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If the critical field in a Gunn diode oscillator is 3.2 KV/cm and effective length is 20 microns, then the critical voltage is:

If the critical field in a Gunn diode oscillator is 3.2 KV/cm and effective length is 20 microns, then the critical voltage is: Correct Answer 6.4 V

Critical voltage of a Gunn diode oscillator is given by the expression lEc where l is the effective length and Ec is the critical field. Substituting the given values in the above equation, critical voltage is 6.4 volts.

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