rsdio writes: Although this circuit does not directly show how, you can use the same idea to make a 3.3 V tolerant I/O pin control an LED with a much higher voltage. The caveat is that all of the current still passes through the I/O pin, but at least you can control several LEDs in series to get more brightness without the necessary higher voltage posing a problem.
A Darlington transistor pair should allow the current to be increased beyond the I/O pin limits, thus breaking free from both voltage and current limits. The latter is left as an exercise well beyond the app note’s content.
Although many LED-driver ICs operate at relatively low voltage, this application note shows how to operate the MAX6974 at higher voltage by adding a common-base-transistor level shifter, allowing it to drive multiple LEDs in series.
So where would the darlington go in relation to this diagram?
Jason: normally you’d connect the base of the Darlington to the IO pin and connect it between the LED and its current limiting resistor and ground. A bit like this diagram, except with the LED and current limiting resistor replacing the solenoid.
Of course, if you do that you don’t need the level shifting transistor as well – since you’re doing low-side switching your switching circuit doesn’t see any of the voltage across the LED or resistor. In fact, this circuit is a lot easier to understand than the one in the app note!
The reason this app note does need a level shifter is because the LED driver chip they’re using is a current-controlled sink that needs to see and control the current through the LED chain. I think this means some of the voltage powering the LED chain is across the driver circuitry.
I don’t think I understand what this circuit accomplishes that isn’t done by a simple low-side NPN transistor in a traditional low-side switching setup? (though it seems to be pretty common for magazine “design idea” submissions to be a lot more complex than they need to be…)
“The LED driver chip they’re using … needs to see and control the current through the LED chain.”
Oops. That makes sense. I should read more carefully before I post.
The difference between this and a standard low-side drive is that the IC output is connected to the transistor’s emitter, not the base. This is a common-base amplifier that has a current gain of almost exactly 1, but a large voltage gain.
If you connected a voltage-level output to the base and the emitter resistor to ground, this would be a standard drive, but it would only be a switch. The emitter resistor, the Vbe of the transistor, and the supply voltage would set the current. Even if you do PWM, all of those factors come in as multiplicative factors. In this circuit, the LED current is exactly the current sunk by the driver IC minus the (tiny) base current.
I tried to research a detailed answer to Jason’s question, and first discovered that most Darlington transistors have only 3 terminals. I was originally thinking along the lines of the 6N138 optoisolator with 4-terminal Darlington output, or at the very least a discrete Darlington transistor pair.
In the midst of my research, I realized that a simple two-stage transistor circuit would be required to allow both current and voltage to exceed the I/O pin limits. The configuration would probably not actually work as a Darlington, and would certainly require discrete transistors or a hybrid with more than 3 terminals.
As others have pointed out, this particular circuit is constrained by the fact that the controlling circuit affects the current directly, rather than merely acting as a binary switch.
In other words, sorry for my misleading suggesting that a Darlington pair would make any improvement to the circuit!