# Diode (Electronics Questions)

1. Describe the types of breakdown current caused by excessive reverse voltage across a diode.

2. Describe the ideal diode, second approximation and third approximation

3. Describe how to test a diode using a Digital Multimeter and Digital Voltmeter

4. Describe how half-wave and rectifiers works

5. Describe the relationships among the base, emitter, and collector currents of a bipolar junction transistor

1. The breakdown reverse voltage of a diode is defined as the reverse voltage at which the reverse current through the diode (which in normal conditions is very small) begins to increase exponentially, hence the diode become conductive in the reverse conduction region. There are two types of breakdown in diodes first is Zener breakdown and second is avalanche breakdown.

In the two types of breakdown an exponentially increasing number of carriers are generated: in Zener diodes (which are heavily n or p doped) the carriers generated by the breakdown are due to the increasing electric field whereas in avalanche multiplication breakdown the carriers  (which this time are pairs electron-hole) are generated by the interaction of a carrier having enough energy with the crystal lattice (or in other words with the covalent bond in the lattice) which release a pair electron-hole, which multiplies further and so on.

2. There are three diode approximations. The ideal approximation considers the diode as a single closed switch in the forward region of conduction and an open switch in the reverse conduction region.

The first approximation takes into account the fact that there is an opening voltage threshold for the diode in the direct conduction region of about 0.7 V (for Si) and considers the diode as an ideal open (closed) switch in series with a reverse voltage source of 0.7 V.

The second approximation takes into account also the fact that the n and p regions (and also the barrier region) of the diode have finite conduction resistances (even in the direct conduction region) and considers the diode as an ideal switch in series with a voltage source (0.7 V for Si) and a resistance equivalent to the resistance of the n and p regions in direct conduction. This way the voltage drop on the diode in the direct conduction region increases with the direct current.

3. Using an analog multimeter (Volt Meter) to test a diode, switched to measuring resistances one should get a finite resistance (of some hundreds ohms) of the diode in the direct conduction region and an infinite resistance (of some tens mega ohms) in the reverse conduction region. Using a digital multimeter just switch the multimeter to diode test position and you should get a voltage drop between 0.4 and 0.9 V in the direct conduction region (depending on the diode material) and an OPEN CIRCUIT reading on the reverse voltage conduction region.

4. A half wave rectifier is the simplest rectifier that exist. It is composed by a single diode connected in series with the AC circuit. This way one half of the alternation current passing through the diode (when the diode is in the direct conduction region) and the other half of the AC wave is stopped by the diode. this way after passing through the diode one will have only the positive alternates of the AC current, hence a DC non filtered current.

A full rectifier consists of four diodes connected into a bridge. One diagonal of the bridge is connected to the AC voltage source while on the other diagonal one gets the DC voltage. During one half of the AC wave two diodes are open while during the other half of the AC wave the other two diodes from the bridge are open.

5. Let IC be the collector current, IB the base current and IE the emitter current.

One has the relation $I_E = I_C+I_B$ where $I_C = \beta*I_B$, where beta is the so called DC current gain.

Hence $I_E = I_B+\beta*I_B = I_B(\beta+1) = \mu*I_B$

$\alpha$, the current amplification factor, is defined as  $\alpha = IC_/I_E$

Hence $I_B = I_E-I_C$  or $I_B = I_E -\alpha*I_E$  that is $I_B = I_E(1-\alpha)$

$\beta = I_C/I_B = I_C/I_E/(1-\alpha) =\alpha/(1-\alpha)$

Usually $\alpha$ is less than 1 about 0.99, hence beta is about 99.