Hi all,
Here's a doubt that sprang up while I was doing my circuits lab at college.


If we an ac signal is given across a series combination of resistor and diode and the output is seen across the diode in forward biased condition.(resistor in between ac signal source and the diode,negative terminal of ac source and the cathode of the diode is grounded)the resulting circuit acts as a clipper but is it possible to find out the voltage value of the ac signal at which the diode starts conducting if the value of resistance of the resistor is know and peak to peak voltage of ac signal is known.?


what we did at lab?
I connected a function generator (ac signal generator) which produces
a sine wave of peak to peak voltage of 20V with frequency of 12 kHz in series with a 300 resistor and IN4007 diode in forward biased condition
The above circuit acted as a clipper.
Is there a way to calculate at what value of input voltage the diode will start conducting?

-Surya.
ECE

 

Look at the datasheet for a 1N4007. You will see that the forward voltage depends on the current through the diode.

 

Look at the datasheet for 1N4001-1N4007 diode and find the I-V characteristic curve.
Your diode current is approx. 20V/300Ω = 0.066A.
From the curve you will see that the diode forward voltage = 0.7V

 

exactly I know the cut in voltage is 0.7V too,but the question is,at what value of the "Input voltage" will the diode start conducting,because since there is a resistor too,there will be a drop voltage drop across it too,so AT WHAT VALUE OF THE INPUT VOLTAGE will the voltage drop across the diode be 0.7V?
I hope I am clear in my question..
do please clear my doubt..
-sincerely,
surya.

 

The diode starts conducting when the voltage is greater than zero volts.

The diode voltage will be 0.7V when the input voltage is 20V.

 

hey how do u say 20V why not 15V be more specific and scientific plz..

 

Because you said 20V.

 

The graph shows 50 ma through the diode at .7 volts of diode voltage. Your 20 volt, peak to peak voltage is 10 volts in each polarity. Because we are calculating for 50 ma with a .7 volt loss, the 50 ma must also flow through the resistor.

E=IR
E=.05 * 300
E = 15 for the resistor.

The peak voltage must be 15.7 volts to get the diode to have .7 volts across it.

 

And another datasheet for the 1N4007 shows 0.7V at 10mA, which would require an input voltage of only 9.7V. So you can see that even the variations in the datasheets for the same part mean that expecting an exact answer is unrealistic. Through in the strong dependence of these curves on temperature and you start to see what assuming that the diode always has 0.6v or 0.7V or even 0.5V or 1V across it when forward biased yields sufficiently accurate results for nearly all situations.

So you can take your 20Vpk sinusoid and shift it down 0.7V and then remove everything below 0V and you will have a waveform that is within ~0.1V everywhere.