I need it to be at least 3W power disipation and Ic > 0.5A
Do you know a widely used transistor with those specs?
It's just for driving 20x20mA led array
Digital circuits only need 'switching' transistors. They are either fully off or fully on (saturated), so dissipate very little power at all. ULN2803 are convenient to use if you are switching several arrays.
Otherwise there are many discrete transistors that will switch 400mA. Remember to provide adequate base drive to ensure saturation.
If you really want to operate in the linear region and dissipate 3W, you choose a package that can use a heat sink.
You are aware that you will only have 3W on the transistor when, and only when you have 6V across it and 0,5A going through it? (6V*0,5A=3W)
I have not seen your design but I imagine that you are only going to use the transistor as a simple switch so you will have a very small voltage drop across it so you will have much much less power dissipated than 3W?
A quick look in the datasheet of a BC337 (in TO-92 plastic package) actually reveals that it should be able to handle it, although it will get quite hot. So I would probably prefer a transistor in a TO-220 package because of better thermal design... So look at what you have/can get in a TO-220 package, I will almost guarantee you it will stand up to the load, no matter what you choose!
A ZTX689B will have a reasonable hFE at 400mA and is in a TO92 case.
Note that a saturated transistor will only dissipate 80mW @ 400mA for a VCE(sat) of 0.2V. If you are multiplexing a display, the transistor will dissipate more power when changing state. (Like your AVR does)
Why not just use a FET ?
Some comments and clarification here, just to put it in one place:
1) Power dissipation in a switching device is NOT (Load current) * (supply voltage). It is predominantly (load current) * (saturation voltage). For a broad range of devices (both bipolar and FET), the saturation voltage for a switching device rated at a half-amp or so, is in the range of 0.3V, more or less (Darlington transistors being ignored, here). If this is true, then the peak power dissipation is more like 0.15W = 150mw, and the average is more probably less than half that if the transistor is on less than half the time.
2) Bipolar transistors tend to require a series current limiting resistor in the base when driven from a logic-level source. The value is a trade-off between switching speed + low saturation voltage and load current from the driving source. FETS tend to be easier to drive at low frequencies but require attention to gate threshold voltage; FETs typically do NOT need a series gate resistor.
3) For a given voltage and current rating and saturation voltage, FETs tend to cost more than bipolar transistors. This may be an issue in high volume applications but should be less of a consideration in one-off applications.
Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net
Hi guys, I'll be having a 430mA Ic current.
Vcc will be 10V and Vb will be digital output. I'm not using a resistor in the base, is it wrong?
So, Ve will be 5-0.7 = 4.3V
There is a 10 Ohm resistor (2W) on emitter.
I thought that P(transistor) = Ic * Vce, so, 430mA * (10-4.3) = 2.45W
Am I right?
I've found the BD169
I don't understand those two power dissipation values, 1.25 and 20 watts.
I am horrified. Surely you can just copy existing schematics? Use Google for tutorials, pay attention in class, buy a textbook ...
You always need a base resistor. You also need to be aware of the hFE. The BD169 is appalling in most respects. Most importantly, you will need a massive base drive current. The necessary current to saturate the BD169 is more than the AVR is allowed to source.
With the correct base resistance you just damage your AVR. With no resistor, you simply destroy it.
lol, I've seen the schematic on a university book. And there was no base resistor in it.
But i'll research a bit more on this topic and digital output currents.
Thanks for the tip :)
by the datasheet I see avr will need about 40mA output current in the base to allow those 430mA
So I need another transistor, for example a BC547 to drive this current on the BD139 base. Right?
Find a suitable schematic and post a link to it.
I am assuming that you are an university student.
Transistors can be connected in several different ways. When you say you are using a NPN transistor, we would guess that you are using 'common emitter'. e.g. emitter to GND, base to AVR o/p via a resistor, collector connected to LED array common. The LEDs have their own resistors.
However the conventional multiplexing arrangement has a PNP transistor sourcing the common anodes, and AVR pins sinking each row cathode via a resistor.
Google is your friend. Copy an existing design.
Yes. If the hFE is too low, you add another transistor. But it is simpler to choose a transistor with an hFE of 300 in the first place. Or to use an FET (with care) in the first place.
The 'common collector' (emitter follower) topology will not need a base resistor, but the load is between emitter and GND. You still need a limiting resistor, and the topology does not really suit a 5V AVR and you get a larger VCE.
p.s. a common-cathode LED array will use an NPN to sink cathodes and AVR pins to source each row anode. No way can his use a 12V supply.
I would recommend to make a circuit similar to this
I googled this one, so you will need different resistor values and probably another transistor, and of course you will supply 10V (like you said) instead of 5V...
With a setup like this and everything dimmensioned correctly you should have a Vce of somewhere between 0.1V and 1V depending on which transistor you use.
Look for Vce(sat) - Voltage Collector-Emitter saturation - in your transistor datasheet to find what kind of voltage you will have across it.
Voltage * current = power... And if you have a good datasheet you should be able to find values for "temperature junction to ambient" in Â°C/W, from there it should be quite easy to figure out what kind of temperature your transistor will be operating at (ie. will it burn or live).
You need the resistor! So look for the Hfe and make sure you pick a transistor where you will have a base current of less than 20mA (maximum of the AVR). The higher the Hfe the better.
You COULD use another transistor to drive the base of your LED driving transistor if the Hfe is not large enough This is called a "darlington pair" - google it..! I'm not an expert on darlingtons or transistor couplings.
But I would recommend getting the right transistor from the start, if you are at a university they should have a large rack of different transistors, at least they have that around here!
If you have no base resistor the transistor will blow on the first or if you are lucky the second time you try to switch it on. the Base-emmitter junction is nothing more then a diode, so you will be trying to put AVR_VCC over a diode with a drop of 0,8V max. This WILL smoke your transistor ( if it does not literally explode) almost immediately. Also between the base resistor and the base connection of the transistor I suggest placing another resistor to ground(emmiter) to make sure that the transistor is OFF during the reset stage of the AVR. Else your LED will flicker very shortly when power is applied
I used a resistor on emitter (emitter polarization), that's why the transistor didn't blow up.
But still got very hot.
So tried Geronimo's suggestion and it's working great now :)
© 2020 Microchip Technology Inc.