In volume III Semiconductors topic Diode Check its state that if using a very low test voltage multimeter( TO LOW TO FULLY COLLASEP THE DEPLITION REGION OF PN JUNCTION) to test a diode using ohmeter in correct forward biased direction of diode you will get a very high value of resistance. Its not clearly explain the reason why you will get a high value of resistance ?
Principle Of Low Voltage Tester
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http://All About Circuits E-book.com All About Circuits E-Book (Volume III- Semiconductors) chapter 3 topic diode and rectifiers ( meter check of diode )
Thanks
Allied
Thanks
Allied
A meter not having a diode check function may not have a high enough voltage out on the leads to put a P-N junction into conduction. The device might appear bad, as it would check open in both directions.
If you have a Simpson 260 with some age to it, you really won't be able to use it to check diodes and transistors. The dodge back in the 1960's was a test fixture called an octopus (because of the number of leads). It used half the secondary winding of a filament transformer and a resistor network to give a graphic display of the P-N junction conduction on the CRT of an oscilloscope.
If you have a Simpson 260 with some age to it, you really won't be able to use it to check diodes and transistors. The dodge back in the 1960's was a test fixture called an octopus (because of the number of leads). It used half the secondary winding of a filament transformer and a resistor network to give a graphic display of the P-N junction conduction on the CRT of an oscilloscope.
Independant of the issue of measurement, the resistance value associated with a diode is related by Ohm's Law: I = V/R over the range of the diode I-V characteristic.
An important point to note is that the I-V characteristic of the diode is related by the Eber's Moll Equation which shows the relationship is an exponential, therefore the resistance varies over the applied voltage range.
For a silicon diode, over the voltage range 0 to ~ 0.6V the change in current is negigible (because of the diode depletion region), think of Ohm's Law again: I = V/R - if V = 0-0.6, for I to remain close to 0, then R must be high.
For a silicon diode, for the voltage range above 0.65V (upto around 0.9V where the diode destroys itself) the change in the current is large for small changes in the input voltage, think of Ohm's Law again: I = V/R - if V ~ 0.65-0.9, for I to have a large change for small V, then R must be low.
Put it another way, the derivative of the Eber's Moll Equation is the conductance, dI/dV, where the resistance can be derived as: R = 1/Conductance.
Dave
An important point to note is that the I-V characteristic of the diode is related by the Eber's Moll Equation which shows the relationship is an exponential, therefore the resistance varies over the applied voltage range.
For a silicon diode, over the voltage range 0 to ~ 0.6V the change in current is negigible (because of the diode depletion region), think of Ohm's Law again: I = V/R - if V = 0-0.6, for I to remain close to 0, then R must be high.
For a silicon diode, for the voltage range above 0.65V (upto around 0.9V where the diode destroys itself) the change in the current is large for small changes in the input voltage, think of Ohm's Law again: I = V/R - if V ~ 0.65-0.9, for I to have a large change for small V, then R must be low.
Put it another way, the derivative of the Eber's Moll Equation is the conductance, dI/dV, where the resistance can be derived as: R = 1/Conductance.
Dave
