Lecture 1 – Semiconductor Devices: By Karl Lebkuchner

Lecture 1 – Semiconductor Devices: By Karl Lebkuchner Suleyman Demirel University January 2013

Semiconductors 1. A semiconductor is a material that has four electrons in it’s outer shell. These four electrons will be shared with four other atoms. 2. Silicon (si 14) and germanium (ge 32) are the most common electronic semiconductors. 3. P-type semiconductor is made by adding some material (Boron atoms) to the semiconductor which has only three electrons in it’s outer shell. So one outer electron is missing. 4. N-type semiconductor has some atoms added to the semiconductor which have five electrons in it’s outer shell. One outer electron is extra.

PN Junction 1. An A PN Junction is made when N-type semiconductor is joined next to P-type semiconductor. The PN junction is where they touch. When they first touch, outer electrons from some of the N-type atoms will move into the P-type atoms. This will make the N-type semiconductor side more positive, and the P-type semiconductor more negative 2. A PN junction can be modeled using an ideal diode, a series voltage source and resistor (rd) , rd = Δ vd / Δ id . These values are read from the diode graph.

Ideal diode (PN Junction): A diode can be use like a switch. This is what we call an ideal diode. (Fig.1). For an ideal diode, when it is forward biased , the current will flow and there will be no voltage on the diode. This is like a closed switch. For an ideal diode, when it is reverse biased , the current will not flow and there will be voltage on the diode. This is like an open switch.

Forward biased diode (PN Junction): Connecting the positive terminal of a voltage source to the P-type side (anode) of the diode and the negative terminal of a voltage source to the N-type side (cathode) of the diode is called a forward biased connection (like Fig.2). As the forward biased voltage increases, the current in the diode will increase, but in a non linear change like in Figure 3. So the current slowly increases as the voltage increases to about 0.7V for si and 0.3 for ge. The current quickly increases when the forward voltage is bigger than ~ 0.7V

DC Diode Load Line Example: Given this circuit, With a 1/2W diode that has a forward bias graph shown. Given Vs = 5V, R1= 1K. What is the current in the circuit? Using load line analysis, start with Vd = 0V, then Id = 5V / 1K = 5ma. Then use Id=0, then Vd must be 5V. Id mA Draw a load line from Id=5ma to Vd=5V. Draw a dashed line from where the load line hits the diode curve to the Id axis Id=4ma.

DC Diode Approximate Example: Given this circuit, With a 1/2W diode that has a forward bias graph shown. Given Vs = 5V, R1= 1K. What is the current in the circuit? Using approximate diode analysis For a silicon diode, start with Vd = 0.7V. (for germanium use Vd=0.3) Then Id=(5V – 0.7V) / 1K so Id = 4.3mA Id mA

Reversed biased diode PN Junction: Connecting the negative terminal of a voltage source to the P-type side (anode) of the diode and the positive terminal of a voltage source to the N-type side (cathode) of the diode is called a reverse biased connection (like Fig.4). As the reverse biased voltage increases, the current in the diode will increase, but in a non linear change like in Figure 5. The reverse current slowly increases (in uA not mA) as the voltage increases. For the ideal diode, the reverse current is 0 (open circuit).

Diode Example 1: Given this circuit, with V1 = 10V, V2 = 0V, R1=1K Find Vr using approximate diode models? V1 = 10V so D1 is forward biased, and Vd1 = 0.7V. V2 = 0V so D2 is reverse biased, and Vd2 = open. So Vr = 10V – 0,7V = 9.7V Given V1=0V, V2=0V, R1=1K Find Vr? V1 = 0V so D1 is reverse biased, and Vd1 = open. V2 = 0V so D2 is reverse biased, and Vd2 = open. So there is no voltage on resistor. Vr=0V. This circuit performs OR logic.

Diode Example 2: Given this circuit, with V1 = 10V, V2 = 0V, R1=1K Find Vr using approximate diode models? V1 = 10V so D1 is reverse biased, and Vd1 = open. V2 = 0V so D2 is forward biased, and Vd2 = 0.7. So Vo= 0,7V Given V1=0V, V2=0V, R1=1K. Find Vo? V1 = 0V so D1 is forward biased, and Vd1 = 0.7V. V2 = 0V so D2 is forward biased, and Vd2 = 0.7V. So Vo=0.7V. This circuit performs AND logic.

Bipolar NPN Transistor 1. An NPN transistor is two PN junctions that are connected but in opposite ways, like in figure. 2. Emitter is one of the N-type semiconductors. 3. Collector is the other N-type semiconductor. 4. The Base is the P-type semiconductor, so it needs to have the base to the emitter junction forward biased to make current flow. The Base needs to be about Vbe = 0.7V larger than the emitter to make current flow.

NPN Transistor Model: The symbol for the NPN transistor is in figure 6 · There are two important current equations for bipolar transistors: Current summing: Ie = Ic + Ib Current gain: Ic = β x Ib where the β is set when the transistor is made, and it is usually about 100. · A full model can often be used in the place of a bipolar transistor. This model is in figure 7 · A simple model can sometimes be used, which removes the base (rb) and collector (rc) resistances

Common Emitter Bipolar Transistor This graph is for a bipolar transistor. It shows the relationship of the collector current (Ic) to collector voltage (Vce) for a grounded emitter. The section in red, shows the cutoff region of the transistor, when base current is zero (Ib=0). Transistor should be off. The section in blue, shows the saturation region. This is when the collector to emitter voltage is very small, and the collector current is changing fast. Use Vce=0 Active region is where collector current changes linearly.

DC NPN Transistor Example: Given this circuit, With with Va = 5V, R1= 1M, Vf = 10V, R2= 10K, and β is 100. What are the currents Ib, Ic and Ie? What is Vce (collector to emitter voltage)? Starting with Vbe = 0.7V, then Ib = (5V – 0.7V) / 1M = 4.3uA. Ic = β Ib = 430uA Ie = 434.3uA Vce = Vf – R2 Ic = 10 – 10K x430uA = 5.7V

Bipolar PNP Transistor 1. An PNP transistor is two PN junctions that are connected but in opposite ways, like in figure. 2. Emitter is one of the P-type semiconductors. In Figure 8. 3. Collector is the other P-type semiconductor. 4. The Base is the N-type semiconductor, so it needs to be about Vbe = 0.7V smaller than the emitter for current flow.

DC PNP Transistor Example: Given this circuit, With with Va = 4V, R1= 1M, Vf = 12V, R2= 10K, and β is 100. What are the currents Ib, Ic and Ie? What is Vce (collector to emitter voltage)? Starting with Vbe = 0.7V, then Ib = (-4V + 0.7V) / 1M = -3.3uA. Ic = β Ib = -330uA Ie = -333.3uA Vce = Vf – R2 Ic = -12 – 10K x-330uA = -8.7V

TTL NAND Gate: Given this circuit, with with Va = 5V, and Vb = 0V. What is Vout? Q1 is off, because Va=5V Using all transistors as switches. If Vbe > 0.7V then transistor is ON. Or it is OFF. Q2 is on, because Vb=0V Since Q2 is on, then Q3 is off, because ib3=0. Q4 is on, because Q3 is off and ib4>0. Q5 is off, because Q3 is off and ib5=0. Vout = high, because Q4 is on and Q5 is off.

Emitter Coupled Logic (ECL): 1. Emitter coupled logic was made to increase the speed of logic circuits. It uses a small difference in voltage (0.7V) to turn on and off transistors. 2. ECL gates often use negative voltage supplies. So logic high would be –0.7V while logic low would be –1.4V. 3. ECL gates usually use more power than other types of logic circuits.

ECL Gate: Given Vcc=0V, Vee=-5V, Va= -0.7V, and Vb = -1.4V, Q2 is on, because base current comes from R3. Vce=0.7V Using all transistors as switches. If Vbe > 0.7V then transistor is ON. Or it is OFF. Q1 is on, because base current from Q2. For Q1, Ve = -1.4V Qa is on, because base to emitter is 0.7V Qb is off, because base to emitter is 0V Vout = off, because Qa took away Q3 base current.