Clipper circuit in OpAmp

Question

I came across this clipper circuit:

I was analyzing this circuit. I could not very clearly understand what is the purpose of resistance, R here.

I understand that if we do not include this resistor, it would mean that we are actually assigning a fixed voltage to Vo.

What I do not understand is how its presence is solving 'some' issues. How is the voltage at Vo being affected by this resistance?

Answer 1

This specific circuit solution has long been explained... and probably many more after that. But for all of them to be truly understood, we need to see the general idea behind them.

Let's play around with the circuit in CircuitLab and learn some wisdom about this kind of op-amp circuits with "dynamic negative feedback".

STEP 1: Op-amp follower undisturbed

In a voltage follower, the op-amp effortlessly accomplishes its task of keeping its output voltage equal to the input voltage.

^{simulate this circuit – Schematic created using CircuitLab}

STEP 2: Op-amp follower disturbed

But its task becomes more complicated if some disturbance occurs between its output and the load. Now it has to make an extra effort to overcome it.

Resistance disturbance

Resistance is the simplest and most common disturbance to an op-amp. Then let's put a 1 k resistor in the op-amp output to see how it will react to our intervention. Let's also connect two voltmeters - before (V2) and after (V3) the resistor.

To simplify the circuit, let's apply a little trick that CircuitLab allows us to do - instead of connecting a load, make the voltmeter V3 perform its function. To do this, open the V3 parameters window and set a relatively small resistance (1 k), i.e., deliberately "worsen" the voltmeter. So remember that V3 is actually a 1 k resistor (just "voltage-visualized"). With the same success, we can replace R with a 1 k "bad" ammeter, but let's not overdo it :-)

^{simulate this circuit}

We see that with a -1 V input voltage applied, the op-amp is forced to produce twice the output voltage (-2 V) in order to have a -1 V voltage across the "load" (V3). Thus of that 2 V, 1 V is lost across the resistor, and 1 V is left for the load.

From the DC sweep simulation below, we see that the op-amp is actually forced to amplify 2x before the disturbance to follow 1x it.

This is actually the idea of ​​a non-inverting amplifier that we arrived at in a natural way (not as in the textbooks given ready-made). Let's state it and put it in the "bag of circuit tricks": A non-inverting amplifier is a "deliberately disturbed" follower.

Voltage disturbance

Now we can "provoke" the op-amp with various disturbances to see how "reliable" it is. Thus, the resistance above was linear (proportional, multiplicative) disturbance. Let's now try with some nonlinear (constant, aditive) disturbance, for example with a constant voltage source. It is "floating"; so let's use a battery with the same voltage (1 V) as the voltage drop across R above (I have chosen illustrative convenient values ​​of the quantities).

^{simulate this circuit}

Now, the op-amp steps down its output voltage by an additional constant value of 1 V to compensate for this constant voltage disturbance.

Low diode disturbance

Diodes, like batteries, maintain a constant voltage across themselves when forward-biased. Then let's replace the battery with a diode and thus get even closer to the OP's circuit. For simplicity, I have used an "ideal" diode from the CircuitLab library and ajusted its forward voltage to the favorite 1 V.

^{simulate this circuit}

So, when the input voltage is negative, everything is the same as above because the diode behaves as a "battery".

High diode disturbance

But at a positive input voltage, the diode is reverse biased and the op-amp's task becomes overwhelming.

^{simulate this circuit}

In an effort to compensate for this gigantic disturbance, it continuously raises its output voltage until it reaches the supply voltage and stops there. The output voltage at point 3 is zero since the load connects it to ground.

Let's see it over the entire input voltage range (-5 V ÷ 5 V).

So the circuit behaves as a precision zero voltage clipper.

Current disturbance

Thanks to these experiments, we gradually realized that an op-amp with negative feedback behaves like an almost "ideal" voltage source because connecting a grounded load does not change its output voltage. And how will it react if we connect the load to some voltage, for example 1 V, instead of ground?

^{simulate this circuit}

It turns out that the op-amp does not care about this and its output voltage exactly follows the input voltage.

Aha... we gradually start to figure out what is going on here: When connected in parallel with a real voltage source (with Ri = 1 k), the "ideal" voltage source (with Ri = 0) imposes its voltage (0 V) as stronger; when the ideal source us disconnected, the weak real voltage source sets the output voltage at point 3 through the resistance R equal to its voltage (1 V).

Op-amp clipper

So that was the idea behind this op-amp clipper:

• If Vin < Vref, the diode switch connects the op-amp output ("ideal voltage source") to the load (real voltage source) and imposes its voltage to the output 3).

• If Vin > Vref, the diode switch disconnects the op-amp output from the load supplied by Vref and the latter determines the output voltage (3).

^{simulate this circuit}

Passive diode clipper

The same idea is exploited by the passive version of this clipper circuit only there the diode forward voltage is not compensated and must be taken into account. Also the current of the reference voltage source flows through the input source.

^{simulate this circuit}

If you set VF = 0 V in the D parameters window, the graph will be the same as in the op-amp version above (that is, the op-amp makes the real diode ideal).

STEP 3: The general idea

Finally, let's summarize the idea:

• If we connect an ideal voltage source to a real one in parallel, the ideal source determines the voltage across them.

^{simulate this circuit}

• If we disconnect the ideal voltage source from the real one, the real source determines the voltage.

^{simulate this circuit}

In this way, two voltage sources can be switched to a common load with a simple "on-off" transistor (SPST) switch. In the straightforward solution, they are switched by the more complex SPDT switch.

Of course, this convenience comes at a cost - a higher output resistance in one state. An analogy can be made between these two configurations and the two types of transistor amplifier stages - with a collector resistor and "push-pull".

STEP 4: Implementation

The problem is that we usually only have ideal (perfect) sources. Then we have to artificially worsen an ideal source to make it real (imperfect).

For this purpose, we connect a resistor in series to an ideal voltage source. Thus the combination of the two elements acts as a real voltage source.

STEP 5: What is the resistor for?

It's time to answer the OP's question (a little late indeed, after more than three years :-)

I could not very clearly understand what is the purpose of resistance, R here.

The resistor is needed to "worsen" the reference voltage source. Without it, a conflict between the two ideal voltage sources would result.

Answer 2

If the value of the input voltage Vi is less than the value of the reference voltage Vref, then the diode D1 conducts. Then, the circuit given above behaves as a voltage follower. Therefore, the output voltage V0 of the above circuit will be same as that of the input voltage Vi, for Vi < Vref

If the value of the input voltage Vi is greater than the value of reference voltage Vref, then the diode D1 will be off. Now, the op-amp operates in an open loop since the feedback path was open. Therefore, the output voltage V0 of the above circuit will be equal to the value of the reference voltage Vref, for Vi > Vref

Answer 3

Without the resistor no current would flow through the diode (the current consumption of the opamp input is incredibly small). Its also there as a pull down resistor, without it the output voltage would just go up and stay there, like in a "Peak detector circuit" (look on google images) where the resistor replaced with a capacitor.

Another circuit that's a little similar is the diode voltage reference, which is a resistor and diode in series

Answer 4

Think of the diode as a component that has very low ON resistance \$R_{D(on)}\$, and very high OFF resistance \$R_{D(off)}\$. Also, assume that resistance R's value is chosen so that

$$ R_{D(on)} \lll R \lll R_{D(off)} $$

Next, note that two separate signals are vying for control of the voltage at node \$V_O\$: the op amp's output signal \$V_{O}^\prime\$ and \$V_{ref}\$. Voltage \$V_{O}^\prime\$ connects to \$V_{O}\$ through the diode's resistance \$R_{D}\$. Voltage \$V_{ref}\$ connects to \$V_{O}\$ through resistance \$R\$.

Case 1:  The diode is forward biased (it is ON).
In this case the diode's ON resistance is several orders of magnitude less than the resistor's resistance, i.e., \$R_{D(on)} \lll R\$. Current takes the path of least resistance. Therefore, voltage \$V_{O}^\prime\$ has the path of least resistance to node \$V_{O}\$, and \$V_{O}^\prime\$ (not \$V_{ref}\$) has the greatest influence in determining the voltage \$V_{O}\$. In this case, the resistor's job is to isolate voltage source \$V_{O}^\prime\$ from voltage souce \$V_{ref}\$ when the diode is ON.

Case 2:  The diode IS NOT forward biased (it is OFF)
In this case the diode's OFF resistance is several orders of magnitude greater than the resistor's resistance, i.e., \$R \lll R_{D(off)}\$. Current takes the path of least resistance. Therefore, voltage \$V_{ref}\$ has the path of least resistance to node \$V_{O}\$, and \$V_{ref}\$ (not \$V_{O}^\prime\$) has the greatest influence in determining the voltage \$V_{O}\$. In other words, with the diode OFF, the voltage at node \$V_{O}\$ is pulled up to voltage \$V_{ref}\$ through resistor R.