# ECE 240: Electronic Circuits ## Diodes A **diode** is a two-terminal device that only allows current to flow in the direction of the arrow.

(Source: Wikimedia Commons) 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.

(Source: Wikimedia Commons) 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** $\Delta V$.

(Source: Wikimedia Commons) 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 $V_B$.**

(Source: Wikimedia Commons) ## 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**.

(Source: Wikimedia Commons) - 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

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.

(Source: Wikimedia Commons)

(Source: Wikimedia Commons) 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

(Source: Wikimedia Commons) Where $V_{out}=V_{DS}$:

$|V_{ds}|>|V_{gs}|$ 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 $r_0=\frac{1}{\lambda I_D}=\frac{V_A}{I_D}$:

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.

(Source: Wikimedia Commons) 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.