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SuperFaradCapacitor
modified 4 years ago

Simple Adjustable Constant Current Buck Converter - LED driver

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03:50:36
Sadly, it does not work very well in real life. A simple adjustable constant current circuit that uses only an inductor, capacitor, shunt, diode, mosfet, two op amps, some resistors, and a potentiometer. Does it buck, though? I don't know. Notes: The voltage dependent voltage source is in place of an op amp in a differential amplifier configuration. If you start the circuit simulation with the power switch closed, it will say "cannot find solution". You need to start the circuit first, then close the switch. After that the constant current circuit will work properly. The comparator now has some hysteresis due to a positive feedback resistor. This makes it buck instead of be a linear regulator. How it works: The comparator turns the mosfet on because the noninverting input voltage is higher than the inverting input voltage (which is 0 because there is no voltage difference across the resistor, which makes the output of the "differential amplifier" 0) This causes current to rise through the inductor, shunt resistor, and LED. The diff amp raises its output voltage to the point that the comparator turns the mosfet off because the diff amp voltage is higher than the potentiometer voltage. This causes current to fall as the inductor's magnetic field collapses. This causes the diff amp output voltage to fall, which makes the comparator turn on the mosfet, and the cycle repeats. I've breadboarded this circuit to drive a COB LED I scavenged from a light, and it works pretty well. It's more efficient than the linear version without hysteresis because the mosfet does not heat up as much when the current is set to about half.
published 4 years ago
PrathikP
4 years ago
"EC doesn't allow this Switch action" that's not true. What you have built is a LINEAR current regulator and not a switch-mode one. Simply adding an inductor, diode and a capacitor doesn't make it a switch-mode regulator.
PrathikP
4 years ago
Take a look at this https://everycircuit.com/circuit/5009203532660736
mlira
4 years ago
Don't see why it wouldn't work in real life but your op amp is acting more like an indirect voltage follower than a comparator. It automatically changes the resistance of the mosfet until you get the right value across the shunt. Here's a simpler version that's doing the same thing https://everycircuit.com/circuit/6314562665840640
SuperFaradCapacitor
4 years ago
I built the original circuit, and the op amp that I thought was going to act like a comparator acted like a indirect voltage follower, and just put out a constant voltage to the gate of the mosfet.

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