Figure 1 demonstrates a one-way switched-capacitor gain stage circuit, performing a voltage gain of AO = C1/C2 only during the decreasing trail of the input signal Vi (with VJ = 0). During the positive signal variations, the output doesn't change, thereby implementing an incremental peak-detector and accumulator circuit.
This device can be used to measure the amplitude of noisy periodic signals with low peak-to-peak voltages. The input signal is accumulated for a certain number of cycles (NC) to obtain a proper output voltage (VO). The amplitude of the input signal can be derived by simply dividing VO by the number of cycles. This reduces any noise in the measurement.
After the reset transistors M1 and M2 are turned off, the circuit works as a normal inverting gain stage for every negative input variation. During positive voltage variations, the output node VO doesn't change and the feedback is guaranteed by diode D1—avoiding op-amp saturation. The signal path occurs along the feedback path C2-D2, where C2 holds the charge packets and D2 prevents removal of the integrated signal from C2 during the positive input-voltage variations. This operation can be repeated until a proper output signal change is obtained. The total output voltage variation can be expressed as:
where NC is the number of cycles applied to the input and VPP is the negative peak-to-peak voltage variation of the input signal within each period.
Figure 2 shows the circuit's output-voltage variation for 10 cycles of a sinusoidal input signal with an amplitude of 10 mV and a frequency of 100 kHz. The measured output variation is 200 mV, which is in accordance with the above equation. Note that the higher the number of accumulation cycles (NC), the lower the noise contribution at the output voltage.
This circuit can also be used as a phase detector between two signals (VI and VJ) having the same amplitude. When the two signals are applied at inputs VI and VJ, the amplifier accumulates only the negative difference between the two signals. Figure 3 shows the circuit's response to different phase-shift conditions. For a 0° phase shift, the resulting output voltage is 2A (where A is the signal amplitude of VI and VJ), which is the maximum anticipated signal. For 180° of phase shift, the output voltage is zero, as expected.
Note that having a capacitor input to the virtual earth point of an opamp means the signal source will have to drive a very low impedance at high frequencies (so best put a buffer amp in front of it), and makes it difficult to maintain stability and a good high frequency response (so it will be sensitive to choice of IC). The circuit probably has advantages of simpler peak detectors when it comes to low amplitude inputs though.
Mark Aitchison -February 24, 2006 (Article Rating: )
I like the peak detector circuit concept but there are two Vo on Figure 1 and is causing confusion. One Vo is at the AD823 output and the other is at C2-D2 junction. It appears to me the output signal as shown on Figure 2 should be for Vo at C2-D2 junction and not for op-amp output. Please review.
If this is true, then the Vo at C2-D2 needs to be bufffered with another op-amp in order to be useful.
Peter Lee -November 08, 2004
Hi, my name is Don Jennings and I am not spam! I read your mag Electronic Design all the time, but in the 10-28-04 issue page 68 is the last part of the single op-amp peak detector article and the last graph labled "C" bears a striking resemblance to graph "B." In fact, its the same graph. Graph C is supposed to indicate an output of zero with an input phase shift of 180 degrees giving zero volts out and zero degree phase shift giving 188 mV and 90 degrees giving 133 mV out and the readings for B and C indicate 133 mV, so the C image is missing, just showing a dup of B. Thats my 3.14159 cents worth for today! Still read your mag, and I like that circuit and the Sync'd Led one too. Looking at the data points though, maxing out at zero degrees at about 0.2 V dc and going to 0.0 V dc at 180 degrees, for the reading to be linear at 90 degrees, I would expect something like 0.09 V dc not 0.133 as is seen in the graph. That tells me it isnt very linear which is what one would expect from such a simple circuit. There are undoubtedly some feedback tweeks that could make the response more linear. 0.133 V dc is out about 40 from a perfect circuit. --Don Jennings, Inplane Photonics, South Plainfield, NJ. P.S., we only view electrons as secondary here anyway:)
Don, Thanks for alerting us to this production error. The correct Figure 3c is now posted on this page. --Editor
Don -November 04, 2004
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