3
\$\begingroup\$

I'm looking to drive a bank of eight FOD4216 random phase triac drivers. I'm looking specifically at the (insanely expensive, $3!) Fairchild FOD4218, which needs 1.3mA to switch. 8 * 1.3 = 10.4 mA total draw for the bank of 8.

My uC can only sink/source 5mA, so I need some way to drive my optocouplers. I'd prefer to keep as low powered as possible, but a focus on minimizing the number of components leads my concerns. Somehow I need to get my uC from sourcing the whole 1.3 mA per optocoupler, down to sourcing no more than .625mA (5mA/8units = .625/unit) of the 1.3mA drive. Some headroom past that for the uC would be good. What would be good ways to create these 8 optocoupler drivers?

Not sure if it's helpful in doing the maths here, but forward voltage drop in the emitter appears to be between 0.9 - 1.2V across the full range of temperatures.

\$\endgroup\$
  • 6
    \$\begingroup\$ Out of interest, what's the microcontroller you're using? \$\endgroup\$ – Rocketmagnet Feb 22 '13 at 20:26
  • 1
    \$\begingroup\$ How can a microcontroller only sink/source 5 mA when most microcontrollers nowadays can sink/source 10-20 mA per pin? \$\endgroup\$ – m.Alin Feb 23 '13 at 8:58
  • \$\begingroup\$ @m.Alin Perhaps it is more of a system power budget issue than a microcontroller drive ability? Otherwise it does seem a bit odd. \$\endgroup\$ – W5VO Mar 25 '13 at 14:05
  • \$\begingroup\$ There are lots of micros with poor pin drive. It was really the PICs and AVRs who brought big drivers to the scene. You still have to be careful, as there is a pin drive capability as well as a port drive maximum. If you decide to drive 20mA with one port pin, you might find you have no drive budget left for any of the other pins on that port. \$\endgroup\$ – akohlsmith Aug 13 '13 at 21:28
1
\$\begingroup\$

Look at the ULN2003 and related parts within the group darlington arrays, they're good for driving LEDs, small relays, etc. well over your current and power dissipation requirements. They're probably even overkill for such little currents.

Check out the use of a octal buffer or transceiver such as the common 74HC244 though instead of "HC" you may want to use a different part depending on your MCU's Voh/Vol logic levels and supply voltages vs. the required LED supply voltage. If the minimum needed LED anode voltage is sufficiently high (e.g. higher than the Vio/Vcc rail of your MCU / buffer IC) you'll want to choose a buffer that can have perhaps a higher output voltage (>= LED anode voltages when they are off) while maintaining logic level compatibility with the MCU LVCMOS or whatever output levels.

You could also use an application specific standard part which is a purpose built multi channel LED driver, for instance, the TLC59282 or numerous other 8 channel or 15 / 16 channel parts intended for the purpose. Many of those will be able to save MCU I/Os by interfacing to the MCU via a shift register or similar system if that might be useful.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Answer: use quad and octo buffers.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ Perhaps, you're correct. At the same time, it's customary on EE.SE to post answers, which are a bit more elaborate than a single half-line imperative sentence. Try to elaborate a bit more, please. \$\endgroup\$ – Nick Alexeev Aug 13 '13 at 21:25
  • \$\begingroup\$ A buffer chip is going to buffer, using it's own supply to power the outputs. 74xx chips- take for example the 74LVC125A quad buffer- will often sink/source 50ma in this fashion. How further can I assist you? \$\endgroup\$ – rektide Sep 22 '13 at 23:49
0
\$\begingroup\$

If your microcontroller has an I2C interface, then you can just route two wires to an I2C I/O port expander chip. Using I2C commands, you just tell it what pins to turn on and off. There are numerous such chips; data sheet diving will reveal a suitable one which can source all the necessary current.

If it can meet all your other unstated requirements, it could be an ideal solution. You're going to need at least one external component to solve the problem. If it's an I2C port, then you're saving pins on your uC, too, and simplifying the circuit board; you're not having to route 8 uC outputs to 8 current-sourcing buffers.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ You're a funny man: GPIO can't power enough, so use a peripheral bus hooked up to an i/o expander? This does happen to violate my "unspecified" constraints by being a clunky slow plodding roundabout hack, that'll give me no where near the responsiveness of using straight GPIO. \$\endgroup\$ – rektide Sep 22 '13 at 23:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.