Can someone please explain what this circuit does. Or even tell me the name of this circuit.

dannyf

Joined Sep 13, 2015
2,197
The opamp on the right detects the current going through the LED, and the opamp on the left integrates the current.

Once the integral reaches a threshold set by r1 and r2, the opamp on the left turns off the MOSFET thus discharges the integrator.

So it is a pwm controller, with its duty cycle controlled by r1 and r2.
 

OBW0549

Joined Mar 2, 2015
3,566
The opamp on the right detects the current going through the LED, and the opamp on the left integrates the current.

Once the integral reaches a threshold set by r1 and r2, the opamp on the left turns off the MOSFET thus discharges the integrator.

So it is a pwm controller, with its duty cycle controlled by r1 and r2.
Ummm... no.

It could be a PWM controller, if there were something to force the MOSFET on the right to function as a switch instead of just a linear pass element, AND if there were some hysteresis mechanism built into the feedback loop. Without these, however, this is merely a linear constant current regulator which will hold the LED current at a level such that the voltage drop across the 1Ω sense resistor is equal to the voltage at the junction of R1 and R2.
 

AnalogKid

Joined Aug 1, 2013
8,682
OBW is on a roll. The opamp on the right is performing a function similar to the "DC servo" in audio amplifiers that have no coupling capacitors.

ak
 

johndeaton

Joined Sep 23, 2015
63
Without these, however, this is merely a linear constant current regulator which will hold the LED current at a level such that the voltage drop across the 1Ω sense resistor is equal to the voltage at the junction of R1 and R2.
What is the purpose of R3 and the capacitor?
 

OBW0549

Joined Mar 2, 2015
3,566
What is the purpose of R3 and the capacitor?
R3 and capacitor C, along with the op amp whose (-) input they are connected to, form an integrator which serves as the error amplifier portion of the closed-loop current controller.

The error amplifier, in turn, controls the gate voltage of the MOSFET and thereby adjusts its drain current, keeping the current at such a level as to make the voltage being applied to R3 the same as the reference (i.e., setpoint) level established by R1 and R2.

The op amp on the right, along with its network of resistors, operates as a unity gain differential amplifier which translates the differential voltage across the 1Ω current sense resistor into a single-ended (i.e., ground-referenced) voltage for input to the integrator.
 

johndeaton

Joined Sep 23, 2015
63
R3 and capacitor C, along with the op amp whose (-) input they are connected to, form an integrator which serves as the error amplifier portion of the closed-loop current controller.

The error amplifier, in turn, controls the gate voltage of the MOSFET and thereby adjusts its drain current, keeping the current at such a level as to make the voltage being applied to R3 the same as the reference (i.e., setpoint) level established by R1 and R2.

The op amp on the right, along with its network of resistors, operates as a unity gain differential amplifier which translates the differential voltage across the 1Ω current sense resistor into a single-ended (i.e., ground-referenced) voltage for input to the integrator.

That makes sense, but could the left op amp, R3, and C be replaced with a simple comparator circuit that just turns off the MOSFET drive voltage when the output of the differential op amp is higher than the voltage at R1 and R2? If not, why?
 
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OBW0549

Joined Mar 2, 2015
3,566
That makes sense, but could the left op amp, R3, and C be replaced with a simple comparator circuit that just turns off the MOSFET drive voltage when the output of the differential op amp is higher than the voltage at R1 and R2? If not, why?
Not really-- at least, not and have the circuit work very well.

What would happen is that the comparator would oscillate extremely rapidly at an uncontrolled frequency, turning the MOSFET on and off as fast as it could (depending on the comparator used, this could range well into the low RF frequencies).

This would waste a lot of power, make the MOSFET hotter than a stolen handgun, generate tons of RF interference that would radiate all over the place, and end up doing a rather crappy job of regulating the LED brightness.

If you're aiming to create an LED drive circuit that will avoid the power wastage associated with linear controllers, the best thing to do would probably be to start with some sort of current-feedback switching regulator and go from there.
 

johndeaton

Joined Sep 23, 2015
63
Not really-- at least, not and have the circuit work very well.

What would happen is that the comparator would oscillate extremely rapidly at an uncontrolled frequency, turning the MOSFET on and off as fast as it could (depending on the comparator used, this could range well into the low RF frequencies).

This would waste a lot of power, make the MOSFET hotter than a stolen handgun, generate tons of RF interference that would radiate all over the place, and end up doing a rather crappy job of regulating the LED brightness.

If you're aiming to create an LED drive circuit that will avoid the power wastage associated with linear controllers, the best thing to do would probably be to start with some sort of current-feedback switching regulator and go from there.
Thanks OBW. Great explanation as always.
 

OBW0549

Joined Mar 2, 2015
3,566
One example of a current-feedback switching regulator such as I mentioned above would be Linear Tech's LT3474 1 Amp Step-Down LED Driver chip. It's a basic buck-type stepdown switching regulator; but instead of controlling its output to a constant voltage, it controls its output to maintain a constant LED current. Figure 1 on page 7 of the data sheet shows a block diagram of the chip's innards.
 

ci139

Joined Jul 11, 2016
1,696
_Draft-led-ctrl.png
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@2017CE :: Ran with different OpAmp -- the circuit first seemed an idiot proof one , but as my psychiatrist has given up , i managed to prove an opposite for the LT6240 (chosen because likely the best one they got ?)
cc_Draft_CC0100.png
just FYI -- i won't be discussing the subj. here
 
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OBW0549

Joined Mar 2, 2015
3,566
What are the formulas around sizing R3 and C in this circuit? Anyone have a good link to more information?
I doubt the values are critical; R3 = 10 kΩ and C = 0.1 μF would probably work, or any other combination that would yield an RC time constant of somewhere around 0.1 to 100 milliseconds.

But as AnalogKid indicated in post #2, this circuit is way more complex than it needs to be; I would never use it, not in a million years.
 

OBW0549

Joined Mar 2, 2015
3,566
Hopefully some teacher was using it as a learning example instead of a practical circuit :)
Agreed.

The way the circuit is designed, it looks to me like it was intended to supply the LED with an extremely precise current that would be absolutely constant. I've seen constant-current control circuits like that, but never for a trivial task like regulating LED current. If I wanted constant current for an LED, I'd probably use something like the attached:
 

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