Reduce Component advice

[johnnyfp]

Hi,

I'm putting together a SPI based 24channel PWM chip with a load of High powered LED Drivers. The TI SPI based PWM chip has an open drain output for each channel but the TI driver I'm using needs a high on it's input to switch it on.

I've put together the attached analog inverter as per say and was wondering if there's a more elegant way or less component count way of attaching the two devices.

Cheers.

[David_AVD]

What about a pull-up on the PWM chip output and feed the LED driver directly?  You'll need to invert the PWM data of course.  (100% > 0% thought to 0% > 100%)

[johnnyfp]

Yeah, but I would like the initial state to be low, So that if the PWM driver or the MCU is in a uninitiated stated, the lights are not on.

[David_AVD]

Hey I didn't say it was perfect!   

[johnnyfp]

not perfect!!! NOT PERFECT!!!!. I expect nothing but Perfection. Except when it's not perfect!

 

No worries. The funny side of this is that I forgot to put the resistor on the base, and when I first switched this on, I had a nice puff of smoke from the Transistor. What's really annoying is I've had 100 of these boards made up with this missing resistor, hence a redesign, hence asking for advice from people that know more than me.

[Hypernova]

How about still going with the direct drive idea except the 24v rail pulling the LEDs is controlled by a MOSFET? That way the LEDs stay off until you got the PWMs all set up.

[johnnyfp]

Mmmh. Yes that would work. But that would also require an extra pin to enable/disable it which I was trying to avoid.

What about if I replace the PnP with a Mosfet instead? Could I then lose a couple of resistors?

[codeboy2k]

How about a single pullup on each open drain output into real inverter chips?  4 hex inverters in an so-8 package will give you 24 channels, and the pullups can be resistor packs.  Try to avoid 24 discrete FETS and the extra resistors if you can.

Your transistor solution was switching  24 volts into the drivers inputs. Do you need that much into the driver's front end?  I would expect a good driver just uses logic level pwm at the front end and has a separate voltage input pin for the output stages.

If you don't want to use inverters to feed the drivers, then you can put a single pull up on each PWM output, and then directly driving the inputs to the driver IC. All those pullups tie together to a single P-channel MOSFET, with source to your V+, drain to the pullup network.  The gate is connect to another GPIO pin to enable the Vpullup . At this point, you have the smallest parts count, but you have to invert your PWM in the software. And you can enable Vpullup once you have setup all the PWM outputs correctly

[johnnyfp]

 I like your inverter idea, but I need to work with the driver board. Adding an invert or enable pin will mean  a change to the motherboard that these drivers plug in to which I want to avoid for now. But I will adopt for version 2 if this proves too hard to achieve.

[David_AVD]

Maybe derive the enable pin from a simple (delayed on) power good circuit.

[johnnyfp]

I think that I will have to live with the number of components for my first run and then look at revising the motherboard with an inverter and do away with the Transistor and resistors on the driver boards.

Now, to find a 24channel inverter

[codeboy2k]

I think that I will have to live with the number of components for my first run and then look at revising the motherboard with an inverter and do away with the Transistor and resistors on the driver boards.

Now, to find a 24channel inverter

Ah.. you've already done it the first way.  ok, Rev 2 will be your improved version.

Maybe derive the enable pin from a simple (delayed on) power good circuit.

I like this idea. It's simple and avoids the extra enable from the microcontroller. As long as the micro has finished setting up the PWM outputs before the pullups are enabled, this will work nicely.