# How can the diode conduct in negative half cycle of AC signal in clamper circuit?

Could you kindly help me understand the clamper circuit. I watched a couple videos and I'm confused. My confusion is in positive clamper circuit. How can the diode conduct during the negative half cycle of ac signal? Isn't diode supposed to conduct only by positive voltage?

Second thing is how can during the positive half cycle of AC signal the negative side of capacitor accept the positive voltage? I will provide the picture that will hopefully make it easy to understand.

Picture was used from this site

https://electronicscoach.com/clamper-circuits.html

• Have a look at here. Commented Jul 22 at 11:58

Despite this circuit having only two (or three, if you include the resistor) elements, it is hard explain without sticking strictly to the algebra and KVL, and separating the roles of current and voltage. After you "see it" the first time, it seems obvious, but I can't seem to find an analogy to describe it. Perhaps there's none, and that's why the internet doesn't have any decent "explain like I'm 5" type treatments.

This circuit performs what is known as "DC restoration", a name which will make more sense towards the end. I'll try to keep it intuitive, but you'll need to closely follow the algebra.

Forgetting the resistor for the moment, and assuming that the capacitor is initially discharged, figure out what will happen in the following two scenarios:

simulate this circuit – Schematic created using CircuitLab

The ±9V input signal is just some arbitrary amplitude I've chosen for illustration. It could be any amplitude.

On the left, where the top node IN is held at potential $$\V_{IN}=+9V\$$, which represents the positive half of the input cycle. Current must flow from the higher potential to lower, and since 0V (which is lower) is at the bottom, current direction wwould be downwards. I say "would be" because that's not permitted by the diode, and in this state, current is $$\I=0\$$.

On the right, $$\V_{IN}=-9V\$$, representing the negative half of the input cycle. Here current still must must flow from the higher potential to lower, but now it's upwards. The diode won't oppose that and $$\I>0\$$ upwards.

By this reasoning, the answer to your question "how can the diode conduct during the negative half cycle of ac signal?" is that it can only conduct during the negative half of the cycle. I can only assume that you've somehow got current direction confused.

In what follows, the signs may get confusing, and you'll have to follow very closely. I must be consistent here, for the algebra to work. The voltage $$\V_{C1}\$$ across capacitor C1 is the potential difference between nodes OUT and IN, like this:

$$V_{C1} = V_{IN} - V_{OUT}$$

By that convention, a positive value of $$\V_{C1}\$$ represents the state where its upper terminal (node IN) has the higher potential. Conversely, if $$\V_{C1}\$$ is negative, then this implies that C1's lower terminal (node OUT) has the higher potential. Another way of writing that would be:

$$V_{OUT} = V_{IN} - V_{C1}$$

Let's see what happens to the voltages, again assuming that the capacitor is initially discharged, $$\V_{C1}=0V\$$. On the left, there's no current. With no current flowing through the capacitor, it doesn't charge or discharge, and the voltage $$\V_{C1}\$$ across it, which started at $$\V_{C1}=0V\$$, will not change:

$$V_{OUT} = V_{IN} - V_{C1} = (+9V) - (0V) = +9V$$

This state of affairs won't last very long, and you'll never see it again, after the input $$\V_{IN}\$$ becomes negative, as shown above right. To analyse that condition, I'm going to assume an ideal diode, that has no impedance when it conducts, and therefore zero-volts across it, just like a wire. I'll ignore that real diodes have 0.7V across them when they conduct; that's a can of worms that nobody should open on a Monday.

When $$\V_{IN}=-9V\$$ (shown above right), lots of current flows, as we established before, because the diode is forward biased. This will rapidly charge C1, to the full −9V potential difference. Intuitively, when you picture the conductive diode as a short circuit in this state (effectively a wire), the lower end of C1 (node OUT) will quickly end up at 0V, while the upper end (IN) is being held at −9V by our source of input (whatever that may be). The capacitor will have charged to this potential difference:

$$V_{C1} = V_{IN} - V_{OUT} = (-9V) - (0V) = -9V$$

Since we are trying to analyse the output of this circuit, it's more interesting to arrange that to give $$\V_{OUT}\$$ as the subject:

$$V_{OUT} = V_{IN} - V_{C1} = (-9V) - (-9V) = 0V$$

Note: While it's not too relevant to this answer, it might also help to realise that since positive current is flowing upwards through C1, it's the lower end of C1 (OUT) that must be rising in potential with respect to the upper end (IN), in accordance with passive sign convention. Check for yourself that this is in line with the condition $$\V_{IN}=-9V\$$, $$\V_{OUT}=0V\$$.

Now for the good stuff. What happens if we start to raise $$\V_{IN}\$$ again, slowly back to +9V? Remembering that C1 now holds a charge, keeping its lower terminal OUT to be 9V higher in potential than its upper terminal IN, then by raising $$\V_{IN}\$$ we also raise $$\V_{OUT}\$$. OUT will follow IN, but always 9V higher in potential. Algebraically:

$$V_{OUT} = V_{IN} - V_{C1} = V_{IN} - (-9V) = V_{IN} + 9V$$

Critically, since as we raise $$\V_{IN}\$$ we are also causing $$\V_{OUT}\$$ to rise above 0V, diode D1 becomes reverse biased, passing no current. With no current through C1, its charge and potential difference remain constant at $$\V_{C1} = -9V\$$. In fact, this will be the case the whole time $$\V_{OUT}\$$ remains above 0V. Effectively, C1 is now "level-shifting", always adding 9V to $$\V_{IN}\$$. I reiterate:

$$V_{OUT} = V_{IN} + 9V$$

As long as $$\V_{IN}\$$ never again falls below −9V:

• $$\V_{OUT}\$$ can never fall below 0V

• the diode can never become forward biased

• no current can ever flow

• $$\V_{C1}\$$ cannot change

• Input/output relationship remains $$\V_{OUT} = V_{IN} + 9V\$$

As long as $$\V_{IN}\$$ never again falls below −9V, this circuit will "elevate" the input signal potential by 9V, such that it never goes negative. That is how this circuit "restores DC". That is how the output will be a copy of the input, but sitting upon the horizontal 0V axis on a graph.

The purpose of the resistor in your original circuit is to discharge the capacitor slowly, so that the circuit can (over the long term) adapt to keep the output "troughs" sitting on 0V, even as input amplitude changes.

• Makes perfect sense.Thank you very much.The reason why I'm I confused cause the information on ytb or Google it varies in explanation of this circuit. Commented Jul 22 at 19:40

Yes the diode will conduct when voltage over the diode is positive.

But please note the orientation of the diode.

During negative input cycle the diode voltage is positive, so the current flows through diode.

During positive input cycle the diode voltage is negative, so current does not flow through diode.

• How when its negative voltage.Negative half cycle is negative voltage ? Commented Jul 22 at 11:42
• Let's say the positive peak voltage is 5 volts negative peak is 5 average is 0.And for the sake of argument let's say diode needs 0.7 volts to conduct.Explain me this where did you get positive volage in that negative voltage.? Commented Jul 22 at 11:51
• @MarjanOvcar Do you know which way diodes conduct and which way they block? If you flip the diode upside down to conduct only downwards, it sees positive voltage for positive input voltage. But in that circuit the diode is pointing upwards so it sees positive voltage for negative input voltage and conducts upwards. Same thing if you measure 9V battery with multimeter, it's +9V in one orientation and -9V in the other orientation. Commented Jul 22 at 11:55
• Oke that make some sense.How about the capacitator part? Commented Jul 22 at 12:01
• @MarjanOvcar The capacitor can charge only when diode conducts, so on negative cycle. Therefore the left side of capacitor has lower voltage than right side, just as depicted by the sine wave diagrams. Commented Jul 22 at 12:25

# Basic idea

The idea behind the so-called "clamper circuit" is very simple:

A floating constant voltage source is connected in series to the AC input voltage. Thus their voltages are added in a series manner and the resulting voltage is applied to the load. The constant voltage source is implemented by a capacitor that is periodically charged through a diode switch. Ideally (no load) the diode switch turns on intermittently until the capacitor is charged and then remains off. In the real case (there is a load), the diode switch is turned on periodically for a short time to recharge the capacitor.

From this viewpoint, this circuit, which textbooks treat with such awe, is essentially just... a "battery" connected in series with the input voltage!

# Diode-capacitor circuits

There is nothing special about this diode-capacitor circuit because it is actually the well-known half-wave diode rectifier with the positions of the diode and capacitor swapped. Let's for the sake of good understanding examine the two circuit solutions in parallel.

## Half-wave rectifier

Here the voltage across the capacitor is of interest and it is applied to the load.

simulate this circuit – Schematic created using CircuitLab

As we can see from the Time-Domain Simulation, after the capacitor is quickly charged, its voltage remains almost constant.

With this high-resistance load of 1 MΩ, the diode turns on only briefly to recharge the capacitor. So only small current peaks are visible on the graph.

## Clamper circuit

In this (OP's) case, after the capacitor is rapidly charged, its voltage is added to the input voltage and the resulting pulsating voltage is applied to the load. As a result, the input voltage is "lifted" by the value of the DC voltage across the capacitor.

simulate this circuit

As can be seen from the graph below, the amplitude of the voltage is doubled. This is the simplest voltage doubler known as Villard circuit.

RL = 1 MΩ

We can see the same small current peaks are on the graph.

RL = 20 kΩ: If we increase the load (decrease RL), the current peaks become larger because the capacitor discharges more and requires a larger amount of charge to recover.

# Conceptual circuits

The operation of these circuits can be illustrated in a simple way if we replace the input source with a constant voltage source, the charged capacitor with a battery and the diode with a switch. This will allow us to explore the circuit operation through a DC Live Simulation by simply hovering the mouse over the circuit elements.

## "Half-wave rectifier"

Vin < 0: During the negative input half wave, the "capacitor" is charged by a negative input voltage source to -10 V.

simulate this circuit

Vin > 0: During the positive input half wave, the charged "capacitor" is disconnected from the input source; so the voltage across it (in ideal case) stays at -10 V.

simulate this circuit

So ideally the circuit is just a charged capacitor.

simulate this circuit

## "Clamper circuit"

Vin < 0: During the negative input half wave, the "capacitor" is charged by a negative input voltage source to 10 V with the polarity shown in the schematic below.

simulate this circuit

Vin > 0: During the positive input half wave, the total voltage (Vin + Vc) is applied to the load.

simulate this circuit

So ideally the circuit is a voltage source and a charged capacitor in series.

simulate this circuit