EE 315 Introduction to Electronic Analysis & Design
Lecture 12: Diodes II: Constant-voltage diode model, Small-signal diode model, voltage regulators, iterating solutions
Diode I-V Curve
| Condition | Behavior | Current | Voltage |
|---|---|---|---|
| Forward: $v_D \approx V_\text{ON}$ | Voltage Source/Low $R_D$ | $i_D=I_S(e^{\frac{v_D}{V_T}}-1)$ | $v_D \approx V_\text{ON}$ |
| Forward: $0\leq v_D<V_\text{ON}$ | Nonlinear Resistor | $i_D=I_S(e^{\frac{v_D}{V_T}}-1)$ | $v_D = V_T\text{ln}(\frac{i_D}{I_S})$ |
| Reverse: $V_{ZK}\leq v_D< 0$ | Small Current Source | $i_D \approx -I_S$ | $v_D = V_T\text{ln}(\frac{i_D}{I_S})$ |
| Reverse: $v_D \approx -V_\text{ZK}$ | Voltage Source | Must be limited | $v_D \approx -V_\text{ZK}$ |
Note that $i_D=I_S(e^{v_D/V_T}-1)\approx I_Se^{v_D/V_T}$
Solve a simple voltage - resistor - diode series circuit.
- Makes a voltage divider
- Find voltage across the diode.
- Find current through the diode.
- I. Simplified, open circuit or short circuit model
- Determine if the diode is forward-biased or reverse-biased.
- Is a net-positive current entering the positive diode terminal (anode)?
- Is the voltage drop $v_D$ from positive terminal (anode) to negative terminal (cathode) give a positive voltage via KVL?
- Find current through the diode.
- Forward-biased: $0\leq v_D<V_\text{ON}$ model leads to diode as a short circuit $i_D=\frac{V_S}{R}$
- Reverse-biased: $V_{ZK}\leq v_D<0$ model leads to diode as an open circuit $i_D = 0 A$
- II. Constant voltage model: $v_D = 0.7V$
- Determine if the diode is forward-biased or reverse-biased.
- Is a net-positive current entering the positive diode terminal (anode)?
- Is the voltage drop $v_D$ from positive terminal (anode) to negative terminal (cathode) give a positive voltage via KVL?
- Find current through the diode.
- Forward-biased: $0\leq v_D<V_\text{ON}$ model leads to diode as a voltage source in series $i_D=\frac{V_S-0.7V}{R}$
- Reverse-biased: $V_{ZK}\leq v_D<0$ model leads to diode as an open circuit $i_D = 0 A$
- III. Nonlinear model:
- Determine if the diode is forward-biased or reverse-biased.
- Is a net-positive current entering the positive diode terminal (anode)?
- Is the voltage drop $v_D$ from positive terminal (anode) to negative terminal (cathode) give a positive voltage via KVL?
- Find current through the diode.
- Forward-biased: $0\leq v_D<V_\text{ON}$ gives $i_D=I_S(e^{\frac{v_D}{V_T}}-1)\approx I_Se^{v_D/V_T}$
- Reverse-biased: $V_{ZK}\leq v_D<0$ gives $i_D \approx-I_S \approx 0 A$
- $i_D=I_Se^{v_D/V_T}$ from the diode equation
- $i_D = \frac{V_{DD}-v_D}{R}$ from KVL
- $I_Se^{v_D/V_T} = \frac{V_{DD}-v_D}{R}$ which is not algebraic
- Solve graphically by plotting each side of the equation and find their intersection
- Iterated to find the solution
Load line intersection to describe a diodes operating point (bias point/quiescent point)
Iteration to move along the exponential curve
Using diode voltage steps:
$$\Delta v_D = v_{D2} - v_{D1} = V_T \text{ln}\bigg(\frac{I_{D2}}{I_S}\bigg) - V_T \text{ln}\bigg(\frac{I_{D1}}{I_S}\bigg)$$
$$V_T \text{ln}\bigg(\frac{I_{D2}}{I_S}\bigg) - V_T \text{ln}\bigg(\frac{I_{D1}}{I_S}\bigg) = V_T \text{ln}(I_{D2}) - V_T \text{ln}(I_S) - V_T \text{ln}(I_{D1}) + V_T \text{ln}(I_S)$$
$$v_{D2} - v_{D1} = V_T \text{ln}\bigg(\frac{I_{D2}}{I_{D1}}\bigg)$$
$$v_{D2} = v_{D1} + V_T \text{ln}\bigg(\frac{I_{D2}}{I_{D1}}\bigg)$$
Using diode current steps: $$v_{D2} - v_{D1} = V_T \text{ln}\bigg(\frac{I_{D2}}{I_{D1}}\bigg)$$ $$\frac{v_{D2} - v_{D1}}{V_T} = \text{ln}\bigg(\frac{I_{D2}}{I_{D1}}\bigg)$$ $$e^{\frac{v_{D2} - v_{D1}}{V_T}} = \frac{I_{D2}}{I_{D1}}$$ $$I_{D2}=I_{D1}e^{\frac{v_{D2} - v_{D1}}{V_T}}$$
Ideal vs. constant-voltage drop diode model vs. iteration (Using diode voltage steps)
Small-signal approximations
$$\begin{align}
\frac{d i_D}{d v_D}|_\text{q-point} &=\frac{d}{d v_D} I_S(e^{\frac{v_D}{V_T}}-1) \newline
&=\frac{I_S}{V_T}e^{\frac{v_D}{V_T}} \newline
&=\frac{I_S e^{\frac{v_D}{V_T}}}{V_T} \newline
&\approx \frac{i_D}{V_T} \newline
\end{align}$$
$$r_D = \frac{V_T}{i_D}$$
Voltage regulators
- Diodes can be used as simple voltage regulators
- Think of the inverse of an amplifier
- Amplifiers take small changes at the input and translate as large changes at the output
- Voltage regulators reject small changes at the input to keep a stable voltage as an ouput