clamping circuits.......mechansim

[mj1709]

hey, i just wish to clarify the working principles of a clamping circuit.......
Negative Clamping Circuit:-
1. Like when the capacitor is charged in the first half cycle of the
input signal(wudnt it take long time as the time constant of the capacitor used is large?)

2. how does the sinusoidal waveform of the input signal retain itself when the voltage of the
capacitor(which acts as the supplier of voltage) becomes constant,after getting charged (due to large time constant)?

3. what do we mean when we say that clamping circuits shift the dc level of the input signal?

 

[mj1709]

plz sumone reply.....

 

[duke37]

Give me time to get up and open one eye.
A capacitor passes AC but blocks DC so if a diode is placed between the output and ground, the waveform will be raised (or lowered) so that one extreme of the waveform is at ground potential.
A high value resistor is used to keep the waveform clamped ot the right level.

Duke

 

[mj1709]

@duke

thnx.... well basically i have a problem at theory................it wud be nice if u could tell me what do we theoretically mean when we say
1. capacitor allows AC to pass while it blocks DC................?
2. plz see the attachment........

 

[Digital_Angel_316]

plz sumone reply.....

 

[duke37]

The AC signal on the input varies around the reference point. The capacitor passes this waveform but it cannot go below 0V because of the diode, therefore the capacitor is charged by the diode current to raise the voltage above the reference level.
The capacitor must be large enough to pass the current without significant attenuation.

Duke

 

[mj1709]

@DUKE all fine but my simple point is that when you charge the capacitor by the diode current..the load does not get any current right...since it is shorted?

 

[(*steve*)]

No, because as charge flows into the capacitor its voltage rises. This charge can be used to power the load.

 

[duke37]

The output voltage on the first negative half cycle will be attenuated but once the circuit is running, the output is as shown in my little diagram.

Duke

 

[mj1709]

@steve wud it wont hold true if the diode is ideal right.......?

 

[Digital_Angel_316]

The AC signal on the input varies around the reference point. The capacitor passes this waveform but it cannot go below 0V because of the diode, therefore the capacitor is charged by the diode current to raise the voltage above the reference level.
The capacitor must be large enough to pass the current without significant attenuation.

Duke

The voltage doesn't really 'shift', the signal is 'rectified' by the diode, conducting only on positive half cycles (in a half-wave rectifier - which is what a one-diode circuit provides) as shown in an earlier post. To achieve full rectification another diode must be added as in the diagram below:

Note that this circuit uses the center tap of the transformer to achieve the results. Many full wave rectifier circuits are connected across the entire transformer to get full voltage, and require a 'bridge' circuit of diodes as shown in the diagram below:

The output capacitor stores the output voltage in accordance with the charge-rate capacitor characteristic equations which are exponential (e) and a function of R, C and Time. The output includes a discharge element because of the time element of the charging/discharging, and is referred to as 'ripple':

 

[BobK]

@steve wud it wont hold true if the diode is ideal right.......?

If the diode and capacitor are ideal, the capacitor would come to the same voltage of the AC input instantly (drawing infinite current), so, since it spent no time doing this, the load would always be drawing current.

With a real diode and capacitor, the diode will drop a certain voltage depending on the current thorugh it (about 0.6V), and the capacitor would only be able to draw a finite amout of current due to its internal resistance and the current from the AC input would be divided between the capacitor and the resistor just as it would be two resistors.

Bob

 

[(*steve*)]

@steve wud it wont hold true if the diode is ideal right.......?

I have no idea what you're asking.

A perfect diode isn't going to have any real effect on what we're saying.

 

[mj1709]

i got it..........the capacitor starts to discharge as soon it has some path available....................it does not wait......... thnx @steve,bob and duke.......

 

[(*steve*)]

Yeah, You can kinda think of the capacitor as a tank. The transformer and rectifier diodes drop buckets of water into it, and the tap on the tank waters your garden.

Whilst the bucketloads of water are discontinuous, the water supply from the tank is continuous, varying only slightly in pressure between the buckets of water being dumped into it.

 

[Merlin3189]

Clamping

I'm not sure we have addressed the issue of clamping. This circuit is used (well, was used in pre-digital days) to establish an exact DC level for AC signals. The prime example in my experience was TV circuits: the varying modulation (AC) needed to be superimposed on a fixed bias Voltage (DC).

Now, as to how it works, you are nearly right in your thoughts about the diode, but remember the Voltage across the capacitor is in series with the signal source. When the capacitor is charged to the peak Voltage of the AC signal, then the diode is never forward biased at any part of the signal. For example:
If sig = +1V and cap = 0V, then diode is reverse biased, so 1V across Resistor.
If sig = -1V and cap =0V , then diode is forward biased, it conducts, no V across R.
But this has charged up the capacitor to 1V, so
now sig = +1V and cap = +1V, so +2V across diode, again reverse biased, so +2V across Resistor
now sig = -1V and cap = +1V, so 0V across diode and resistor.
If now sig = 0V and cap= +1V, then +1V across diode (reverse bias) and +1V across R.

So input varies from +1 to -1 and output varies from 0 to +2.
The varying part of the signal (AC) remains the same, but now the average Voltage is +1V, rather than 0V.

Depending on the resistance of the source and the diode during conduction, it could take several cycles for the capacitor to reach full charge.
Depending on the load resistor, some of the charge on the capacitor must "leak" through the load, so that it needs partial recharge each cycle.
In practice we design with a high load resistance and low source resistance to make these effects negligible.
To allow for the non-zero forward Voltage drop of the diode and to allow precise adjustment, we can take the non-signal end of the diode to a DC voltage provided by a pair of resistors (or potentiometer.)

In case it might help, I've done an Excel simulation of your circuit and plotted graphs of the voltages. These are attached.
Clamp1 shows reasonable values. Clamp slow has bad values, so that the capacitor charges slowly and discharges too much through the load resistor.
The Voltage plots have a consistent scale, but the current plots are arbitrarily scaled to fit.

How ever you understand capacitors, it is reasonable to think of them as blocking DC and as passing AC.
You can see from the graphs, the exponential charge of the capacitor from 0 to -1V, this being the DC voltage developed by the diode from the 1V p-p AC signal.
The Voltage across the load resistor can easily be seen as the AC signal added to this DC Voltage on the capacitor.
( If one actually solves the differential equations for circuits like this, the solutions come out having two parts, one exponential function and one sine function.)