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Wht is the purpose of this op amp

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Winsu

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Hi All,

I'm not sure if this thread should be in Analog or Digital but here it goes.

I'm trying to find out why the next circuit was done. It is an op amp configured as a voltage follower. The input of the op amp is an PWM and the output is connected to the control voltage of an LED driver ( to control the string current). I wonder why this is done. It is just for pure isolation?, or there are other reasons?. Thanks in advance

PWM-OP AMP-control voltage for Led driver.PNG
 
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Away from the remaining circuit it is difficult to be sure but the most likely reason is because R18 and C25 form a filter and without the high input impedance of the op-amp following it, the load would prevent C25 charging. Don't forget PWM is a pulsed signal, it probably wants converting to something nearer an analog voltage by the filter.

Brian.
 

It is a simple voltage follower configuration.

In my opinion, if one is going to use an opamp anyways, it is better to implement with it a higher order filter.
Far less ripple for a given cutoff frequency.
 

Hi,

Low pass filter, first order, 38Hz
Voltage follower to reduce impedance
I assume R36, R26 are just there to ensure zero output voltage in case of disconnected input signal, or no supply voltage.

Klaus
 

Thank all for answering.

The voltage follower to reduce impedance, why Do I need to reduce impedance?
 

The voltage follower to reduce impedance, why Do I need to reduce impedance?

You know what the load impedance is, we can't determine if the lower driver impedance is actually necessary.
 

PWM -> DC value converter ... with lower output impedance to drive the secret load ( lower than the 4k7 resistor used on the filter ...)
 

The output voltage of that Op amp is connected to an LED driver. The level pf that voltage will determine how hard the LEDs are driven. If I get this right I think the output impedance would just be R35 as It is just a track connected to the Led Driver.

I don't actually understand why the PWM is converted in DC. Is it converted in DC because of the filter or because of the Op amp ( I bet for the filter but I not 100% sure).


Later updated:

I guess the filter transform the PWM signal into a DC signal and then the op amp reduces the impedance ( as the filter sis using a 4k7 resistor). I think this now make sense. Thanks everyone.
 

You got it!
The filter is to average the PWM pulses so the voltage across it is proportional to their on/off ratio. If loaded directly, the voltage would be dropped too low by the out-flowing current to the LED driver. The op-amp is there to maintain the same voltage out as it sees at its input but without significantly loading the filter.

Brian.
 

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