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Topic: TLC5940 + high current (Read 12022 times) previous topic - next topic

lqbert

Rocketgeek:

Thanks for the reply... I'm a little bit fuzzy on para 6 of your reply.  I've included a diagram (thanks original poster) of an P-Channel MOSFET which has what I think was the config you described.  What I don't fully understand is what values to use for resistors 'A' and 'B' in the drawing.

If there was a formula included in the documentation you provided to come up with 'A' & 'B', I'm sorry I must not have understood what I was looking at.

Also, what value should be used for the TLC's constant current resistor?  How does this affect the circuit.

And just to clarify, with your configuration suggestion, will I be able to make the connected LEDs go completely dark?

Drawing below:

Code: [Select]


                         5v           18v
                         |             |
                          |            0.9k
                         |             |
                        'B'k        LEDs (20mA)
                         |            /
                         |           S
OUTx -----'A'k----------------------G||
                                     D
                                      \
                                       |
                                       |
                                      GND






Thanks!!!!

rocketgeek

lqbert: with a P-channel MOSFET, the load goes between the MOSFET and the ground rather than how you have it shown. Otherwise, good.

First, determine the gate voltage you need. This will give you the ratio between A & B, since they form a voltage divider. Since your supply voltage is 18V and -9V below supply is well into saturation for any reasonable MOSFET, you can just set them equal to each other. That's easy, and it means you'll switch at similar speeds on and off.

Second, you need to determine the size of the resistors required to keep from toasting the TL5940. Since it's a constant current output, I don't think you have to worry too much about toasting it during turn-on -- just don't force it to limit the current too much during the on period, since that dissipates heat. If you set your peak current to 20 mA, then V=IR gives you a total resistance of 900 Ohms... or a pair of 470 Ohm resistors (since that's a size that exists).


lqbert

Quote
with a P-channel MOSFET, the load goes between the MOSFET and the ground rather than how you have it shown. Otherwise, good.

Did you mean that I need to just swap where the 18V and GND are, or should I swap everything that's connected to the drain with everything that's connected to the source?

Quote
First, determine the gate voltage you need. This will give you the ratio between A & B, since they form a voltage divider. Since your supply voltage is 18V and -9V below supply is well into saturation for any reasonable MOSFET, you can just set them equal to each other.

So, basically what I want is enough voltage to put the MOSFET into saturation, and to do this, I have to supply some % of the supply voltage (18V in my case)?  Makes sense enough.

Quote
Second, you need to determine the size of the resistors required to keep from toasting the TL5940. Since it's a constant current output, I don't think you have to worry too much about toasting it during turn-on -- just don't force it to limit the current too much during the on period, since that dissipates heat. If you set your peak current to 20 mA, then V=IR gives you a total resistance of 900 Ohms...

I'm assuming you were referring to setting resistor A and B to be equal to 470 Ohms each, and set the current limiting resistor of the TLC to 2*470 Ohms?

... and finally this configuration will also be able to make the LEDs go completely dark?

Thanks so much, and sorry for the late reply... I'm in the process of purchasing a new truck... I hate car shopping.  :'(

Slisgrinder

#33
Nov 13, 2009, 06:19 am Last Edit: Nov 13, 2009, 06:21 am by slisgrinder Reason: 1
It should be 18v --> Source --> Drain --> LEDs --> resistor --> GND

Hope that made sense...

And yes, I am confused about how to get the values for resistor "A" and "B" as well  :-/...
Making the big brother (sort of) of Arduino Mega...with included goodies and awesomeness and the extra, coolness...

Grumpy_Mike

Quote
I am confused about how to get the values for resistor "A" and "B" as well

Resistor A is not needed at all, current through the TLC will be limited by resistor B.
Resistor B is acting as a pull down resistor for the FET and acting as a load for the TLC, virtually any value will do from 1K to 100K.
The current resistor of the TLC will not matter much and can be set to something like 3K3.
You do not use the constant current output of the TLC in this configuration you are only using the PWM bit of it.

lqbert

Quote
First, determine the gate voltage you need. This will give you the ratio between A & B, since they form a voltage divider. Since your supply voltage is 18V and -9V below supply is well into saturation for any reasonable MOSFET, you can just set them equal to each other. That's easy, and it means you'll switch at similar speeds on and off.


In my diagram, I have 5VDC going through Resistor 'B' to the gate... it seems from this that you thought 18VDC would be going through resistor 'B'...

Given that I will be applying 18VDC to the Source and 5VDC where indicated, if I want to run 20mA through 4 LEDs which drop 3.2V per LED, what value of current limiting load resistor should I use?  I don't know how to determine what Vds will be either.

From my understanding:  

18V = 3.2V + 3.2V + 3.2V + 3.2V + V(load res) + Vds

If this is true, how do I know what Vds... after which I could determine the load resistor value (1 equation and 1 variable).

Also, how can I determine what values to use for Resistors 'A' and 'B' for a particular mosfet?  

A run-down on how to calculate these things using values found in a mosfet datasheet would be awesome!

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