Simple circuit consisting of a source, a resistor, and a diode. For forward bias we have 2 models for the diode. 1) Ideal Diode: Short Circuit 2) Exponential Model In this section, we will: 1) Address the suitability of these two models. 2) Develop new models. This will help us to: 1) Analyze diode circuits efficiently. 2) Provide a foundation for modeling trasistor operation. R ID + VD - VDD 30-Jul-19 Carl Leuschen

A) The Exponential Model 1) The most accurate model. 2) The hardest to analyze. Let’s analyze a circuit using the exponential model. IS, n, VT are know quantities. We have two equaition with two unknowns. Q. How do we solve for ID and VD? + VD - R VDD ID 30-Jul-19 Carl Leuschen

a) Plot the two equations on the i-v plane. Graphically a) Plot the two equations on the i-v plane. Eq. 1: diode characteristic curve. Eq. 2: load line. b) The solution is the intersection, Q, of the two curves and is referred to as the operating point. Graphical analysis becomes very difficult for complex circuits. Too difficult to be justified for practical use. Diode Characteristic VDD/R -1/R Operating Point (Q) ID Load Line v VD VDD 30-Jul-19 Carl Leuschen

Remember the two equations: Diode Characterisitc Load Line Iterative analysis Remember the two equations: Diode Characterisitc Load Line Q. What do we know about a diode in forward bias? A. The voltage across the diode is between 0.6 and 0.8 Volts. Approach: 1) Assume VD is 0.7 Volts. 2) Solve for ID using circuit equation (load line). 3) Caluculate new value of VD or ΔVD using the Diode Characteristic curve. 4) Go back to step 2 and repeat until the answer converges. 30-Jul-19 Carl Leuschen

1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. Example: 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. VDD = 5 V, and R=1 kΩ. R ID + VD - VDD 30-Jul-19 Carl Leuschen

1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. Example: 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. VDD = 5 V, and R=1 kΩ. R ID + VD - VDD 30-Jul-19 Carl Leuschen

However, this approach can be easily iterated using computer analysis. For complex circuits this approach is very time consuming to do by hand, because you must re-analyze the circuit for each iteration. However, this approach can be easily iterated using computer analysis. Conclusions the graphical and iterative solution methods using the exponential model are inefficient. For effective circuit analysis a simple model for the junction diode is needed. 30-Jul-19 Carl Leuschen

The Piecewise Linear Model. Consider the i-v relationship for the following configuration. An ideal diode, a battery, and a resistor. F.B. → VD(ideal)=0 → short → ID(ideal)≥0 D: Ideal VD0 VD0 rD rD 30-Jul-19 Carl Leuschen

R.B. → ID(ideal)=0 → open → VD(ideal)≤0 rD rD 30-Jul-19 Carl Leuschen

The Piecewise Linear Model. An ideal diode, a battery, and a resistor. D: Ideal i VD0 rD v VD0 30-Jul-19 Carl Leuschen

the non-linear on/off behavior. voltage source: Approximate the diode exponential i-v relationship using a model containing: ideal diode: the non-linear on/off behavior. voltage source: to model the potential needed to turn the diode on. resistor: to model the exponential slope. The model is called the: Battery plus Resistance Model or Piecewise Linear Model v VD0 D: Ideal VD0 rD 30-Jul-19 Carl Leuschen

1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. Example: 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. VDD = 5 V, and R=1 kΩ. For these characteristics, VD0=0.65 V and rD=20 Ω. R ID R ID D: Ideal VD0 rD + VD - VDD VDD 30-Jul-19 Carl Leuschen

1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. Example: 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. VDD = 5 V, and R=1 kΩ. For these characteristics, VD0=0.65 V and rD=20 Ω. R ID R ID + VD - VDD VDD VD0 rD 30-Jul-19 Carl Leuschen

Q. How are values for VD0 and rD determined? A. Minimize error within a certain range of operating conditions. A. Taylor Expansion – 1 point and a slope. A. Given 2 points we can find a line. 30-Jul-19 Carl Leuschen

Battery plus Resistance Model rD and VD0 is chosen for a given range of operating values. Q. What if we zoom out to a very wide range of values? Q. What does the exponential i-v relationship look like? For the Battery plus Resistance model, the resistance goes to zero. Constant Voltage Drop Model A forward bias diode exhibits a constant drop of 0.7 Volts. One of the most frequently used diode model for initial designs. Can be used without detailed information about the diode. i i D: Ideal VD v v VD0 VD 30-Jul-19 Carl Leuschen

1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. Example: 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. VDD = 5 V, and R=1 kΩ. VD=0.7 R ID R ID + VD - D VDD VDD VD VD 30-Jul-19 Carl Leuschen

1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. Example: The Ideal Diode Model 1mA @ 0.7 V, and ΔV=0.1V for every x10 change in current. VDD = 5 V, and R=1 kΩ. R ID R ID + VD - D VDD VDD 30-Jul-19 Carl Leuschen

Exponential Battery Plus Resistance Constant Voltage Drop Ideal Complex Simple Exponential Battery Plus Resistance Constant Voltage Drop Ideal D: Ideal VD0 rD D: Ideal VD VD=0.762V ID= 4.237mA VD=0.735V ID= 4.26mA VD=0.7V ID= 4.3mA VD=0.0V ID= 5mA 30-Jul-19 Carl Leuschen