Hi Thiago,

My guess is it is an active rectifier (no direct drop voltage loss) with peak detector.

Keys parts: 1) A diode (D1) rectifies, that is it allows to pass electrons in one direction e.g.: only positive voltage, but it has a drop voltage due to the junction direct voltage 2) Something in the feedback in the minus (-) input of an opamp (A1) is magically compensated by the gain ot the opamp, here the direct drop voltage is compensated. 3) Peak detector is made by diode D1 and the capacitor C, capacitor is like a tank you can fill with electrons. 4) A2 works as a isolation buffer, it allows low impedance charges to not affect to the discharge of the capacitor C, it is only discharged by the high impedance input of the opamp.

Let me know if I am wrong!

Hey Manel,

Thanks for the explanation, helped a lot! I actually do know how capacitors, resistors, and diodes work, my problem is with the amp op and with the peak detector, also with putting diodes with a ampop, it cofunses me.

Thanks,

Thiago.

Hi Thiago and Manel, What Manel has posted is right. This video might help you understand what exactly is happening.http://youtu.be/jllsqRWhjGM I don't think the code is exactly what I want. The Peak detector detects the peak of the amplitude of a signal. But doesn't depend on the frequency ( I think!).

Hey Anitta,

Thanks for the link, this guy explains very well! I edit the answer a little. I believe it look mor like what you wanted now, but I am not sure... Please comment here if it is not. What kind of LCD are you using? Do you have code that you started to write? We could have a look, maybe we can help more...

With the added control of the reset transistor to Thiago's code. I'm assuming that you want to loop around resetting, waiting for a short period and then taking a reading.

I've also added some code to convert from ADC reading to voltage, 3.2 should be about the right scale factor but you may need to make some measurements and change that a little if you want it to be accurate. I've also added some code to output to the virtual serial port so that you can see the values on a PC. Handy for troubleshooting.

Serial pc(USBTX,USBRX);
DigitalOut myled(LED1);
DigitalOut resetFET (p21);
AnalogIn ain(p15);
 
//Declare LCD class here
 
int main() {
    
    resetFET = 1; 
    uint32_t actualAnalogValue = 0;
 
    while(1){
       wait(0.01); // give it time to fully drain the capacitor. (will depend on component values)
       resetFET = 0; // disable the reset FET

        /*Waits for the signal in the output of the peak detector to stabilize*/
        wait(1);  // should be a couple of times your expected input frequency of interest.
        
        /*Reads the value in the output of the peak detector*/
        actualAnalogValue = ain.read_u16();

        resetFET = 1; // start the reset now

        /* convert to a voltage by multiplying by 3.2 and dividing by 2^16. */
        float voltage = actualAnalogValue * 3.2 / (1<<16); 
        pc.printf("Peak is %.2f\r\n", voltage);
        
        /*Displays it on the LCD Monitor*/
        //here
   
        }  
}

I've assumed that you have an n channel FET from the capacitor to ground so a high on the gate will turn it on and reset the peak. Make sure you put a resistor in series with the FET otherwise you're effectively shorting the op-amp output to ground which isn't good for the parts. Small values will reset quicker but put more of a strain on the parts.

I've included a short wait at the start of the while loop to give time for the reset to take place. Depending on the components used the time taken to output to the display may introduce a long enough delay that you don't need that extra wait in there.