and the current ratings are how much can be allowed to flow through it. Power = Voltage x Current but as you can see from the data sheet, there is more to it than saying a devices power rating is the maximum voltage times the maximum current as that ignores the amount of heat it would have to dissipate.

So your questions are impossible to answer based on the information you give and the obviously wrong data sheet for that transistor.

If you are talking about audio power output from an amplifier using one of the devices, it still depends very much on the amplifier configuration but if I assume a single ended class A stage driving a low impedance loudspeaker, these would be 'very rough' guesses at highest output you might get:

90W transistor and big heat sink = about 15W
BD139 on a big heat sink = about 5 W
625mW transistor on a big heat sink = 0.2W

I stress though, the power rating of a device is not a measure of how much power it can produce if used as an amplifier.

Brian.

Mnt said:

1) how much power i can get with 90W transistor and a big heatsink. 45W ? or 75% ?
2) how much power i can get from BD139 with some heatsink ?
3) how much power i can get from 625mW transistor without heatsink ?

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These power ratings are heating Watts, not output Watts. The transistor creates heat that must be dissipated away. The amount of heat is the voltage across the transistor multiplied by the current through it. The datasheet says 90W when its case is held to 25 degrees C somehow. The BD139 is allowed to heat with 8W when its case is held to no more than 70 degrees C somehow. The 625mW transistor is usually used without a heatsink so it must not exceed its maximum allowed temperature.

Im not interested in output power.
Here is an example. I want to get the maximum for this BD without a heatsink and also with a small heatsink, not big. same for Q1 but only without heatsink. These watmeter outs are incorrect because there are signal. Correct ones are 50mW and 340mW.
Now for power transistors - I need them to dissipate 37W and want to do this with only one pair of them.

Your Multisim schematic is covered in chicken pox dots and has many wires zig-zagging all over the place. Its many meters have many extra wires and are covering some of the wires and parts.
I see only one BD140 medium power transistor but I do not see anything that will make it hot. The datasheet for a BD139 (your BD140 is probably the same) says that if it is mounted on a "perfect" heatsink (that does not get even slightly warmer than the surrounding 25 degrees C air) then it will be at its maximum rated temperature (not recommended) with 12 Watts. With a pretty big heatsink maybe you should limit the heating to 8 Watts.

I heard to use 1/4 for 625mw transistor. (150mW)
Is this true for BDs ? (313mW).
Im not sure about this, thats why Im asking.
Also i dont know what is the maximum with a very small heatsinking for the bd.
Datasheet says nothing.

A 625mW transistor will be at its maximum rated temperature when dissipating 625mW in 25 degrees C air. I would use it to maybe 470mW (3/4 of its max temperature).
A BD139 or BD140 needs a pretty big heatsink then can dissipate 8W. The datasheet says its thermal resistance from the ambient to its chip is 100 degrees C per Watt so without a heatsink then the chip will be at its max rated temperature with 1.2W when the ambient air is 30 degrees C. I would use it to 0.9W.

To use a heatsink then the datasheet says the thermal resistance from its chip to its mounting base is 10 degrees C per watt, (thermal grease is an additional 0.5 degrees C).

Ok, thanks.

Audioguru said:

A 625mW transistor will be at its maximum rated temperature when dissipating 625mW in 25 degrees C air. I would use it to maybe 470mW (3/4 of its max temperature).
A BD139 or BD140 needs a pretty big heatsink then can dissipate 8W. The datasheet says its thermal resistance from the ambient to its chip is 100 degrees C per Watt so without a heatsink then the chip will be at its max rated temperature with 1.2W when the ambient air is 30 degrees C. I would use it to 0.9W.

To use a heatsink then the datasheet says the thermal resistance from its chip to its mounting base is 10 degrees C per watt, (thermal grease is an additional 0.5 degrees C).

Click to expand...

This is not true, a BD gets very hot at 500mW Pd.

This is not true, a BD gets very hot at 500mW Pd.

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The rated power of 1.2 W without heatsink involves a junction temperature of 150 °C and about 140°C case temperature. The hand calculation also gives > 70 °C case temperature for 0.5 W. If you touch it, you'll still find it "very hot". So what's not true, particularly?

I want it only warm because of thermal runaway.

All I want to show is that the problem can be calculated, no need for guessing. 1/4 of rated power corresponds to about 30 degree of junction temperature raise, respectively 57 mV Vbe variation, or nearly factor 10 increase of collector current.

Add an emitter resistor then thermal runaway will be reduced a lot.

Mnt said:

Bonus question -> how much power i can get with 90W transistor and a big heatsink. 45W ? or 75% ?
another bonus one -> how much power i can get from BD139 with some heatsink ?
another bonus one -> how much power i can get from 625mW transistor without heatsink ?

This is so confusing.

Click to expand...

It clearly depends on the job it is being asked to do. A simple switch does not dissipate much power but can control (or deliver) lot of power.

The power you are getting from the heatsink is *useless* power- simple dissipated heat- good only for heating the room in winter.

Even while the transistor is working in the safe operating area (in terms of the current and voltage), the energy dissipation depends on the exact point (value of current and voltage).

Your question is badly asked. That is perhaps the real confusion...

The American products datasheet tells you everything you need to know but an Oriental datasheet might have some details lost in the translation. The TIP35 by Texas Instruments tells you it can dissipate 125W if its case is held to 25 degrees C by a perfect heatsink that does not exist. Maybe with a huge heatsink with a high speed fan and air conditioning.

The chip inside the transistor must not exceed its maximum listed temperature but its case must be cooler by the amount of thermal resistance listed. The heatsink must be cooler than the case because there is a small gap. The air must be cooler than the heatsink. They all have a thermal resistance then the amount of heatsink is calculated and a heatsink is found with the results of your calculation.