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Topic: amplifying the output channels of a TLC5940 (Read 1 time) previous topic - next topic

davivid

I need to amplify the output channels of a TLC5940 upto 12v @ 5A.

Are there any off the shelf products that will do this?

dc42

You could use one TIP107 PNP darlington pair per 5940 output pin. Configure the 5940 to sink about 20mA per output. The TIP107s will drop up to about 2 volts, so if the load needs 12v then you may need a 14v supply. If it is for driving LEDs then you will need a series resistor or resistors between the collector and the LED, unless the LED modules are already designed to work from 12vDC.
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davivid

Ok thanks. Yes for driving lengths of LED strips, rated at 12v with a max current of 5amps.

Similar to this:
http://www.ledlightszone.co.uk/led-tape/55-flexible-led-strips-non-waterproof-72w-30pcs.html

Since health and saftey may be an issue - are there products/modules already designed for this?


davivid

#3
Aug 24, 2011, 11:26 am Last Edit: Aug 24, 2011, 01:57 pm by davivid Reason: 1
Do I simply supply ~14v to the Emitter, connect the output of the TLC5940 to the Base, and ground the LED from the Collector?

Also how would I best compensate if the power supply is 15v?

CrossRoads

Check the TLC5940 datasheet - I believe it needs to sink at least 5mA on the output pins to operate stably.
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dc42


Do I simply supply ~14v to the Emitter, connect the output of the TLC5940 to the Base, and ground the LED from the Collector?


+14v to the emitter, TLC5940 output to base, collector to +ve side of LED module, -ve side of LED module to GND (i.e. Arduino GND and the -ve side of the 14v supply). The outputs of the TLC5940 are current limited, so you don't need the usual series resistor to the base.

Also how would I best compensate if the power supply is 15v?


That depends on the tolerance of your 15v supply vs. the tolerance of your LED modules. If the LED modules can accept whatever variation you might have in the 15v supply, you can just use a resistor in series with each module.
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Grumpy_Mike

The output of the TLC5940 is only 5V, that is not enough to hold a PNP transistor off, you need another transistor to make it that big.
However, note that if you invert the output of the TLC5940 you will never be able to turn off the LEDs fully. This is because the TLC5940 never turns on the LED fully only 4095 out of 4096 clocks. Of course when it is on you can't see it but on for one cycle in 4096 can be seen.

dc42

#7
Aug 24, 2011, 11:47 pm Last Edit: Aug 24, 2011, 11:51 pm by dc42 Reason: 1

The output of the TLC5940 is only 5V, that is not enough to hold a PNP transistor off


Mike, I didn't think I would ever have reason to contradict your advice because it is usually so good. However, if you look at the data sheet, you will see that the TLC5940 LED output pins are constant-current sinks rated up to +17v. So connecting each output to the base of a PNP darlington whose emitter is held at +12 to +15v is fine.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

dc42

#8
Aug 25, 2011, 12:35 am Last Edit: Aug 25, 2011, 12:37 am by dc42 Reason: 1
I just realised I need to amend my advice. If you drive 1 output of the 5940 at 20mA into the base of a TIP107 with the emitter at around 14v, the 5940 will dissipate about 250mW for that pin. If you drive all 16 at once, that's 4W, which is much too high for the device. So, to reduce the power dissipation to a reasonable level, I suggest:

1. Reduce the output current per pin to 10mA - that should still be enough to drive 5A per TIP107. This reduces the power dissipation to about 125mW per pin.

2. If you intend to drive more than about 4 channels at once, insert a 1K resistor in series with the base of each TIP107. The resistor will dissipate 100mW of the 125mW, so the 5940 now dissipates only 25mW or so per pin.
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davivid

great stuff thanks.


2. If you intend to drive more than about 4 channels at once, insert a 1K resistor in series with the base of each TIP107. The resistor will dissipate 100mW of the 125mW, so the 5940 now dissipates only 25mW or so per pin.


I intend to drive all 16 channels at once, with 5A being more than required (I estimate most channels will need 2-4A). By inserting a 1k resistor in series with the base of each TIP107, will this not drastically lower the TIP107 output?

dc42

#10
Aug 25, 2011, 04:12 pm Last Edit: Aug 26, 2011, 12:08 pm by dc42 Reason: 1

I intend to drive all 16 channels at once, with 5A being more than required (I estimate most channels will need 2-4A). By inserting a 1k resistor in series with the base of each TIP107, will this not drastically lower the TIP107 output?


The TIP107 (or TIP105 or TIP106) has a minimum current gain of 1000 at 3A, and 200 at 8A. Therefore the minimim drive current needed for a 3A load is 3mA. I'm suggesting you use 10mA drive current, which should be enough for 5A with some margin. At 10mA the 1K resistors will drop 10v, but since you have around 12v available, they won't reduce the drive current - but they will dissipate most of the power that otherwise the 5940 would have to handle.

The other possibility I considered was to use P-channel MOSFETs instead of the TIP107s. You could connect each output pin of the 5940 to the gate of a IRF9540 mosfet and through a 1K resistor to +12v, connect the source of the mosfet to +12v, and the drain to the +ve side of the LED module. The advantage of using mosfets is that they drop much less voltage than darlingtons when on, so they run cooler and you only need 12v rather than 14v or so. The disadvantage is that with 1K driving resistance, they will take several microseconds to switch, which is OK if your PWM rate is not too high, but might otherwise cause them to get hot after all. You can avoid this using either a mosfet driver ic or a pair of small-signal transistors per mosfet, but this adds complexity.
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Grumpy_Mike

Quote
I didn't think I would ever have reason to contradict your advice

Yes sorry I wasn't thinking straight. :smiley-red: I forgot about the higher that Vcc input capability (it is quite rare)

I would also put a resistor from the emitter to the base to ensure the transistor (or FET) is turned off.

If you are using a FET the there is no need for a series resistor at all because any gate charge current excess is controlled by the constant current output. Note that with the constant current output you will not get 20mA (or what ever you set) if the load and voltage will not produce it. Instead it will just be turned hard on to ground.

dc42

#12
Aug 25, 2011, 10:30 pm Last Edit: Aug 26, 2011, 12:10 pm by dc42 Reason: 1

I would also put a resistor from the emitter to the base to ensure the transistor (or FET) is turned off.


The TIP107 darlington pair I recommended already has these built in, that's why I didn't suggest it. In the case of the mosfet circuit I mentioned, that function is served by the 1K resistor.
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davivid

ace.

I will order in booth parts and experiment. Hopefully the TIP107 will work out for simplicity. Cheers.

dc42

Bear in mind that the TIP107s will need a big heatsink (5A times about 2v = 10W). The mosfets will be ok with a smaller heatsink (they should dissipate less than 5W when on) unless the switching time is a significant proportion of the total time. If you haven't yet ordered the mosfets, I see that the IRF9Z34N has a slightly lower on resistance than the IRF9540, has a lower gate charge, and is also slightly cheaper - so better all round.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

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