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Basic 2N3904 question. Max Current and Duty Cycle

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mayhem2408

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I am building a project involving 96 LEDs, an ATMega328, a few 2N3904 transistors and a few other bits, to create a programmable led light pattern generator. My question is probably simple.
I know the Max current of the 2N3904 is 200mA. But is this at 100% duty cycle or does the duty cycle not matter.
Can I pull 240mA at 12.5% duty cycle, or even 480mA at 25% duty cycle. I am currently running it at 180mA at 8% duty cycle to keep it under 200mA be safe, but the more current and longer the duty, the brighter I can get the LEDs. Will the 2N3904 be able to handle it. If not, what transistor do you recommend?
 
Max current is only based on max current density handling. Power dissipation within device determines any duty cycle limits. Depending on package type it will be about 200-300 mW's max.

If device is fully saturated at 200 mA of collector current with about 0.25 vdc of collector to emitter voltage then there will be 0.25v x 0.2 amps = 50 milliwatts of dissipation. It can be ON forever at this level of power dissipation.

If not totally saturated, say having 1 vdc of Vce then there will be 1.0v x 0.2 amps or 200 mW dissipation. This is near the max allowable.
 
For good reliability, you should never exceed a Max. rating on a semicondutor.
 
In your situation, the way it works is this:
The maximum driving current into the base of the transistor is limited by the microcontroller. - 20mA
To get the transistor to drop the lowest voltage between the collector and emitter, you need to fully saturate the transistor. It does not matter if you are doing this at 100% or 1% - the transistor will perform the same . The only difference is the wattage dissipated by the transistor and if it heats up too much, the performance will drop.
But for simplicity the results will be the same.
Now put the transistor into a circuit where the current through the transistor is 200mA. Measure the voltage between the collector and emitter.
Now put 250mA through the transistor and measure the collector-emitter voltage drop.
The results will be something like this:
At 200mA the voltage c-e will be about 0.2 to 0.5v It should be in this range as this shows the transistor is fully saturating and working within its specifications and doing its job.

At 250mA the voltage may be 1.1v This is telling you the transistor cannot handle the current. The transistor will still work but the voltage across the collector-emitter leads will be higher than expected and the LEDs will not be illuminated to their fullest.
 
A 2N3904 has a max allowed collector current of 200mA. All of them work well up to 50mA but some work poorly above 50mA.

A 2N4401 has a max allowed collector current of 600mA. All of them work well up to 200mA but some work poorly above 200mA.
 
nicht kleckern, sondern klotzen!

Use a BC337. It's good for continuous 1A collector current.

Boncuk
 
Use a BC337. It's good for continuous 1A collector current.
The Philips datasheet for their BC337 has a max allowed collector current of only 500mA. Then if the base current is 50mA, its max saturation voltage is 0.7V which heats it pretty badly.
Its pins are backwards compared to American transistors.
 
I just built a test circuit with the 2N3904 for a test and ran 170mA through it for about 30 seconds and it got so hot I started to smell the plastic case cooking (well part of the smell was possibly the 1/4 watt resistor I was using). The transistor package was very hot to the touch. I know these are supposed to handle 150C but given that they don't have a way to attach a heat sink I don't think I would ever run this much current through them on a continuous basis. The 2N3904 didn't fail and handled the 170mA without problems but given how hot it got I feel the practical limit to how much current you show allow through this transistor is much less than the rated 200mA if it's going to be ON for an extended period of time.

Anyone have thought's on how hot you should allow this transistor to heat up? Unfortunately I don't have a way to measure exactly how hot it got other than my finger. It was way too hot to keep your finger on it.
 
Heating is a simple calculation caused by VOLTAGE across it times the current in it.
Its datasheet says it will be at its maximum allowed temperature when it heats with 625mW.
If it is a saturated switch with 0.5V across it then its heating is 0.5V x 170mA= 85mW and it will be barely warm. But if it has 3.5V across it then its heating is 3.5V x 170mA= 595mW and it will be almost at its maximum allowed temperature.
If it has 5V across it and a current of 170mA then its heating is 5V x 170mA= 850mW and it will probably be destroyed.

Post your schematic then we can simply calculate its temperature.
 
I used a 12V 1A power source connected to the transistor's collector with a 68ohm resistor. The emitter was grounded. I used my multi-meter to measure the current flowing from emitter to ground.
The transistor's base I connected to bench supply via a 1K ohm resistor and I slowly ramp'ed the voltage up from 0V to about 3.5V until 170mA was flowing.
I didn't realize that voltage would effect the heating (I'm a software guy by education but know enough EE to be dangerous).
 
I didn't know what this specification meant:

Total Device Dissipation
@ TA = 25°C 625mW
Derate above 25°C 5.0mW/°C

Total Device Dissipation
@ TC = 25°C 1.5W
Derate above 25°C 12mW/°C

Using Ohm's Law I can see I was running 2.1W through the transistor. Does the above mean the the max sustained current is really only 52mA? (using I = P / V) , 625 / 12
 
You did not post a schematic and you did not say how and where the 68 ohm resistor is connected.

170mA in the 68 ohms resistor produces a voltage across the resistor that is 170mA x 68 ohms= 11.56V. Then the remainder of the 12V will be 12V - 11.56v= 0.44V was across the transistor. Then the transistor heated with only 170mA x 0.44V= 75mW which is almost no heating. I suspect that you did not have a 68 ohm resistor in series with the collector of the transistor and the transistor had a current of 170mA and 12V across it and heated with 170mA x 12V= 2040mW.

The specification says that when the Ambient air temperature (TA) is 25 degrees C then the transistor's chip will be at its max allowed temperature of 150 degrees C when the transistor is heating with 625mw. If the ambient temperature is higher than 25 degrees C then the amount of heating in the transistor must be reduced by 5mW for each degree C that is higher. The spec also talks about having the temperature of the Case (TC) cooled to 25 degrees C somehow but how can you attach a heatsink??

The base voltage is about 0.8V and you used a 3.5V supply feeding a 1k resistor to the base of the transistor. Then the base current was (3.5V - 0.8V)/1k= 2.7mA which is much too low for the transistor to conduct more than maybe 50mA. The spec shows that for it to saturate (turn on well) then the base current must be 1/10th the collector current.

Why do you want to cook the transistor? If you want 2W of heat then use a resistor value of 68 ohms in series with the collector of the transistor. Then the transistor saturates with a voltage across it of 0.5V and the resistor has the remaining 11.5V, a current of 11.5V/68 ohms= 169mA and heats with 11.5V x 169mA= 1.95W. But the transistor needs a base current of 170mA/10= 17mA.
 
How about using a power logic-level MOSFET?
 
Okay, to clarify things a bit. I was running an experiment to see how much current I can run through a 2N3904 transistor. (I have several dozen of them on hand).
I know a power MOSFET is better suited for this application (I have some on order), but, I wanted to know how much current a 2N3904 will allow. The spec indicates 200mA but without any mention of voltage at this current. Max voltage for this part is 40V so I (incorrectly) assumed I could run 12V at 170mA with no problem. Apparently Power comes into play here. I am a software engineer by trade and electrical engineering is a hobby (learning experience) for me. So I don't really understand much of what is given in the datasheet. Analog circuitry is like black magic. :) LOL

The TA = 25°C 625mW spec. apparently indicates a maximum current at a given voltage. A max voltage/current graph would have been very helpful to me but I don't see anything like that in the spec.

So after the helpful responses by "audioguru" I now know that at 12V around 52mA is the max current. Correct?

P.S. I would post the schematic but I don't have a hosting site.
 
I was running an experiment to see how much current I can run through a 2N3904 transistor.
Some of them might survive 220mA. Why are you making a heater? A transistor can be used as a switch or as an amplifier, not as a heater.

I know a power MOSFET is better suited for this application (I have some on order), but, I wanted to know how much current a 2N3904 will allow. The spec indicates 200mA but without any mention of voltage at this current. Max voltage for this part is 40V so I (incorrectly) assumed I could run 12V at 170mA with no problem. Apparently Power comes into play here. I am a software engineer by trade and electrical engineering is a hobby (learning experience) for me. So I don't really understand much of what is given in the datasheet. Analog circuitry is like black magic. :) LOL

The TA = 25°C 625mW spec. apparently indicates a maximum current at a given voltage. A max voltage/current graph would have been very helpful to me but I don't see anything like that in the spec.

So after the helpful responses by "audioguru" I now know that at 12V around 52mA is the max current. Correct?

625mW is 40V at 15.6mA or 3.1V at 200mA or anything in between. You never told us what the 68 ohm resistor connects to and why you are making a transistor into a heater. I never use a transistor at its maximum allowed temperature because it will fail on a hot summer day or if it is in an enclosure or if the 12V is a car battery charging at 14.4V.

I would post the schematic but I don't have a hosting site.
We do not use another site for hosting a schematic, THIS is a hosting site. When you are typing a post see the "upload a file" box? I copy then paste parts into PAINT program then add a few lines to make a schematic, then I upload its file into a thread like this one.
 
The transistor power dissipation equals the voltage across the transistor and the current through it at the same.
So if you are switching 12V to a load and you fully turn on the transistor so it's ON voltage is no more than a volt, than 200mA through it will dissipate 200mW or less,+ which is within it's rating.
The maximum current rating basically assumes that the transistor is fully on (saturated with a base current of a least 1/10 the collector voltage).
If the transistor is not fully on, then you need to take that into account when calculating the power dissipated which, in turn, will limit the maximum current at that voltage.

Make sense?
 
I know the Max current of the 2N3904 is 200mA. But is this at 100% duty cycle or does the duty cycle not matter.
Can I pull 240mA at 12.5% duty cycle, or even 480mA at 25% duty cycle.

You can dig one hole in an hour. If you work twice as fast, can you dig the hole in 30 minutes then rest 30 minutes then dig another hole etc for the whole day?

No. Working twice as fast will KILL YOU.

That the same with a transistor. The higher current for the short duration will produce a voltage across it that will be more than twice as much as when it is operating at half the current. Thus the dissipation (heat produced) when it is operating, will be VERY VERY high and when you spread this heat over the full cycle, the transistor will be HOTTER.
 
When the transistor is fully conducting your load, measure the voltage between collector and emitter. Multiply this by the load current and Voila!, you have the power dissipation.
 
The datasheet shows that the performance of a 2N3904 is poor above 50mA and does not show any specs above 100mA. If you want higher current in a little transistor then use a 2N4401 that has a max current of 600mA and some important specs are shown at 500mA.
 
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