Op amp as peak detector

Answer 1

Peak DetectorA peak detector determines the peak voltage on an ac wave and converts it into a dc voltage. A diode puts the peak voltage on a capacitor. The second op amp is a buffer. This means it holds the same voltage input and output, but it allows current to be used at the output without altering the voltage. An op amp buffer is created by connecting the output to the inverting input.The purpose of the buffer in this instance is to tell the first op amp what the voltage is on the capacitor. In simple designs, this is done by simply connecting the inverting input of the first op amp directly to the capacitor. But the buffer allows a second diode to be used in the feedback loop without altering the capacitor voltage. The diode then pulls against the output of the buffer instead of the capacitor.The purpose of the feedback diode is to prevent the output from saturating on the down-swing. Saturation means the output goes to the supply rail, while excess current is drawn through the input, which can create undesirable effects. The feedback diode pulls the inverting input down to the same level as the noninverting input, which means the output is never more than 0.6V lower than the noninverting input or the bottom of the voltage swing on the ac input wave. The feedback resistor allows a difference in voltage to be created between the inverting input and the output of the buffer.The design challenge of a peak detector is in the fact that the diode pulls the voltage up but not down. Something needs to bring the dc voltage on the capacitor back down, either when the voltage being measured decreases, or when an over-voltage spike is created for some reason such as noise, contact of leads or turn on of the device.Leakage current through the diode when reverse biased will be enough to control the capacitor voltage when stable measurements are being made. At other times, more current is needed to bring the capacitor voltage down. A resistor can be used for that purpose. But it will then create some ripple on the capacitor, because it will be pulling voltage down while the diode is not pulling it up. The ripple is greatest for lowest frequencies, because then there is more time during each cycle for voltage to decrease. Therefore, low frequencies require some compromise between ripple on the dc and rate of equilibration.Here's how to calculate ripple on the capacitor. Start with this formula: V/S = I/C. It says volts per second equals current over capacitance. Say there is one volt on the cap. Divided by 10M of resistance equals 0.1µA of current. Divided by 0.47µF equals 0.213 volts per second. If the frequency is 10Hz, there is 0.05 seconds for each half of the wave. So 0.213V/S times 0.05S equals 11mV of ripple. If the frequency is increased to 1KHz, a half wave is 0.5mS, and ripple is 0.0005 × 0.213 = 0.11mV. If the cap voltage is increased to 2V, the ripple is 0.21mV, or twice as much, but the same percent.The rate at which the voltage equilibrates is not linear. At 1V on the cap, it is 0.213 volts per second; but at 2V, it is 0.426V/S. So it would take about 3-5 seconds to equilibrate when 1V too high. Pulling it down instantly with a push button switch is so easy that it is better than compromising the design or waiting for equilibration.