Here you go: http://forum.allaboutcircuits.com/blog.php?b=116Looking forward to seeing it sport, is there a link?
What are esoteric components?Looks a nice tidy design, without using esoteric components.
Little used, difficult to understand, difficult to obtain, etc.What are esoteric components?
This is not the best design, its an easy and cheap design i made just for hobby. I know it could be done better.The half-wave rectifier charges the coupling capacitor when it shouldn't. It should use two diodes, not one.
With a single rectifier diode then the output coupling capacitor C3 becomes charged then the rectifier doesn't work anymore. R8 helps but its value is too high. Adding a diode discharges C3 on each cycle so the rectifier continues to work.What do you mean by the above comment you said?
How should i connect this two diodes?
The diode you propose to use is useless because the output of the bandpass filter swings almost between 0 and 5 volts, so this diode never contacts. The capacitor discharges through the op amp. The ouput signal of the bandpass filter swings around 4.3 V and C3 in combination with R8 form a high pass filter with a cutoff frequency of 1 Hz as to remove the DC component from the output signal of the bandpass filter. Thus, we have only an AC signal across R8 and the rectifier works.With a single rectifier diode then the output coupling capacitor C3 becomes charged then the rectifier doesn't work anymore. R8 helps but its value is too high. Adding a diode discharges C3 on each cycle so the rectifier continues to work.
Why does your schematic have measles dots all over it? Multisim?
I thought of it more and you are right because the discharge time is greater than the charging time so we need to put the diode you suggested to reduce the discharge time. That's why i like this site, we all learn form each other.In your circuit, C3 gets charged more and more with the signal until it doesn't work properly anymore.
In my circuit, the added diode discharges the extra DC across C3 on every cycle so it continues to rectify properly.
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