# Why design a clamper circuit with battery in series with a resistor?

I have been trying to design a circuit that clamps the positive peaks of an ac signal to +6 V. We can use batteries, resistors, and capacitors of any value desired in addition to Zener or conventional diodes. We are allowed 0.6 V for the forward drop.

The solution that the Professor has given me looks like this:

I understand how the first branch (with the Zener and the Silicon diode) is resulting in -6V when V-in is greater or equals to 6V during the positive half cycle, the diodes will conduct and the voltage at that branch will be the output voltage. I also am aware that the capacitor voltage during this period will be the difference between the Input Voltage and the diode Voltages.

However, I do not understand the use of that battery voltage in series with the resistance. How does it contribute to the design?

From Electrical Engineering:Principles and Applications, International Edition or 7th Edition: Principles & Applications Book by Allan R. Hambley. Chapter 9, Exercise 9.18, Solution shown in Figure 9.36.

• That circuit does not do what you describe. It will give a full amplitude AC signal at the output with the negative peak sitting at 6V. Commented Oct 29, 2023 at 20:29
• @KevinWhite Actually, the solution is also given in the book called Electrical Engineering:Principles and Applications(International Edition or 7th Edition, Book by Allan R. Hambley). Commented Oct 29, 2023 at 20:32

# The problem

Your professor should not give you complete circuit solutions as if they came from God but should formulate the problem and show you the way to solve it step by step. And it is very simple - figuratively speaking, to "move" itself an AC voltage with some value, e.g. 5 V. The same task is solved in bias circuits, voltage doublers, etc.

# DC implementation

## "Shifting" voltage source

To solve it, we need to add a 5 V constant voltage to the input AC voltage. So, we connect a 5 V DC voltage source in series to the input source Vin.

simulate this circuit – Schematic created using CircuitLab

Run Time-Domain Simulation and you will see that Vin is moved up by 5 V.

This is a conceptual circuit where anything is possible. In real circuits, however, it is not convenient to use such "floating" voltage sources because the power supply is grounded; charged capacitors are used instead. For purposes of understanding, however, it is convenient to initially think of capacitors as "rechargeable batteries". This allows us to explore circuits with the convenient CircuitLab Live DC Simulation.

## Ideal diode

Let's start from the very beginning with the problem that is solved by the simplest clamping circuit - "self-shifting" AC voltage by the value of its amplitude (for example, 1 V).

Vin > 0: During the positive half wave, the "capacitor" is quickly charged through the diode D to Voffs = Vin (for simplicity, I have used an "ideal" diode with VF = 0 V from the CircuitLab library).

simulate this circuit

Vin < 0: During the negative half wave, the "capacitor" voltage Voffs is added to Vin (I have vertically flipped the input source to see the polarities better), and the sum Vout = -Vin - Voffs = -2 V appears at the output. The diode is reverse biased and has no effect (as if it is not there).

simulate this circuit

The result of this is that the input voltage appears at the output shifted down by 1 V (by its amplitude), and for some reason beyond my understanding, this is called "clamping".

## Ideal "Zener diode"

We continue to follow the circuit evolution by improving the circuit...

In practice, it has become necessary to set this offset with a given value. Then they may have noticed that VF of the "bad" diodes they used was subtracted from Vin in the loop and the capacitor charged to a lower voltage; accordingly, Vin shifted less.

Vin > 0: So if we raise the voltage across our diode to 6 V (in the CircuitLab parameters window), the "capacitor" will charge up to 5 V, and the input voltage will drop that much. But wait... the polarity is reversed! So the input voltage will shift up with 5 V.

simulate this circuit

Vin < 0: During the negative half wave, the input voltage (1 V) is subtracted from the "capacitor" voltage Voffs (5 V). As a result, the negative input voltage appears shifted up to 4 V at the output. This voltage is less than the "Zener" voltage (6 V); so the "Zener diode" has no effect (as if it is not there).

simulate this circuit

## Real Zener diode

Real Zener diodes have the undesired (in most cases) feature when forward biased to behave like regular diodes (with VF = .7 V). We can solve this problem by connecting a forward-biased Si diode in series to the Zener diode (selected so that their total voltage is 6 V). This will also help us to get the exact 6 V value.

The operation of the circuit is the same as the one above.

simulate this circuit

simulate this circuit

# AC implementation

Although the circuit seems to work fine let's make sure by passing an initial (bias, quiescent) current through the diodes produced by an additional voltage source Vref and a resistor R in series. We also replace the emulating source with a real capacitor. Thus we get the OP circuit and decide to investigate it.

The capacitor needs to be recharged from time to time. If the load has a high resistance, this will happen very briefly when the diodes conduct.

simulate this circuit

But if the load has a low resistance, the capacitor will discharge more and require a longer charging time. As a result, the shape of the signal is distorted.

simulate this circuit

## The role of Vref-R

However, I do not understand the use of that battery voltage in series with the resistance. How does it contribute to the design?

The Zener diode (and the whole D-Z network) acts as a voltage stabilizer; we can think of it as a 5.4 V (6 V) voltage source.

simulate this circuit

Problem: But it is not a perfect "voltage source"; its IV curve is not strictly vertical but exponential. I have obtained it by the help of the DC Sweep Simulation applying a variable (sweeping) DC input voltage.

So when the current through it is too small or varies, the voltage across it is also small and varies, and this is the case here. As you can see in the unloaded circuit above, only a small charging current periodically flows through the D-Z diode network.

Solution: To keep its voltage constant, the (Zener) diode needs a more significant constant current. It is provided by the network Vref-R that acts as a simple 1 mA current source. But instead of "philosophizing" a lot, let's see it in practice by removing the network... how easy it is here!

simulate this circuit

The result confirms our assumptions - there is no current flowing, Vin and Vc are almost equal, there is no shift...

# Simplification

Finally, we ask ourselves the question, "Why is the diode necessary at all?" We come to the conclusion that this is only for obtaining a total voltage drop of 6 V. In the CircuitLab library, we find a suitable Zener diode that provides this voltage by itself, and thus simplifies the circuit.

simulate this circuit

As you can see, the graphical results are the same as above.

# Generalization

The circuit of the diode clamper reminds us of something familiar - the diode rectifier. And indeed, if we compare them, they turn out to be the same device consisting of three elements in a loop - an AC voltage source Vin, a diode D and a capacitor C. In both, the AC input voltage is rectified by the diode and the capacitor is charged to its amplitude. The diode is switched on only for short moments (when the capacitor is recharging) and has almost no effect on the load.

## Diode circuits

Clamper: Here, the load is connected in series to the capacitor.

simulate this circuit

As a result, the input voltage is "shifted" by the value of the capacitor voltage (AC + DC voltage is applied to the load).

Rectifier: In this case, the load is connected in parallel to the capacitor.

simulate this circuit

As a result, only the capacitor DC voltage is applied to the load.

## Zener diode circuits

Clamper: By changing the (Zener) diode forward (backward) voltage, we can adjust the "shifting" voltage.

simulate this circuit

As a result, the input voltage is "shifted" by a higher value.

Rectifier: In this case, the voltage drop (VF = .7 V) across the diode is undesired.

simulate this circuit

... since a lower voltage is applied to the load.

• I understood quite a lot, but I have one question. Must I be familiar with quiescent current to understand the reason of the 15 V that is in series with the resistor? My Professor yet did not introduce us to quiescent. It will take a week or two. Commented Oct 31, 2023 at 19:27
• I understood everything now, Thanks a lot. "As you can see in the unloaded circuit above, only a small charging current periodically flows through the D-Z diode network. To keep its voltage constant, the (Zener) diode needs a more significant current. It is provided by the network Vref-R that acts as a simple current source. But instead of "philosophizing" a lot, let's see it in practice by removing the network... how easy it is here!" This was the main headache of mine but now I understood! Thanks a lot Commented Oct 31, 2023 at 19:40
• @RK Eshat, Thanks for the response. A piece of advice from me that will serve you well in life: When you want to understand something, don't use common professional terms; they have another purpose. Think simply and express yourself in the simplest words. When you understand it, you can allow yourself to express yourself "scientifically" in order to be respected:-) It is humor but there is also some truth in it. So here you have to pass a current through a (Zener) diode to make its voltage strictly constant. See the schematic 5.3 and the explanations I added for you. Ask more questions. Commented Nov 1, 2023 at 9:44

The battery and resistor are used to set the quiescent bias conditions for the diode + zener branch. The diode + zener has to be biased to have well-defined characteristics, such as voltage across terminals, and internal impedance.

Quiescent conditions: consider the case when vin is zero ie: no signal. Assume R = 9k. The voltage across the R will be 9V (15V - 6.0V) so the current flowing in the loop formed by the battery, R, and the diode + zener, will be 9V/ 9k = 1mA.

If a positive signal is applied, then the diode + zener will appear to be a short-circuit to that signal and most of the signal current will flow in the diode + zener leg, so the result will be Vo will be unchanged from the bias state.

If a negative signal is applied, then the signal current will cause the diode + zener current to reduce from 1mA toward zero. When this current reaches zero, the voltage at Vo+ is no longer clamped to 6V, and so can move downward to follow the signal. The signal current will now be determined by the input signal voltage, the battery, and the resistor - the diode blocks any current flowing in the diode+zener leg.

Note that the circuit as shown does not include the internal impedance of the signal source. To get full understanding of this circuit, one must consider both the internal impedance of the signal source (Vin), and the impedance of the coupling capacitor C. The battery also has an internal impedance, but we can assume this is insignificant compared to the value of R. However, to obtain a basic understanding, we can assume all of these impedances to be zero for the intended operating frequency.

Hope this helps.

• Good answer although it doesn't show well the idea of ​​the circuit during the positive half wave. Commented Oct 30, 2023 at 6:19
• @Circuitfantasist What I wrote about that was: "If a positive signal is applied, then the diode + zener will appear to be a short-circuit to that signal and most of the signal current will flow in the diode + zener leg, so the result will be Vo will be unchanged from the bias state." | What more is there to be said? Commented Oct 30, 2023 at 21:55
• Fabio Barone, This is true, but my thought was that it would be good to say that, during the positive half-wave, this is how the capacitor recharges; this is the essential thing in this case. The "diode short-circuit" actually means that the input source is connected to the capacitor and charges it. Commented Oct 31, 2023 at 5:15
• I am sorry but I am quite unfamiliar with quiescent bias conditions. I think this what we will learn in two weeks. Is this something that is necessary to understand this circuit? Commented Oct 31, 2023 at 19:04
• @RKEshat Yes, it is quite important to understand the concept of " quiescent bias conditions". This is the condition when the signal being applied to the circuit is zero. This allows us to greatly simplify the circuit to its "small-signal equivalent model", where we can simplify the real-world complex behaviour into quite simple (linear) behaviour, which allows an intuitive understanding. It seems you are a student of electronics, so yes, I encourage you to read up on "small-signal analysis" and "small-signal model". Enjoy! Commented Nov 1, 2023 at 3:41

Let's rearrange the circuit into this schematic:

simulate this circuit – Schematic created using CircuitLab

We assume ideal components. An actual built circuit will have certain deviations.

The static case is simple. Apparently D is conducting with a forward voltage of 0.6V, and ZD maintains its zener voltage of 5.4V. This sums up to 6V at V_o. C is charged to any voltage difference between V_in and 6V.

Now let the voltage raise at V_in. C tries to maintain its charge, but D and ZD are "shorting" it and take any current from C, maintaining 6V at V_o. The voltage across C is always the difference between V_in and 6V. This continues as long as the voltage at V_in rises.

When V_in stops rising, there is no more change in the charge of C. We have the static case again.

Now let the voltage fall at V_in. C tries to maintain its charge, and because its right leg's voltage falls, D and ZD do not conduct any more. Consequently, the current through R goes into C, changing its charge and therefore changing its voltage. V_o follows the fall of V_in, but the values of R and C determine how exactly. This continues as long as the voltage at V_in falls.

You ask about the contribution of the "battery" and the resistor. In the static case, they provide the source for D and ZD to maintain 6V.

In the dynamic case, R builds with C a high-pass filter for the AC voltage.

• Very well... but what's the point of all this? Commented Oct 30, 2023 at 8:31