Diodes

What is the rectifier circuit?

-A circuit which converts AC voltage to DC ideally without loss -During the positive half cycles, the vI will cause current to flow through the diode in its forward direction. It follows that the diode voltage VD will be very small - During the negative half cycles, the diode will not conduct and Vo will be 0 -

What is a bridge rectifier?

-An alternative implementation of the full-wave rectifier is a bridge rectifier. -When the instantaneous source voltage is positive, D1 and D2 conduct while D3 and D4 block -When the instantaneous source voltage is negative, D1 and D2 block while D3 and D4 conduct -No need for a center-tapped transformer, but Series connection of TWO diodes will reduce output voltage. -PIV = VS-VD -Consider the circuit during positive half- cycles. The reverse voltage across D3 can be seen via the segment D2-R-D2 and can be written: V D3(reverse)= vo+VD2but VD2=VD and at the maximum level vo= VS-2VDhence PIV=VS-VD.

What happens if the rectifier with a filter capacitor has a resistor in series?

-One must now consider the discharging of capacitor across load.

How does a rectifier circuit work

1) increase/decrease rms magnitude of AC wave via a power transformer 2)convert full-wave AC to half-wave DC (still time - varying and periodic) via the rectifier 3)employ a low-pass filter to reduce wave amplitude by > 90% 4)employ a voltage regulator to eliminate ripple 5)supply dc load

What is the diffusion current under open circuit terminals?

As the concentration of holes is high in the P region and lower in the n region. Holes diffuse across the junction. A similar thing happens with the electrons which adds up to create a current ID which flows from the p side to the n side.

How is the reverse bias region approximated

If v is negative and a few times larger than VT (25 mV) in magnitude, the exponential term becomes negligibly small compared to unity.

What is the ideal diode model?

In applications that involve voltage much greater than the diode voltage drop, we many neglect the diode voltage drop all together while calculating the diode current The result is the ideal diode model that assumes the slope of Io vs VD is vertical at 0 volts

What is the characteristic of the currents under open circuit?

In open circuit conditions: ID = Is The equation is maintained by Vo, as if ID increases then more bound charge will be uncovered on both sides of the junction, the depletion layer will widen, and the voltage across it (V0 ) will increase. This in turn causes ID to decrease until equilibrium is achieved with ID = IS

What is the breakdown region?

In this region the voltage across the diode remains nearly constant while the current varies (i.e., small internal resistance)

What is drift current?

Some of the thermally generated holes in the material move towards the junction and reach the depletion region. There they experience the electric field in the depletion region which sweeps them to the p side - A similar thing happens to the thermally generated electrons - These two current components add together to make the drift current Is, whose direction is from the P side to the n side of the junction - It is not dependent on Vo as it is determined by the number of minority carriers

What is a PN Junction diode?

The PN junction is a diode which contains a P type semiconductor and an N type semiconductor which are in contact. In practice it is a silicon crystal with differently doped sections.

What is the Zener effect?

The Zener effect occurs in a reverse biased p-n diode when the electric field enables tunnelling of electrons from the valence to the conduction band of a semiconductor, leading to a large number of free minority carriers which suddenly increase the reverse current.

How is the breakdown region approximated

The breakdown region is entered when v<Vzk This is normally non destructive.

What is the constant voltage drop model?

This model assumes that the slope of Id vs Vd is vertical at 0.7V Therefore we are approximating the exponential characteristic for a constant voltage which is about 0.7V

How to do graphical analysis using exponential model?

1) Plot relationships of the two equations on a single graph 2) Find the intersection of the two 3) Load line and diode characteristic intersect at the operation point - It is intuitive but the effort required for this analysis for a complex circuit is too great in practice.

What is the current voltage characteristic of the ideal diode?

- If a negative voltage is applied to the diode, no current flows and the diode behaves like an open circuit (Reverse Bias) -If a positive voltage is applied to the diode, there will be a voltage drop of 0 across the diode so it behaves like a short circuit (Forward bias) -the i-v characteristic of the ideal diode is highly nonlinear; although it consists of two straight-line segments, they are at 90° to one another. A nonlinear curve that consists of straight-line segments is said to be piecewise linear. If a device having a piecewise-linear characteristic is used in a particular application in such a way that the signal across its terminals swings along only one of the linearsegments, then the device can be considered a linear circuit element as far as that particular circuit application is concerned.

How is the forward bias region approximated?

-Cut in voltage, when current flow appears to be minimal, when it is smaller than 0.5V. This is a consequence of the exponential relationship -Fully conducting region- where Rdiode = 0 between 0.6-0.8V -At a given temperature, the forward i-v characteristics varies with temperature the voltage drop across the diode decreases by approximately 2 mV for every 1°C increase in temperature

What is the rectifier with a filter capacitor?

-Pulsating nature of rectifier output makes unreliable dc supply. -As such, a filter capacitor is employed to remove ripple. -step #1: source voltage is positive, the diode is forward biased, capacitor charges. -step #2: source voltage is reversed, the diode is reverse-biased (blocking), the capacitor cannot discharge. -step #3: source voltage is positive, the diode is forward biased, capacitor charges (maintains voltage). This is unrealistic Because, for any practical application, the converter would supply a load (which in turn provides a path for capacitor discharging)

A couple of observations of load rectifier with capacitor

-The diode conducts for a brief interval (Dt) near the peak of the input sinusoid and supplies the capacitor with charge equal to that lost during the much longer discharge interval. The latter is approximately equal to T. -Assuming an ideal diode, the diode conduction begins at time t1(at which the input vI equals the exponentially decaying output vO). -Diode conduction stops at time t2shortly after the peak of vI During the diode off-interval, the capacitor C discharges through R causing an exponential decay in the output voltage (vO). At the end of the discharge interval, which lasts for almost the entire period T, voltage output is defined as follows - vO(T) = Vpeak-Vr -When the ripple voltage(Vr) is small, the output (vO) is almost constant and equal to the peak of the input (vI). the average output voltage may be defined as

What is the reverse bias case?

-The externally applied reverse bias Vr adds to the barrier voltage, reducing the number of holes that diffuse into the n region and electrons that diffuse into the p region. -The end result is that Id is dramatically reduced -A reverse bias voltage of about 1 volt will cause Id =(approx)0 and the current across the junction will be equal to just IS In the reverse bias case the pn junction conducts only a very small and constant amount of current equal to Is

What is the forward bias case?

-The externally applied voltage Vf subtracts from the barrier voltage which in turn decreases the barrier voltage -This increases the rate of diffusion of holes and currents -The result is that the pn junction will conduct significant current Id-Is which will cause significant current to flow

What is a Full wave rectifier?

-The full wave rectifier uses both halves of the input. -The key here is center-tapping of the transformer, allowing "reversal" of the current -When the instantaneous source voltage is positive, D1 conducts while D2 blocks When the instantaneous source voltage is negative, D1 blocks while D2 conducts -The direction of current flowing across load never changes (both halves of AC wave are rectified). -The full-wave rectifier produces a more "energetic" waveform than half-wave. -PIV for full-wave = 2VS-VD -This is because diode D2 towards the output will see voltage vo and towards the coil voltage -vs Hence the total voltage difference will be vo+vs. And the maximum is when vs=VS hence PIV=2VS-VD

What is a half wave rectifier?

-The half-wave rectifier-utilizes only alternate half cycles of the input sinusoid -Constant voltage drop diode model is employed. -current-handling capability will be the maximum current that does not destroy the device due to heat. -peak inverse voltage (PIV)-what is maximum reverse voltage it is expected to block w/o breakdown? -Observe that when vs is negative the diode will be cut off and vo will be zero. Hence, PIV is equal to maximum input voltage VS -It is possible to use the diode exponential model in describing rectifier operation; however, this requires too much work -Regardless of the model employed, one should note that the rectifier will not operate properly when input voltage is small (< 1V). .

What is the depletion region?

-The holes that diffuse across the junction into the n region will quickly recombine with the free electrons in that region causing the holes and free electrons to disappear meaning the bound positive charges are no longer neutralized, so now there is a region of positive charge. The same thing happens on the other side causing a region with a negative charge. -This causes a potential difference across the region. -The resulting electric field opposes the direction of holes into the n region and electrons into the p region. The voltage drop acts as a barrier that has to be overcome for them to diffuse. -The current Id is dependant on the barrier voltage

What is the small signal diode model?

There are applications in which a diode is biased to operate at a point on the forward i-v characteristic and a small ac signal is superimposed on the dc quantities. For this situation, we first have to determine the dc operating point (VD and ID) of the diode using one of the models discussed above. Most frequently, the 0.7-V-drop model is utilized. Then, for small-signal operation around the dc bias point, the diode is modeled by a resistance equal to the inverse of the slope of the tangent to the exponential i-v characteristic at the bias point. The technique of biasing a nonlinear device and restricting signal excursion to a short, almost-linear segment of its characteristic around the bias point is central to designing linear amplifiers using transistors,

What is the Avalanche effect?

the Avalanche effect in which charges that are accelerated to high speeds due to the large electric field in the depletion region collide with atoms in the silicon lattice causing charges to be dislodged. In turn, these dislodged charges have sufficient energy to liberate additional electrons. In other words, this avalanche effect is a cascading, ionization process.