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Topic: TLC5940 + high current (Read 11 times) previous topic - next topic

ill_switch

Mike, help me through this please! My knowledge is limited in this area.

From the datasheet, it looks like the 28-PDIP package has a max dissipation of about 2.4w at 25C.

Now, I if I want all 16 channels at 120mA, and the chip is running at 5V, then I can calculate the dissipation:

16 channels at 5v and 120mA is 16*5*.12 = 9.6 watts

But that doesn't seem right, does it? The chip isn't dissipating 9.6w - the whole circuit is, correct? How do you calculate dissipation of the chip?

Grumpy_Mike

#26
Sep 23, 2009, 04:16 pm Last Edit: Sep 23, 2009, 04:17 pm by Grumpy_Mike Reason: 1
Page 13 of the data sheet has the full formula for the power calculations, but basically it is the voltage across something times the current through it. In this case the something is the driver outputs and when pulled down are a quite low voltage but not zero. Unfortunately I can't find this in the data sheet so you will have to just use the formula they give.

Have you seen my tutorial on power calculations:- http://www.thebox.myzen.co.uk/Tutorial/Power.html
You can plug the package thermal properties into that.

ill_switch

Ok, thanks. It looks like my thought of basically having all 16 pins at 5v and 120mA is not possible.

lqbert

I basically want to hook up 4 LEDs (20mA/LED) to each channel of 6 TLC5940s.  I also want to be able to make the LEDs go completely dark.  

I don't understand exactly what I need to do to keep from frying my TLC5940s (i.e. keeping the output voltage within spec. yada, yada, yada).

Would using Rocketgeek's P-channel MOSFET be the way to go for me (I'm hoping he answers)?  

If so, I could use some help picking the right size components for this setup (and understand why... I actually would like to learn, not just mooch) and a simple diagram would be MOST awesome, as wel!!

Thanks in advance for any responses.... they are greatly appreciated!

lq_bert

rocketgeek

lqbert: Got your message -- been distracted by life. I put the circuit described together in order to fade a bunch of common-cathode RGB LEDs. The first thing you should do is go read International Rectifier's excellent intro to power mosfets, AN-1084 Power MOSFET Basics (http://www.irf.com/technical-info/appnotes/an-1084.pdf). It covers entirely N-channel MOSFETs, but it's a good intro and P-channels are mostly the same with negative signs on most of the parameters. Then check out AN-940 (http://www.irf.com/technical-info/appnotes/an-940.pdf) for more on how to handle P-channel MOSFETs. The rest of this message is going to assume you've read them.

You want to select a mosfet with the minimum gate charge (Qg) that can handle the supply voltage with a decent safety factor. 80 mA is so laughably little current for a modern power MOSFET that you probably don't need to worry about that when making your selection.

Now, the easy way to drive the gate of the mosfet off is to connect it, through a protection resistor, to the sink pin of the TLC5940. That will turn the MOSFET on when then pin starts sinking. You then need a pull-up resistor between the gate and the power supply to turn the MOSFET off, as well.

Sizing the gate resistor is important because all of the pins of the TLC5940 turn on at the same time. Therefore, the turn-on current spike of all of the MOSFETs gets soaked up by the TLC5940 at once. Luckily, it's very short.

You want to size the pull-up such that it turns off the MOSFET quickly. However, making it smaller (in order to speed turnoff) has two significant negative consequences. One, it reduces the amount by which you can pull down the gate voltage, since it forms a voltage divider with the gate resistor. The farther you can pull that voltage down, the better the MOSFET conducts while on. Second, you're drawing current through the gate and pull-up resistors in series while that pin is on. As long as you can pull down at least five volts, you should be fine to turn it on.

Keep the on current in mind tho; you want to time-average that with the turn-on current in order to determine the average current through each channel. That will give you the input to the power dissipation equation.

Clear as mud?

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