You most definitely have something wired wrong - there is no way it can produce an output voltage greater than it's supply voltage.
The circuit isn't actually a peak detector, it's nearer to being a precision rectifier and buffer as there is nothing in it to hold the peak level. I presume you are measuring the output voltage with respect to ground and not the negative or positive rails.
The virtual ground of your ICs is at +1.2V. If your input is 1V peak and symetrical around 0V, then at no time will the diode be in play. Your signal is negative with respect to the opamp and always will be.
Look at #8. His input signal is ground based (I'm assuming that it is symetric around ground). His opamp's "mid supply level" is above the maximum signal from his "ground based" sine wave generator. And his input circuit is DC coupled.
Hi, instead of using ampop why you dont use AD8307 ( multistage logarithmic instrumentation amplifier ) ? You get analog out on any signal from DC to 500MHz
Here are some measures:
Power in | Vout
-20dBm 1,678V
-10dBm 1,9237V
-5 dBm 2,057V
-1dBm 2,1502
0 dBm 2,1808V
There are many circuits using AD8307 googalize it. Very low cost easy implementation.
Look at #8. His input signal is ground based (I'm assuming that it is symetric around ground). His opamp's "mid supply level" is above the maximum signal from his "ground based" sine wave generator. And his input circuit is DC coupled.
I'm not sure why "mid supply level" makes any difference in that design. The input voltage (ground) is well within the limits for that device so the problem shouldn't be input overload.