open collector constant-current driver -> MOSFET gate

[damiannz]

hi everyone,

this is my first post here so i hope i'm doing everything correctly!

i have a Texas Instruments TLC5940 constant-current PWM LED driver which has open-collector output pins. normally, these allow an adjustable constant current to flow through an LED connected to a power source up to 18V. but i need 350mA, which the TLC5940 can't deliver; so rather than using this current directly i'd like to use it to open the gate of the MOSFET in this circuit.

i have two questions:

1. since the pins on the TLC5940 are rated up to 18V, do i need the Zener diode there?

2. the TLC5940 includes error-detection circuitry. my initial tests with a multimeter and pull-up resistor indicate that it will not complain here, but could i be doing damage to the device in the long-term?

thanks in advance for any advice!
cheers,
damian

 

[itsmoked]

That would be ill advised. That controller is a high end programmable constant current sink - not open collector.

If you want to run that other circuit just use one of the many nice pull-down chips. Pin drivers, etc. They will cost and fraction of that TI part, are readily available, and come in a dozen flavors.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[electricuwe]

If you have to stick to the TI part because you need the coplex functions one easy way to increase output current would be to use current mirror circuits at the output.

Current mirror circuits are very commonly used in complex IC, but there are also some devices available that just have this function (also from TI).

But you can also build a current mirror from discrete bipolar transistors. Ensure that you include emitter resistors in this case. Googleing "current mirror" will for sure give more detailed information.

 

[damiannz]

thanks for the replies,

yes @electricuwe, i do need the other functions of the TLC5940.

so, let me see if i have this correct. i could use a basic current mirror like this:

           +5V       +12V 
            |         |
            +-----    +-------------
            |    |    |            |
            \    |    /            /\^ LED chain
          Q1 |---+---| Q2          |
            /         \            |
            |         |       ------------
            R1        +-------|LED driver|
            |         |       ------------
       ---------      R2           |
       |TLC5940|      |            |
       ---------      |            |
        |             |            |
        0v            0v           0v

with Q1 == Q2 NPN transistors
and R1 == R2 (?)

then, the constant current LED driver will be turned on (when current flows through the TLC5940) by being pulled +ve by the voltage across R2, and turned off (when the TLC5940 pulls the voltage to 0 making Q1+Q2 turn off), by being pulled -ve.

is that correct?

thanks again for the assistance!

 

[electricuwe]

Your suggestion is more complex than needed:

Just design the current mirror with the proper ratio and use the output directly to drive the LED string.

But ensure that you use emitter resistors. Otherwise current mirrors with discrete transitors do not work properly.

 

[IRstuff]

High side current mirrors use PNP transistors. Your configuration simply turns on the NPN transistors, without regard to their collector currents.

TTFN

FAQ731-376

 

[jimkirk]

Have you thought about paralleling output channels of the 5940? With all outputs on and 1 V across the output pin, each channel looks good for 54 mA minimum. That would require 7 outputs in parallel to get 350 mA. Fewer outputs active allows more current per output, just mind the total dissipation. If you have spare channels, it might work.

 

[OperaHouse]

It may be easier to perform the maximum LED current function with an external three terminal regulator set up for constant current. Then the pulse width output only needs to be a simple on/off driver.

That hobby circuit basically does both those functions, but it inverts the signal. The current sense resistor causes the drive to clamp once the base voltage reaches conduction. Of course, you are back to good old analog heat.

Yes and no, the zener is only there to protect the gate from excessive voltage. If supply voltage is under the gates maximum it is not needed.