Signal rectification is a process in which you process an AC (Alternating Current) signal (often sinusoidal) to get a DC signal or a signal with a DC component.

Why Rectify Signals

        There are many situations in which you want to "rectify" a signal.  Here are some of the most common situations.

• In power supplies the AC supply voltage (60 Hz in the U.S., from the wall plug) is converted to a constant DC value to power electronic circuits that require a constant voltage to operate.  Examples include the following:

• Computers

• Television sets

• Radios

• Stereo Systems

• Guitar Amplifiers & Sound Systems in General

• In signal systems, information coded into the amplitude of a sinusoidal signal must be rectified to extract the information.  Examples include the following:

• AM radios

• Sensors - like some kinds of tachometers - which produce a sinusoidal signal where the amplitude encodes the value of a physical quantity like rotational rate.

Signal rectification is an important concept and is often used in a variety of systems.

        Here is a simple signal rectifier.  When the input voltage is positive, the output voltage is approximately equal to the input.

Using A Capacitor In A Rectifier - Building A Peak Detector

        Here is a circuit.

Here is what happens in this circuit.

• When the sinusoidal input is increasing and the capacitor is uncharged, current flows through the diode to charge the capacitor.  (Remember, current is just charge flowing, so when current flows through the diode into the capacitor, charge accumulates in the capacitor.)

• At some point in time, the input voltage begins to decrease.  At that point, the diode stops conducting and the capacitor/resistor combination is effectively cut off from the rest of the circuit and the capacitor begins to discharge through the resistor.

• As the capacitor discharges, eventually the input voltage gets to a point where it equals the capacitor voltage, and the diode begins conducting again, charging the capacitor in the process.

       There are some things to consider.  The first point to consider is just how far the voltage will droop from the peak until it begins to increase again.  Just past the peak the voltage across the capacitor will decrease for a short while.  What happens is that the input voltage (the sine wave) is decreasing but not decreasing very quickly.  (As you get further from the peak the voltage begins to decrease more quickly.)  When the input voltage is not decreasing very quickly, the charge flowing out of the capacitor can flow through the resistor, and there will still be current flowing through the diode.  (Some of the current flowing through the diode comes from charge on the capacitor, and some flows through the diode.)  However, sooner or later, the input voltage gets to the point where it is decreasing faster than the capacitor voltage can change.  The limit on how fast the capacitor can discharge is determined by the resistor.  For a given voltage, V_out on the capacitor, the current out of the capacitor is limited to V_out/R.  Since the capacitor current is CV_outd/dt, the rate of change of capacitor voltage can be computed as dV_out/dt = -V_out/(RC).  At that point

        There are a few other points to note about this circuit.

• The resistor in parallel with the capacitor may well be the internal resistance of the capacitor.

• The simulation above assumes that the diode is an ideal diode.  That's pictured below with the bold red graph.

 We can add a "bias" to the diode model to get closer to reality.  The voltage-current curve for a better model is shown below.  Click the button to see how closely the model comes to the v-i curve for a "typical" diode.