# 电气工程代写|数字电路代写digital circuit代考|EE365

## 电气工程代写|数字电路代写digital circuit代考|REVERSE BIAS MODELING

In the region where the diode is basically not conducting there are several possible linear models from which to choose. Each of these is based on the principle that the diode has a small leakage current that is fairly constant in the reverse bias region: that is when the diode voltage is between a few negative multiples of $V_t$ and the Zener breakdown voltage ${ }^6$ the diode current is constant at $-I_S$. The two common models are: a current source of value $-I_S$; or a large resistor. These models are shown in Figure 2.12.

The value of the reverse resistance, $r_r$, for the second model can be approximated using one of two techniques: (a) using Equation (2.9) to determine the dynamic resistance about some Q-point, or (b) assuming that the diode achieves its true reverse saturation current at the Zener breakdown voltage. While method (a) allows for an exact dynamic resistance at some point, it is often difficult to choose the proper Qpoint for a particular application. Method (b) is easier to calculate, but is less accurate at any point and underestimates the dynamic resistance for large reverse voltages.
Example $2.9$
Assume the diode of Example $2.6$ (with Zener breakdown occurring at a voltage of $-25 \mathrm{~V}$ ) is connected in series with a $4 \mathrm{~V}$ source and a resistance of $820 \Omega$ so that the diode is reverse biased.
Calculate the diode current and voltage using:
(a) the current source model for a reverse biased diode
(b) the resistor model for a reverse biased diode

(a) If the diode is replaced by a $1 \mathrm{nA}$ current source then all circuit elements carry $1 \mathrm{nA}$ of current and the diode current is $-1 \mathrm{nA}$. The voltage across the resistor is given by:
$$V_r=(1 \mathrm{nA})(820 \Omega)=820 \mathrm{nV} \text {. }$$
Kirchhoff’s Voltage Law applied to the loop gives the resulting diode voltage:
$$V_d=V_r-4=-3.999999 \overline{1} \overline{8} \mathrm{~V} \approx-4.00 \mathrm{~V} \text {. }$$
(b) The reverse resistance can be approximated as:
$$r_r \approx|-25 \mathrm{~V}| /(1 \mathrm{nA})=25 \mathrm{G} \Omega .$$
The diode voltage and current are given the following:
\begin{aligned} V_d &=\frac{25 \mathrm{G} \Omega}{25 \mathrm{G} \Omega+820 \Omega}(-4 \mathrm{~V})=-3.999999869 \approx-4 \ I_d &=\frac{-4 V}{25 \mathrm{G} \Omega+820 \Omega}=160 \mathrm{pA} \end{aligned}

## 电气工程代写|数字电路代写digital circuit代考|LIMITER OR CLIPPING CIRCUIT

Diodes are often used in waveshaping applications. In particular, when used with a DC voltage in series with the diode, the output signal can be limited to the reference voltage level of the DC voltage source. Examples of clipping circuits are shown in Figure 2.13.

The simplified forward bias diode model of Figure $2.11$ can be used to analyze clipping circuits.

The circuit of Figure $2.13 \mathrm{a}$ will be used as an example of this analysis. When the input voltage $v_i \leq V_d+V_{n f}$, the diode is reverse biased (or OFF). ${ }^7$ Therefore, the diode can be thought of as an open circuit. The output voltage in this case follows the input voltage,
$$v_v=v_i .$$
When the voltage $v_i>V_d+V_{r e f}$, the diode is forward biased (or ON). Using the piece-wise linear model of the forward biased diode, a simplified equivalent circuit of the clipping circuit of Figure 2.13a is developed in Figure $2.14$.

The output voltage $v_o$ of the clipping circuit when the diode is forward biased is found by analyzing the circuit in Figure $2.14$ using superposition and voltage division,
$$v_o=\frac{r_d}{R_s+r_d} v_i+\frac{R_s}{R_s+r_d}\left(V_d+V_{\text {ref }}\right) .$$

If $r_d \ll R_S$ then the output voltage is held at a constant value
$$V_o=V_d+V_{r e f}$$
The input-output voltage relationships for the five diode clippers circuits are given in Table $2.1$

# 数字电路代考

## 电气工程代写|数字电路代写digital circuit代考|REVERSE BIAS MODELING

(a) 反向偏置二极管的电流源模型
(b) 反向偏置二极管的电阻器模型
(a) 如果二极管被替换为 $1 \mathrm{nA}$ 电流源，然后所有电路元件进行 $1 \mathrm{n} \mathrm{A}$ 的电流和二极管电流是 $-1 \mathrm{nA}$. 电阻两端 的电压由下式给出:
$$V_r=(\operatorname{lnA})(820 \Omega)=820 \mathrm{nV} .$$

$$V_d=V_r-4=-3.999999 \overline{18} \mathrm{~V} \approx-4.00 \mathrm{~V} .$$
(b) 反向电阻可以近似为:
$$r_r \approx|-25 \mathrm{~V}| /(\ln \mathrm{A})=25 \mathrm{G} \Omega$$

$$V_d=\frac{25 \mathrm{G} \Omega}{25 \mathrm{G} \Omega+820 \Omega}(-4 \mathrm{~V})=-3.999999869 \approx-4 I_d \quad=\frac{-4 V}{25 \mathrm{G} \Omega+820 \Omega}=160 \mathrm{pA}$$

## 电气工程代写|数字电路代写digital circuit代考|LIMITER OR CLIPPING CIRCUIT

$$v_v=v_i .$$

$$v_o=\frac{r_d}{R_s+r_d} v_i+\frac{R_s}{R_s+r_d}\left(V_d+V_{\mathrm{ref}}\right) .$$

$$V_o=V_d+V_{r e f}$$

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