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# ENTC 3320 - PowerPoint PPT Presentation

ENTC 3320. Absolute Value. Half-Wave Rectifier. The inverting amplifier is converted into an ideal (linear precision) half-wave rectifier by adding two diodes. . - 0.6 V. 0 V.

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### ENTC 3320

Absolute Value

• The inverting amplifier is converted into an ideal (linear precision) half-wave rectifier by adding two diodes.

- 0.6 V

0 V

• When E, is positive, diode D1 conducts, causing the op amp’s output voltage, VOA,to go negative by one diode drop (-0.6 V).

• This forces diode D2 to be reverse biased.

• The circuit’s output voltage Vo equals zero because input current I flows through D1.

• For all practical purposes, no current flows through Rf and therefore Vo = 0.

• This causes D2 to conduct.

• The circuit then acts like an inverter, since Rf = Ri, and Vo =+E1.

• Since the (—) input is at ground potential, diode D1 is reverse biased.

• Input current is set by E/Ri and gain by ─Rf/Ri.

• Remember that this gain equation applies only for negative inputs, and Vo can only be positive or zero.

• The most important property of this linear half-wave rectifier will now be examined.

• An ordinary silicon diode or even a hot-carrier diode requires a few tenths of volts to become forward biased.

• Any signal voltage below this threshold voltage cannot be rectified.

• However, by connecting the diode in the feedback loop of an op amp, the threshold voltage of the diode is essentially eliminated.

• For example. in Fig. 7-2(b) let rectifier will now be examined. E, be a low voltage of —0.1 V. E, and R, convert this low voltage to a current that is conducted through D2.

• VOA goes to whatever voltage is required to supply the necessary diode drop plus the voltage drop across R~. Thus millivolts of input voltage can be rectified, since the diode’s forward bias is supplied automatically by the negative feedback action of the op amp.

• Finally, observe the waveshape of op amp output rectifier will now be examined. V~ in Fig. 7-3. When E~ crosses 0 V (going negative), V(~ jumps quickly from —0.6 V to +0.6 V as it switches from supplying the drop for D2 to supplying the drop for D1. This jump can be monitored by a differentiator to indicate the zero crossing. During the jump time the op amp operates open loop.

• The output voltage rectifier will now be examined. Vo equals 0 V for all negative inputs.

• Circuit operation is summarized by the plot of V~ and VOA versus E.

7-1.4 Signal Polarity Separator rectifier will now be examined.

• The following circuit is an expansion of the previous circuits.

• When E, is positive, diode D1 conducts and an output is obtained only on output V0,.

• V0, is bound at 0 V.

• When E, is negative. D2 conducts, V0. = —(—E,) = +E,. and V0~ is bound at 0 V.

Introduction waveshapes.

• The precision full-wave rectifier transmits one polarity of the input signal and inverts the other.

• Thus both half-cycles of an alternating voltage are transmitted but are converted to a single polarity of the circuirs output.

• For example, the absolute values of |+2 | and | shown below.2 | are +2.

• The symbol | | means “absolute value of.”

• In a precision rectifier circuit the output is either negative or positive, depending on how the diodes are installed.

• Three types of precision rectifiers will be presented. shown below.

• The first is inexpensive because it uses two op amps. two diodes, and five equal resistors.

• Unfortunately. it does not have high input resistance.

• ~o a second type is given that does have high input resistance but requires resistors that are precisely proportioned but not all equal.

• Neither type has a summing node at virtual ground potential.

• Thefirst type of precision full-wave rectifier or absolute-value circuit is shown below.

• Figure resistance equal to 7-8(a) shows current directions and voltage polarities for po~iti~e input signals. Diode D~ conducts so that both amps A and B act as inverters, and V0 = +E,.

• Figure 7-8(b) shows that for negative input voltages, diode D.~ conducts. Input rent I divides as shown, so that op amp B acts as an inverter. Thus output voltage V0 positive for either polarity of input E, and V0 is equal to the absolute value of E,