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tip120 driving led array

darren adcock

Sep 26, 2016
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HI all.

I would like some help with a circuit please. I have discalcula so please bare with me.

I have made an array of 120 5mm U.V Led's for an exposure tank. I have divided the Led's into groups of 4 and have a 47ohm resistor before each group. I am taking power from the emitter of a Tip120. The Tip120 is being switched on by a LDR. The LDR is controlled by a 555 monostable timer circuit.

The TIP120 gets hot. I have tried to compensate for this by adding a large heatsink.

My question is is the TIP120 able to drive this many LED's? the forward current on each LED is 20ma. I can't quite get the numbers to make sense to be able to calculate what's required. I did think I could just divide the array into to two with two TIP120's, but then thought I could probably put the TIP120's in series to compensate.

Anyhow. Am a bit out of my depth here so looking for advice.

Best

Darren
 

kellys_eye

Jun 25, 2010
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It boils down to the saturation voltage across the transistors e-c connections. BJTs aren't that good as switches in high-power applications so consider changing it for a MOSFET that can have an 'on' resistance of milli-ohms with a consequent dissipation of near-to-nothing.
 

darren adcock

Sep 26, 2016
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Ok, so after researching I've found the irf540 mosfet. This a good choice?
 

kellys_eye

Jun 25, 2010
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Indeed it is - quite common, up to the job too. Connect as per a transistor but add a resistor from gate to ground to help with switching off.

As ever, always download the datasheet for the devices you are using. It will show you all the parameters and safe operating zones etc.

There are plenty of examples on t'internet of using the IRF540 as a switch - the main factor being the drive voltage to switch it on. Anything over 5V and you're in the ball park.
 

BobK

Jan 5, 2010
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First things first, show us your cicuit. We cannot devine how you connected it.

Bob
 

duke37

Jan 9, 2011
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If you are taking the output from the emitter, then the transistor will not be fully on unless you have a subsidiary higher voltage supply to drive the base or gate.

Ground the emitter/source and use the collector/drain to feed the LEDs. Then a voltage is available above the ground.to drive the transistor inputs.
 

duke37

Jan 9, 2011
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1. You will be able to put two LEDs in series and perhaps three. This will reduce the total current by a factor of two or three. Each leg should have its own resistor so that all legs have the same current independent of the temperature. An overall resistor is not necessary.

2. Put the LEDs in the collector circuit so there is sufficient voltage to drive the base.
 

BobK

Jan 5, 2010
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This is why I wanted to see a circuit.

1. Most beginners connect the load to the emitter of an NPN transistor, as you have. This makes an emitter follower, where the emitter is at a voltage of about 0.6 (or 1.2 for a darlington) less than the voltage driving the base. When you put the load on the collector, and the emitter is connected directly to ground, the transistor can reach saturation of about 0.2V. This has a huge effect on the amount of power dissipated in the transistor.

In your case, with 120 LEDs drawing 20mA each, there is a current of 2.4A. Let's say your LDR divider is producing 6V to drive the base (which is probably better than it actually is.) Then the emitter voltage will be 6 - 1.2 = 5.2V. So there is 9 - 5.2 = 3.8V across the transistor. At 2.4A that is 9W. No wonder your transistor is getting hot.

2. If you had the load on the collector side, the voltage across the transistor will be about 2V. Using a non-Darlington could reduce this to 0.2V.

3. As duke37 said, you can put 2 LEDs in series (and use a resistor for each pair,) and you current will be halved.

4. The combination of 2 and 3 could reduce your power consumption in the transistor to 0.2V x 1.2A or 0.24W, which would hardly get warm without any heat sink. A MOSFET would do even better.

Bob
 

Audioguru

Sep 24, 2016
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An LED part number has a range of forward voltage. Each LED will probably have a different voltage. You have LEDs connected directly in parallel so the one with the lowest forward voltage will hog most of the current and soon burn out. To prevent the burn out you must measure each LED and match the groups of LEDs. Or use a current-limiting resistor for each series group of LEDs so that no LEDs are directly in parallel.
 

darren adcock

Sep 26, 2016
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Thanks Bob and duke and Kelly's eye. This has been a really helpful learning curve.

I got impatient waiting for a irf540 to arrive so went through my circuit junk box and found an 80NF70 with a large heat sink. I got this all wired in yesterday and it seems much happier.

Initially I was connecting the output of a monostable 555 timer circuit to an npn base to use that as a switch. I ended up with a whole load of burnt out 555's. Lesson learned
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Initially I was connecting the output of a monostable 555 timer circuit to an npn base to use that as a switch. I ended up with a whole load of burnt out 555's. Lesson learned

Yes, you need some resistance between the signal source and the supply rail the emitter is connected to.

In your initial configuration with the load connected between the emitter and ground (of an NPN transistor), the load itself is that resistance. However this results in the transistor not switching on fully and (as you found) this causes it to get hotter than it should.

If you tie the emitter to ground and place your load between the collector and the positive power supply rail, you need a resistor in series with the base of the transistor to limit the current (or you may damage the signal source (the 555) or the transistor). In your case you found that the 555 was the weak link.

Deciding the correct value for the resistor is a little tricky. Something between 100Ω and 1kΩ will probably be OK for this transistor in this circuit (it also depends on the size of the load and the magnitude of the signal).

In our resources section we have one on LEDS, one on using transistors to drive a load, one on saturation in transistors, and finally one on "digital" temperature measurements that may be useful for you.
 

darren adcock

Sep 26, 2016
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Ooooo thanks Steve. That's really helpful. I usually stick to low power audio circuits but like a new challenge and I've found it's all relevent.

I wll check out the resources section, I didn't realise there was one.

This has also made me realise i need to pay more attention to temperature of components. When the 555 burned out it would short causing a smoking potentiometer, took me a while to figure this out. A way of checking temperature would have been useful here.

Best

Darren
 

darren adcock

Sep 26, 2016
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Now I've got it working ok. I need to check it's safe (components anren't going to burn up). How do I do this please? As Bob says it's drawing 2.4amps. Do I need a power supply that can deliver this?

Best

Darren
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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If the circuit requires 2.4A then I would be looking for a power supply capable of at least 3A (maybe more if it will be on continuously).
 

darren adcock

Sep 26, 2016
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Ok thanks Steve. It has a zero - three minute timer. I will get ordering a power supply today. Just out of curiosity was it really neccesary to use U.V led's for the exposure unit? I just had loads of em on hand so was convenient, but also thought that this might be best anyhow.
 
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