If you are not familiar with diodes, the following article will give you the knowledge you need to understand this article.

http://www.associatedcontent.com/article/2283189/electronics_the_diode_and_boolean_logic.html?cat=59

If you are not familiar with transistors, the following article will give you the knowledge you need.

http://www.associatedcontent.com/article/2274179/electronics_the_transistor_and_boolean.html?cat=59

Figure One shows a circuit with three inputs and one output. The inputs, A, B and C could be at one of two voltage levels: 1.3 volts or 0 volts. The voltage supply Va delivers 5 volts output. In every problem, the output Vo needs to be calculated.

For this circuit, our calculations will assume the ideal diode. Another words, the forward biased diode has a 0.6 volt drop across it regardless of the current through it. The inclusion of the resistor R2 is deliberate.

The inputs A and B are connected to the anodes of diodes D1 and D2 respectively. Input C is connected to the base of the transistor Q2.

Problem 1:

Input A = 1.3 volts

Input B = 0 volts

Input C = 1.3 volts

R1 = 50 ohms

R2 = 50 ohms

Beta = 50

Solution:

Looking at input A, the current flows through diode D1, through the resistor R2 and through the base emitter junction of Q1 to ground.

Va = Vd1 + Vr2 + Vb1

where Vb1 is the voltage at the base of Q1

Va = 1.3 volts, Vd1 = 0.6 volts and Vb1 = 0.6 volts

The voltage across R2 is

Vr2 = Va – Vd1 – Vb1

Vr2 = 1.3 volts – 0.6 volts – 0.6 volts

Vr2 = 1.3 volts – 1.2 volts

Vr2 = 0.1 volts

The current is

Ir2 = Vr2/R2 = 0.1 volts/50 ohm = 0.002 amperes

Beta = Ic1/Ib1

Ic1 = Beta * Ib1

Ic1 = 50 * 0.002 amperes = 0.1 amperes

The collector resistor R1 is 50 ohms

The voltage across the collector resistor is equal to the product of the collector current and the value of R2

Vr1 = Ic1 * R1

Vr1 = 0.1 amperes * 50 ohms = 5 volts.

Hence the voltage across R1 is 5 volts.

This means that the collector of Q2 is at 0 volts and the base is at 0.6 volts. Hence the base collector junction has become forward biased. The base emitter junction is also forward biased. When both junctions are forward biased, the transistor is in saturation. Hence the voltage from collector to emitter is

Vce = Va – Vr1 = 5 volts – 5 volts = 0 volts

Q2 is turned off because the voltage at the base of Q2 is 1.3 volts and the voltage at the collector of Q2 is the same as the voltage at the base of Q1. The voltage at the base of Q1 is 0.6 volts Hence the base emitter junction of Q2 is reverse biased.

Problem 2:

Input A = 0 volts

Input B = 1.3 volts

Input C = 1.3 volts

R2 = 50 ohm

R1 = 50 ohms

Beta = 50

Vo = ?

Solution:

The solution to this problem is the exact same as the solution to problem one except the Vd1 is replaced by Vd2. If the you understands this, you can proceed to problem 3.

Looking at input A, the current flows through diode D2, through the resistor R2 and through the base emitter junction of Q1 to ground.

Va = Vd2 + Vr2 + Vb1

Va = 1.3 volts, Vd2 = 0.6 volts and Vb1 = 0.6 volts

The voltage across R2 is

Vr2 = Va – Vd2 – Vb1

Vr2 = 1.3 volts – 0.6 volts – 0.6 volts

Vr2 = 1.3 volts – 1.2 volts

Vr2 = 0.1 volts

The current is

Ir2 = 0.1 volts/50 ohm = 0.002 amperes

Beta = Ic1/Ib1

Ic1 = Beta * Ib1

Ic1 = 50 * 0.002 amperes = 0.1 amperes

The collector resistor R2 is 50 ohms

The voltage across the collector resistor is equal to the product of the collector current and the value of R1

Vr1 = Ic1 * R1

Vr1 = 0.1 amperes * 50 ohms = 5 volts.

Hence the voltage across R1 is 5 volts.

This means that the collector of Q2 is at 0 volts and the base is at 0.6 volts. Hence the base collector junction has become forward biased. The base emitter junction is also forward biased. The transistor is in saturation. Hence the voltage from collector to emitter is

Vce = Va – Vr1

Vce = 5 volts – 5 volts = 0 volts

Q2 is turned off because the voltage at the base of Q2 is 1.3 volts and the voltage at the collector of Q2 is the same as the voltage at the base of Q1. The voltage at the base of Q1 is 0.6 volts Hence the base emitter junction of Q2 is reverse biased.

Problem 3:

Input A = 1.3 volts

Input B = 1.3 volts

Input C = 1.3 volts

R2 = 50 ohm

R1 = 50 ohms

Beta = 50

Vo = ?

Solution:

We have two current paths in input to the base of Q1. The first current path is from input A through D1, R2 and the base emitter junction of Q1. We’ll call this first current i1. The second current path is from input B through D2, R2 and the base emitter junction of Q1. We’ll call this current i2.

The previous two problems have told us that each current is 0.002 amperes.

i1 = i2 = 0.002 amperes

The total current i through R2 is the sum of i1 and i2.

i = i1 + i2 = 0.004 amperes

The voltage drop across R2 is

Vr2 = i * R2

Vr2 = 0.004 amperes * 50 ohms = 0.2 volts

Here is where we encounter a problem. The input voltage is 1.3 volts. The voltage drop across D1 equals the voltage drop across D2.

Vd1 = Vd2 = 0.6 volts

Since the voltage drop across D1 equals the voltage drop across D2, we use the term Vd to represent the voltage drop across the diodes.

The voltage Vbe for Q1 is

Vbe = Va – Vd – Vr2

Vbe = 1.3 volts – 0.6 volts – 0.2 volts = 1.3 – 0.8 = 0.5 volts

The transistor Q1 is off because the base emitter voltage Vbe is less than 0.6 volts!

Initial analysis:

The effects of Q2 on the voltage output Vo.

Problem 4:

Input A = 1.3 volts

Input B = 0 volts

Input C = 0 volts

R2 = 50 ohm

R1 = 50 ohms

Beta = 50

Vo = ?

Solution:

The current paths in this case are as follows.

The current flows through D1, R2 and the emitter base junction of Q2. Since there is no resistor connected to the Q2 base, Q2 will turn on hard and the voltage at the base of Q1 will drop well below the 0.6 volts needed to turn on Q1. Hence Q1 will be off and there will be no current flowing through the resistor R1. If there is no current through the resistor R1, then the voltage drop across R1 will be

Vr1 = Ic1 * R1 = 0 amperes * 50 ohms = 0 volts

The voltage at the collector of Q1 will be

Vq1 = Va – Vr1 = 5 volts – 0 volts = 5 volts

Hence the output voltage Vo equals 5 volts.

If input C = 0 volts, the output voltage is 5 volts regardless of the states of the inputs A and B.

I have a Bachelor of Science in Electrical Engineering along with some experience in digital logic design and troubleshooting.

Micro-Electronics

Digital and Analog Circuits and Systems

ISBN 0-07-042327-X