Peak detector circuit.

[Heevy]

Hello Everyone, im new to this forum, so i hope im posting the correct place, also english is not my native tongue, so please excuse my mistakes, should i make any.

Anyway, i'm supposed to build a part for a cycloptic robot that can follow a line on the floor, using a photo-diode and the reflection from some red diodes.I've been given a peak detector circuit with a low pass filter which has as input a 10.8 kHz sine wave coming from our band-pass filter. i'm eventually supposed to give a 5 min presentation on how the circuit works, but i must admit i'm not completely confident how it works. could anyone give me a run down of the circuit? any help is appreciated.

I've attached the schematic.

Thanks

Heevy.

 

[Arouse1973]

Hello Heevy, welcome to EP. Are you sure the waveform is a sine wave? And is the frequency 10.8 KHz? Something's not right with the values. Also the values for the feedback R22 and R23 I would expect to be much larger, they are going to discharge the capacitor very quickly. Also adding the low pass filter to the output forms a Quasi peak detector but it has a -3dB roll off at 106 Hz. Can you check the values and wave form type and frequency.
Thanks
Adam

 

[Heevy]

Hello, and thanks for answering
the following is what i calculated and what was given already. please excuse my ignorance on the subject, im still rather new to this.
tau = 1 ms C15 = 22 nF and Av = 2 is giving in my case

R*C = tau = 1 ms and
→ R22 + R23 = τ / C15 = 1 ms / 22 nF ≅ 45.45 kΩ

→ R22 / R23 = 1 → R22 = R23 → R23 = 45.45 kΩ / 2 ≅ 22.7 kΩ → R22 = 45.45 kΩ / 2 ≅ 22.7 kΩ

that is how i got the sizes for the resistors. regarding the input i've attached a picture of the oscilloscope with the input wave and output.

Regarding:

"Also adding the low pass filter to the output forms a Quasi peak detector but it has a -3dB roll off at 106 Hz"

I have no idea what this mean, is the anywhere in particular i can read up on this subject?

it is stated in my assignment that by adding the low pass filter with a break frequency of 100 hz, it will smooth out the signal. and that the robot still reacts to the signal.

perhaps i should mention, that the PCB has already been tested in the robot and it runs fine.

it is also stated that the different components is only meant for guidance, so improving on the values is only a good thing.

any improvements and further explanation is very appreciated.


Heevy

 

[Arouse1973]

Ok I didn't realise you had been given some specifications already. As you already know the first part of the circuit is the peak detector and has a gain of 2. The output will rise up until the inputs are of the same value. The reason the feedback path is taken from the other side of the diode is so it cancels out the diode drop and gives you an actual x2 output of what's on the input. This output charges up the capacitor C15 through the diode and the 15K. Then as the output fall C15 discharges through the two 22.6K resistors but not the diode, because the diode block current from going the other way. A very small amount of charge will also be lost through the inputs of the two op amps.

The second stage is as you know a low pass filter which will hold the output at x4 of the input for a longer period of time than the period of the input waveform this is because of the time constant of the 100 K and 15 nF. This means the output drops very little as the input changes which helps to keep a smooth level on the output.

The addition of the low pass means the output won't respond very quickly to changes on the input which is some cases is not desirable. It will take time to respond to a change in the input and so if you are relying on this level to make a decision to do something on the robot it may be slow to react.

Comments:

The 15K (R24) in my opinion is too high a value because this can cause an opamp to overshoot, this is the first stage that the opamp could become unstable, too much delay in the feedback loop. This should be removed or lowered to 47R or something around that value. A better approach for the low pass filter would be to have the RC network before the gain stage, this gives what ever the filter is connected to on the output a lower impedance circuit to work with which maybe desirable, especially if you wanted to connect this circuit to an A to D convertor.

There is also the possibility that the filter could be removed completely and the peak detector circuit values changed to give a similar response. This will depend on the maximum acceptable ripple on the output and the response requirements of the circuit.

Thanks
Adam

 

[Heevy]

Thanks alot for the help Arouse! i will take this into consideration!

Heevy