In the diode study it was established that a silicon diode when in the conduction zone presents a voltage drop of 0.7 V.
There are many circuits where this voltage drop is not desired. Thus,
operational amplifiers help to substantially soften this value in some circuits of interest. In the following, we will review some
types of circuits previously studied, but we will use the operational amplifier next to the diodes and study their behavior and performance.
It is noteworthy that when we use an Opamp with the halfwave rectifier function, we are obviously limited in power
supply by the characteristics of the Opamp used. In general, its use is in small signals,
that is, low power.
In the Figure 66-01 we present the circuit of a half wave rectifier using an operational amplifier. Note that this
circuit has two outputs. One only with the positive sinusoid semicycle
(V +) and the other with the sinusoid negative semicycle (V -).
Figure 66-01
Thus, when the positive semicycle of the sinusoid arrives at the negative input of the operational amplifier, it will
appear inverted at the output.
Thus, the D1 diode enters the conduction zone as it is directly polarized. So we have a feedback loop
between the output and the negative input via path 1-a-m. Since the circuit gain is equal to - 1, the positive
semicycle of the input signal appears integrally at the output
V -, offset by 180° relative to the input signal.
And when the sinusoid's negative semicycle reaches the negative input of the operational amplifier, at the output the
signal appears inverted, so the D1 diode enters in the cutting zone and the D2 diode enters
the conduction zone. Thus, the path 1-c-m will be the operational amplifier feedback circuit with a
voltage gain of - 1. So, the input signal appears fully in the output V + out of phase
in 180° relative to the input signal.
Observation - Importantly, if this circuit is designed without one of
the feedback loops, if only one of the outputs is desired, it may happen that the operational amplifier goes into saturation,
significantly reducing the rectifier response.
To achieve a full wave rectifier, we can rely on the halfwave rectifier circuit and use a small artifice to achieve
the goal. We used a second operational amplifier in the inverter configuration and a summing circuit. Thus, by summing
the output of the half wave rectifier with the input signal, we obtain a full wave rectifier. The trick is to make the
output signal of the half wave rectifier amplified by a factor
- 2 while the input signal is amplified by a factor - 1.
What is the purpose of this? According to Figure 66-02,
OP1 displays at the output (point b) the rectified and lagged signal of 180° relative to the input.
And this signal will be added to the non-inverted signal (point c), that is, there will actually be a subtraction
of the two signals. To get a valid signal on the output of OP2, it reverses the result of "subtraction"
and we have - (-2 - (-1)) = +1.
Figure 66-02
Note that only the rectifier output for the positive semicycle of the sine has been used. For the sinusoid negative
semicycle, the rectifier output is null and the OP2 inverts the input signal that appears on its inverter input.
Thus, at the output of OP2 we have the negative input semicycle appearing as a positive semicycle in the rectified signal.
All this transformation can be accompanied by
Figure 66-03.
In the previous chapter we saw some types of limiting circuits. In this chapter we will add a
operational amplifier to the circuit and study its behavior.
In the Figure 66-04 we see a typical simple limiter circuit using an operational amplifier.
Note that in series with the diode we have a battery with voltage equal to V1. Of course,
the voltage of this battery plus the conduction voltage of diode D sets the limit value for the circuit
to operate as a limiter. Let's take a look at yours.
Figure 66-04
While the positive semicycle of the input voltage, Vi, does not reach the value of
VD + V1, the diode D will be in the cutting zone. So the voltage at the output
of the operational amplifier, Vo, is given by:
If Vi ≤ VD + V1⇒ Vo = - (Rf / Ri ) Vi
When the positive semicycle of the input voltage exceeds the value of
VD + V1, the diode D enters in the conduction zone. Then the voltage at
the output of the operational amplifier, Vo, will be limited to this value, but with the inverted
polarity, because the operational amplifier is in the inverter configuration, that is:
If Vi > VD + V1⇒ Vo = - ( VD + V1 )
Figure 66-05
See Figure 66-05 for the circuit transfer characteristic graph. Note that for the negative semicycle input,
the circuit output has a positive value, obeying the slope of the line whose value is given
by - (Rf / Ri) . This inclination is obeyed even when the input voltage is in the positive
semicycle until it reaches the value of VD + V1. For values higher than this the output
voltage is limited to - (VD + V1 ).
We can create a circuit where the output is limited to a positive value by simply reversing the direction of the diode
and the battery. We can see in Figure66-06 a typical example of this circuit.
Figure 66-06
Note that in this circuit there will be no limitation on the entire positive semicycle of the input voltage. In negative semicycle,
as long as the input voltage does not reach the value - (VD + V1 ) output voltage will obey the
ratio - (Rf / Ri). Therefore the output voltage will be given by:
If Vi ≥ - (VD + V1 ) ⇒ Vo = - (Rf / Ri ) Vi
When the input voltage reaches the value - (VD + V1 ) output voltage will be
limited to the value + (VD + V1 ). Therefore the output voltage will be given by:
If Vi < - (VD + V1 ) ⇒ Vo = + ( VD + V1 )
Figure 66-07
See Figure 66-07 for the circuit transfer characteristic graph. Note that for the positive semicycle input,
the circuit output has a negative value, obeying the slope of the line whose value is given
by - (Rf / Ri) . This inclination is obeyed even when the input voltage
is in the negative semicycle until it reaches the value of - (VD + V1 ).
For values lower than this the output voltage is limited to + (VD + V1 ).
Evidently, by joining the last two circuits studied, we can elaborate a limiting circuit double , as shown in the Figure 66-08.
Figure 66-08
Based on what we have already studied, we can easily see that the upper branch is responsible for the operation
in the positive semicycle of the input voltage, while the lower branch is responsible for the operation in the negative
semicycle of the input voltage. In this case, the circuit works exactly as stated above. So, let's not repeat the explanation.
Of course the circuit transfer characteristic graph changes slightly and we present it in the Figure 66-09.
Figure 66-09
Notably we have three possible situations for the output voltage as shown below.
If Vi < - (VD2 + V2 ) ⇒ Vo = + ( VD2 + V2 )
If - (VD2 + V2 ) ≤ Vi ≤ - (VD1 + V1 ) ⇒ Vo = - (Rf / Ri ) Vi
The circuits studied in this item use batteries to achieve the desired limiting point. The use of batteries is only a didactic resource. In practice, we replace it with a resistive network that meets the project objectives. Thus, in Figure 66-10 we can see a circuit that uses resistive dividers in order to achieve the limiting points. Let's look at how this circuit works.
Figure 66-10
As soon as the input voltage starts the positive semicycle, the output voltage starts its negative semicycle because the
operational amplifier is in inverter configuration. And both diodes are in cut. As long as the input voltage does not exceed
the threshold value for conducting D1, the output voltage follows the circuit voltage gain, or
Vo = - (Rf /
Ri ) Vi. To D1 enter in the conducting zone, the output voltage must reach
a negative value such that Va is - 0.7 V or more since the anode in
D1 is connected to ground via positive input from the operational amplifier. Therefore, to determine
the value of Va we can use the superposition theorem . So applying this theorem, we get:
To understand how we get these equations, we can rely on the circuit shown in Figure 66-11, replacing
the voltage of VD by Va.
Figure 66-11
Now we need to determine to what value of Vo the limiting circuit will start acting.
To do so, let us base on Figure 66-11 where we have the output circuit scheme for the boundary
condition where D1 will be conducting . In this case, the diode will not circulate electric
current, as indicated in the figure below and, therefore, the current that circulates for
R1 is the same as that circulating for R2. Based on this information, we can
write the equations of the two meshes to determine I, or:
So the limit for diode conduction, when Vo goes through the negative semicycle, let's call it
Vo-. After an algebraic work in the equation above, we come to:
Note that when Vo reaches the required value for diode D1
enter in the conducting zone, the voltage at point a is fixed at Va = - 0.7 V. And since
V is also constant, so the current over R1 remains constant. Therefore, an increase in
current through the diode must circulate through R2, generating an effect that R2
acts in parallel with Rf and the incremental gain is given by:
Av = - ( Rf || R2 ) / Ri
And when the input voltage is in the negative semicycle, the circuit transfer characteristic can be found identically to the one above. As long as the input voltage does not exceed the threshold value for conducting D2, the output voltage follows the circuit voltage gain, that is, Vo = - (Rf / Ri ) Vi. To calculate the value of Vb we employ the superposition method as previously done. Thus we find:
To find this equation, we rely on the circuit shown in Figure 66-12, with the removal of the battery VD from the circuit.
Figure 66-12
To determine the value of Vo where the limiter circuit starts acting, we repeat the previous process using the same considerations made in the opportunity. Note that when the diode
D2 starts conduction, point b will have its voltage set at + 0.7V . So based on the circuit of Figure 66-12 we can write:
At the limit for diode conduction, when Vo traverses the positive semicycle, we will call Vo + . After an algebraic work in the equation above, we come to:
Here - V is also constant, so the current over R4 remains constant. Therefore, an increase in current through the diode must circulate through R3, producing an effect as if R3 acts in parallel with Rf and the incremental gain is given by:
Av = - ( Rf || R3 ) / Ri
Finally we will present the graph of the circuit transfer characteristic. See the
Figure 66-13 .
