How does this circuit work?

Thread Starter

Frank@Frank

Joined Jul 24, 2021
23
Hi guys.
I saw a led circuit like this.
the led is pulse drived, and there is a photodiode seems to be feedback to the Led driver circuit.
but it is not the standard Voltage follower or inverting amplifier or any basic circuit that I learned in the txetbook.
so I am curious how this circuit work and how to determine the values?

PS: please ignore the value of C,R.
 

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Thread Starter

Frank@Frank

Joined Jul 24, 2021
23
What is your circuit supposed to do?

is that a transmission gate in series with the base drive?
Hi!
I think the photodiode is monitoring the light power of led and feedback to the driver circuit.
But don’t know how it works and how to calculate the parameters of capacitors or resistors
 

Thread Starter

Frank@Frank

Joined Jul 24, 2021
23
Is that a transmission gate in series with the base drive?
You mean the components in series between the output of op-amp and the base of the transistor?
That’s an analog switch.
This is not drew by me.
I config it out from a pcb board and want to know how it works.
 

Alec_t

Joined Sep 17, 2013
14,280
My guess would be that the photo-diode is sensing ambient light. In daylight it passes current, allowing C1 to charge up to a voltage which exceeds the reference, thus causing the op-amp output to go low and turn off the transistor and LED.
At night the photo-diode passes next to no current, so C1 discharges, the op-amp output goes high and turns on the transisitor and LED. The PWM'ed gate/switch controls LED brightness by pulsing the transistor base.
 

Thread Starter

Frank@Frank

Joined Jul 24, 2021
23
My guess would be that the photo-diode is sensing ambient light. In daylight it passes current, allowing C1 to charge up to a voltage which exceeds the reference, thus causing the op-amp output to go low and turn off the transistor and LED.
At night the photo-diode passes next to no current, so C1 discharges, the op-amp output goes high and turns on the transisitor and LED. The PWM'ed gate/switch controls LED brightness by pulsing the transistor base.
Hi Alec!
I measure the output of the op-amp. It was not just high or low. The PS is 3.3V and the output is about 1.x Volt.
 

Alec_t

Joined Sep 17, 2013
14,280
the output is about 1.x Volt.
We don't know whether what I referred to as an op-amp is indeed one, or if it is a proper comparator IC. It's also unknown if that device is a rail-to-rail type capable of driving its output fully low (~0V) or fully high (~3.3V). Depending on the contents of the PWM'ed box of tricks, 1.xV may or may not be enough to turn on the transistor.
 

Irving

Joined Jan 30, 2016
3,845
The opamp with capacitor feedback is an integrator. Its output represents the time integrated photocurrent from the diode which, as you surmised, is related to the light output from the LED. Do you know what LED, photodiode, supply voltage, PWM frequency/duty cycle?
 

Thread Starter

Frank@Frank

Joined Jul 24, 2021
23
The opamp with capacitor feedback is an integrator. Its output represents the time integrated photocurrent from the diode which, as you surmised, is related to the light output from the LED. Do you know what LED, photodiode, supply voltage, PWM frequency/duty cycle?
Hi!
The VCC is around 9V.
PWM frequency is 20kHz/5%.
Don’t know the exact part number of the LED and photodiode.
But the VF of Led is around 1.8V.
 

Thread Starter

Frank@Frank

Joined Jul 24, 2021
23
The opamp with capacitor feedback is an integrator. Its output represents the time integrated photocurrent from the diode which, as you surmised, is related to the light output from the LED. Do you know what LED, photodiode, supply voltage, PWM frequency/duty cycle?
BTW, there is a another resistor in series between VCC and photodiode
So it seems to be an integration circuit. Tried to find the relationship between the output of opamp and the feedback from photodiode but have no clue to write the equations.
 
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