Recall Lecture 8 Clipper – Step 1: Find the clip value by doing KVL at the output branch – Step 2: Set the conditions to know whether diode is on or off – sketch your output waveform Clamper – Step 1: Calculate value of V C by knowing which cycle it is charged to (what is the peak value of that cycle) then confirmed the polarity of the capacitor – Step 2: performed KVL as diode is now reverse- biased to obtain equation of V out. Sketch your output waveform

Multiple Diode Circuits

Final Exam SEM I 2013/2014

DIODEIDID VDVD OFF0 V D < V  ONI D > 0 V D = V  REMEMBER THAT: A pn junction diode will conduct when the p-type material is more positive than the n-type material

OR GATE V1V2VO Vo = voltage across R D1 and D2 off; no current flow,000 D1 off, D2 on, current flow, Vo – V2 + V  = 0 05V ( 1 )4.3V D1 on, D2 off, current flow, Vo – V1 + V  = 0 5V ( 1 )04.3V Both on, using both loops will give the same equation 5V ( 1 ) 4.3V

V1V2VO Both on, using both loops will give the same equation D1 on, D2 off05V ( 1 )0.7 D1 off, D2 on5V ( 1 )00.7V Both are off; open circuit no current flowing through R since no GND destination 5V ( 1 ) 5V AND GATE Vo = node voltage

Chapter 4 Bipolar Junction Transistor

REMEMBER THIS Current flow in the opposite direction of the electrons flow; same direction as holes eee I hhh

Transistor Structures  The bipolar junction transistor (BJT) has three separately doped regions and contains two pn junctions.  Bipolar transistor is a 3-terminal device.  Emitter (E)  Base (B)  Collector (C)  The basic transistor principle is that the voltage between two terminals controls the current through the third terminal.  Current in the transistor is due to the flow of both electrons and holes, hence the name bipolar.

Transistor Structures  There are two types of bipolar junction transistor: npn and pnp.  The npn bipolar transistor contains a thin p-region between two n-regions.  The pnp bipolar transistor contains a thin n-region sandwiched between two p- regions.

 Active Operating range of the amplifier. Base-Emitter Junction forward biased. Collector-Base Junction reverse biased  Cutoff The amplifier is basically off. There is voltage but little current. Both junctions reverse biased  Saturation The amplifier is full on. There is little voltage but lots of current. Both junctions forward biased 3 Regions of Operation

OPERATIONS - npn  The base-emitter (B-E) junction is forward biased and the base-collector (C-B) junction is reverse-biased,.  Since the B-E junction is forward biased, electrons from the emitter are injected across the B-E junction into the base  I E  Once in the base region, the electrons are quickly accelerated through the base due to the reverse- biased C-B region  I C ACTIVE MODE C B E  Some electrons, in passing through the base region, recombine with majority carrier holes in the base. This produces the current  I B + - V BE iBiB

C B E TO ILLUSTRATE - V BE + Imagine the marbles as electrons A flat base region with gaps where the marbles may fall/trapped – recombine A sloping collector region represents high electric field in the C-B region Hence, when enough energy is given to the marbles, they will be accelerated towards to base region with enough momentum to pass the base and straight ‘fly’ to the collector

MATHEMATICAL EXPRESSIONS C B E + - V BE IEIE ICIC IBIB I E = I S [ e VBE / VT -1 ] = I S e VBE / VT Based on KCL: I E = I C + I B No. of electrons crossing the base region and then directly into the collector region is a constant factor  of the no. of electrons exiting the base region I C =  I B No. of electrons reaching the collector region is directly proportional to the no. of electrons injected or crossing the base region. I C =  I E Ideally  = 1, but in reality it is between 0.9 and

Based on KCL: I E = I C + I B I C =  I B I C =  I E I E =  I B + I B = I B (  + 1)  = [  /  + 1 ] I E = I B (  + 1) Now With I C =  I B  I B = I C /  Hence, I E = [ I C /  ] (  + 1) I C = I E [  /  + 1 ] Comparing with I C =  I E

B C E - + V EB OPERATIONS - pnp FORWARD ACTIVE MODE  The emitter – base (E- B) junction is forward biased and the base-collector (B- C) junction is reverse-biased,. I E = I S [ e VEB / VT -1 ] = I S e VEB / VT **Notice that it is V EB IEIE ICIC IBIB Based on KCL: I E = I C + I B

pnp Transistor- Active mode

SUMMARY: Circuit Symbols and Conventions npn bipolar transistor simple block diagram and circuit symbol. Arrow is on the emitter terminal that indicates the direction of emitter current (out of emitter terminal for the npn device) pnp bipolar transistor simple block diagram and circuit symbol. Arrow is on the emitter terminal that indicates the direction of emitter current (into of emitter terminal for the pnp device) Based on KCL: I E = I C + I B

EXAMPLE 4.1 Calculate the collector and emitter currents, given the base current and current gain. Assume a common-base current gain and a base current of. Also assume that the transistor is biased forward in the forward active mode. Solution: The common-emitter current gain is The collector current is And the emitter current is

BJT: Current-Voltage Characteristic I C versus V CE

 The Emitter is common to both input (base-emitter) and output (collector- emitter).  Since Emitter is grounded, V C = V CE  With decreasing V C (V CE ), the junction B-C will become forward biased too.  The current I C quickly drops to zero because electrons are no longer collected by the collector Common-Emitter Configuration - npn Node B 0V

Characteristics of Common-Emitter - npn NOTE: V EC for PNP

Examples EXAMPLE 1 Given I B = 6.0  A and I C =510  A Determine ,  and I E EXAMPLE 2 NPN Transistor Reverse saturation current Is = A with current gain,  = 90. Based on V BE = 0.685V, determine I C, I B and I E EXAMPLE 3 PNP Transistor  = 60, I C = 0.85mA Determine , I E and I B