eifueo/docs/2a/ece240.md

# ECE 240: Electronic Circuits

## Diodes

A **diode** is a two-terminal device that only allows current to flow in the direction of the arrow.

<img src="https://upload.wikimedia.org/wikipedia/commons/b/b4/Diode_symbol.svg" width=300>(Source: Wikimedia Commons)</img>

The current across a diode is, where $I_s$ is a forced saturation current, $V$ is the voltage drop across it, and $V_T$ is the **thermal voltage** such that $V_T=\frac{kT}{q}$, where $T$ is the temperature, $k$ is the Boltzmann constant, and $q$ is the charge of an electron:

$$I=I_s\left(e^{V/V_T}-1\right)$$

!!! tip
    - $V_T\approx\pu{25 mV}$ at 20°C
    - $V_T\approx\pu{20 mV}$ at 25°C

A diode is open when current is flowing reverse the desired direction, resulting in zero current, until the voltage drop becomes so great that it reaches the **breakdown voltage** $V_B$. Otherwise, the above current formula is followed.

<img src="https://upload.wikimedia.org/wikipedia/commons/2/2a/Diode_current_wiki.png" width=500>(Source: Wikimedia Commons)</img>

Diodes are commonly used in **rectifier circuits** — circuits that convert AC to DC.

By preventing negative voltage, a relatively constant positive DC voltage is obtained. The slight dip between each hill is known as **ripple** $\Delta V$.

<img src="https://upload.wikimedia.org/wikipedia/en/8/8b/Reservoircapidealised.gif" width=500>(Source: Wikimedia Commons)</img>

In a simple series RC circuit, across a diode, Where $R_LC>>\frac 1 \omega$, and $f=\frac{\omega}{2\pi}$:

$$\Delta V\approx \frac{I_\text{load}}{2fC}\approx\frac{V_0}{2fR_LC}$$

### Zener diodes

A Zener diode is a calibrated diode with a known breakdown voltage, $V_B$. If the voltage across the diode would be greater than $V_B$, it is **capped at $V_B$.**

<img src="https://upload.wikimedia.org/wikipedia/commons/9/92/Zener_diode_symbol-2.svg" width=200>(Source: Wikimedia Commons)</img>

## Voltage/current biasing

Solving for current for each element in a series returns a negative linear line and other non-linear lines.

- the linear line is the **load line**, which represents the possible solutions to the circuit when it is loaded
- Depending on the base current $I_s$, the diode or transistor will be **biased** toward one of the curves, and the voltage and current will settle on one of the intersections, or **bias points**.

<img src="https://upload.wikimedia.org/wikipedia/commons/2/27/BJT_CE_load_line.svg" width=600>(Source: Wikimedia Commons)</img>

- To bias current, as $R\to\infty$ (or, in practical terms, $R>>diode$), the slope of the load line $\to 0$, which results in a constant current.
- To bias voltage, as $R\to 0$, the slope of the load line $\to\infty$, which results in a constant voltage.

!!! example
    <img src="https://miro.medium.com/v2/resize:fit:432/1*mijJgpHdt7DDmrPsb7tOcg.png" width=200 />
    
    The current across the resistor and the diode is the same:
    
    \begin{align*}
    i_D&=\frac{V_s}{R} \\
    i_D&\approx I_se^{V_D/V_T}
    \end{align*}

If a diode is put in series with AC and DC voltage sources $V_d(t)$ and $V_D$:

\begin{align*}
i_D(t)&=I_se^{(V_D+V_d(t))/V_T} \\
&=\underbrace{I_se^{V_D/V_T}}_\text{bias current}\ \underbrace{e^{V_d(t)/V_T}}_\text{$\approx 1+\frac{V_d}{V_T}$} \\
&=I_D\left(1+\frac{V_d}{V_T}\right) \\
&=\underbrace{I_D}_\text{large signal = bias = DC}+\underbrace{I_D\frac{V_d(t)}{V_T}}_\text{small signal = AC}
\end{align*}

Diodes may act as resistors, depending on the bias current. They may exhibit a **differential resistance**:
$$r_d=\left(\frac{\partial i_D}{\partial v_D}\right)^{-1} = \frac{V_T}{I_D}$$

!!! example
    Thus from the previous sequence:
    
    $$i_D(t)=I_D+\frac{1}{r_d}V_d(t)$$

### Signal analysis

1. Analyse DC signals
  - assume blocking capacitors are open circuits
  - turn off AC sources
2. Analyse AC signals
  - assume blocking capacitors are shorts
  - turn off DC sources
  - replace diode with effective resistor (the differential resistor)

!!! tip
    Most $R$s in the circuit can be assumed to be significantly greater than $r_d$, so $r_d$ can be removed in series or $R$ can be removed in parallel.

!!! warning
    Oftentimes, turning off a DC source to nowhere is actually a short to ground.

## MOSFETs

A MOSFET is a transistor. Current flows from the drain to the source, and only if voltage is applied to the gate.

<img src="https://upload.wikimedia.org/wikipedia/commons/6/69/Mosfet_saturation.svg" width=500>(Source: Wikimedia Commons)</img>

<img src="https://upload.wikimedia.org/wikipedia/commons/9/91/Transistor_Simple_Circuit_Diagram_with_NPN_Labels.svg" width=300>(Source: Wikimedia Commons)</img>

In strictly DC, current passes the gate if the gate voltage is greater than the threshold voltage $V_G>V_t$. The difference between the two is known as the **overdrive voltage** $V_{ov}$:

$$V_{ov}=V_G-V_t$$

At a small $V_{DS}$, or in AC, the slope of $I_D$ to $V_{DS}$ is proportional to $V_G$. The **channel transconductance** $g_{DS}$ represents this slope, which is constant based on the **transconductance parameter** of the device.

$$\frac{I_D}{V_{DS}}=g_{DS}=k_nV_{ov}$$

Before the saturation region, the current grows exponentially:

$$\boxed{I_s=k_n(V_{ov}-\tfrac 1 2V_{DS})V_{DS}}$$

Afterward, it remains constant, based on the overdrive voltage:

$$\boxed{I_s=\frac 1 2k_nV_{ov}^2}$$

### Common-source amplifiers

<img src="https://upload.wikimedia.org/wikipedia/commons/4/4f/N-channel_JFET_common_source.svg" width=200>(Source: Wikimedia Commons)</img>

Where $V_{out}=V_{DS}$:

<img src="https://media.cheggcdn.com/media/b65/b65d59bd-ac35-4d28-b811-0ad1b5cf5bb6/phpCBbhn6" width=700 />

$|V_{ds}|>|V_{gs}|$ indicates AC voltage gain.

The gain can be modelled with Ohm's law:

$$V_{DS}=V_{DD}-I_DR_D=V_{DD}-\frac 1 2k_n(V_{GS}-V_t)R_D$$

At a certain gate voltage:

\begin{align*}
A_V&=\frac{\partial V_{DS}}{\partial V_{GS}} \\
&=-g_{DS}R_D
\end{align*}

### Small signal analysis

The current from the drain to the source is equal to:

$$i_D=g_mV_{gs}$$

For small signals, a transistor is equivalent to, where $r_0=\frac{1}{\lambda I_D}=\frac{V_A}{I_D}$:

<img src="https://i.stack.imgur.com/EZK7K.png" width=600 />

It can be assumed that the differential resistance is always significantly smaller than any other external resistance: $r_o << R_d$.

To solve for the output resistance of the amplifier, turn off all sources and take the Thevenin resistance $R_{DS}$.

### Common-drain amplifiers / source followers

The input resistance of common amplifiers is infinity.

<img src="https://upload.wikimedia.org/wikipedia/commons/3/30/N-channel_JFET_source_follower.svg" width=200>(Source: Wikimedia Commons)</img>

As $V_{gs}$ is not necessarily zero, dependent sources must be left in when solving for output resistance, and so a small test source at the point of interest is required.

### Common-gate amplifiers

These can be represented by either the T-model or pi-model. The gate of the transistor is grounded.

$$
A_{VO}=g_mR_d \\
G_V=\frac{V_o}{V_{sig}}=g_mR_d\left(\frac{1}{1+g_mR_{sig}}\right)
$$

<img src="https://upload.wikimedia.org/wikipedia/commons/9/99/Common_Gate.svg" width=200 />

<img src="https://upload.wikimedia.org/wikipedia/commons/a/a9/Common_gate_output_resistance.PNG" width=400 />

### Differential pairs

These are used at the input of opamps.

In **differential mode,** assuming $Q_1=Q_2$:

$V_{in}^+=-V_{in}^-=\frac{V_d}{2}$, so the current going down from both gates is equal $i_{gs1}=-i_{gs2}$. This means that node before $R_E$ is effectively ground, so the circuit can be split into two common source circuits.

$$G_D=\frac{V_o^--V_o^+}{V_d}=\frac{R_{C1}g_m}{1}=-\frac{-R_{C1}}{r_m}$$

<img src="https://upload.wikimedia.org/wikipedia/commons/3/3a/Differential_amplifier_long-tailed_pair.svg" width=300 />

In **common mode**:

$V_{in}^+=V_{in}^-$

$$G_{CM}=-\frac{R_D}{r_m+R_S+2R_C}$$

The **common-mode rejection ratio** is:

$$\frac{G_D}{G_{CM}}=1+\frac{2R_C}{r_m+R_s}$$

## MOSFET biasing

To bias a MOSFET:

- the transistor must be on: $V_{GS}>V_t$
- the transistor must be saturated $V_{DS} > (V_{GS}-V_t)$

$$V_{GS}=V_G-R_EI_D$$

This is a negative feedback loop that forces a constant $I_D$.

<img src="https://i.stack.imgur.com/Yxslx.png" width=300 />

With two DC supplies ($-V_{EE}, V_{DD}$), having an $R_G$ results in:

$$I_D=\frac{-V_{EE}}{R_S}-\frac{V_{GS}}{R_S}$$