Suppose you have a sinusoidal signal whose voltage varies from -5v to +5v like this signal
Have you ever wondered how can you convert the same sinusoidal signal to another sinusoidal signal whose voltage varies from different voltage points ?
i.e sinusoidal signal which is varying from -5v to +5v to -->> 0v to -10v or 0v to 10v ?
Yes such an electronic circuit is called clamper circuit.
A clamper circuit is an electronic circuit which doesnt change the peak to peak voltage of a signal but moves the signal up/down by adding sufficient DC value to the signal.
Well here is the circuit which moves the Sinusoidal signal down
i.e -5v to 5v signal as depicted above is MOVED DOWN
i.e the output voltage now, varies from 0v to -10v
For vin =-5v , vo=-10v
and
For vin=5v, vo=0v.
So the output voltage peaks from 0v to -10v
Since the output waveform has voltage which is less than 0v.The circuit above is called negative clamper.
Output
Note : The above circuit just shifts the input signal down as shown and doesnt change the shape of the signal
Simulation software used: PSPICE
Have you ever wondered how can you convert the same sinusoidal signal to another sinusoidal signal whose voltage varies from different voltage points ?
i.e sinusoidal signal which is varying from -5v to +5v to -->> 0v to -10v or 0v to 10v ?
Yes such an electronic circuit is called clamper circuit.
A clamper circuit is an electronic circuit which doesnt change the peak to peak voltage of a signal but moves the signal up/down by adding sufficient DC value to the signal.
Well here is the circuit which moves the Sinusoidal signal down
i.e -5v to 5v signal as depicted above is MOVED DOWN
i.e the output voltage now, varies from 0v to -10v
The sinusoidal signal has a amplitude of 5v .The input signal is
To find the output
Step 1) : Find the capacitor voltage.As you can see The diode is forward biased during positive half cycle and maximum input voltage is 5v.So capacitor gets charged to 5v.(we have considered diode as ideal)
Step 2) :After finding the capacitor voltage replace the diode with open circuit and find the output voltage
Apply kvl to the above ckt : Vo+5-vin=0
vo=vin-5v
Note:Every point of input signal voltage is substracted by 5volts.
Lets just consider the peak voltages below
and
For vin=5v, vo=0v.
So the output voltage peaks from 0v to -10v
Since the output waveform has voltage which is less than 0v.The circuit above is called negative clamper.
Output
Note : The above circuit just shifts the input signal down as shown and doesnt change the shape of the signal
Simulation software used: PSPICE
