Circuit diagram:

Waveforms:

When d.c supply $V_{BB}$ is applied.Capacitor 'C' begins to charge through R exponentially toward $V_{BB}$.During this charging emitter circuit of UJT is an open circuit.

The control voltage $V_c = V_E$ is given by,

$$V_{c}=V_{E}=V_{B B}\left(1-e^{-t / R C}\right)$$

The charging time constant is given by

$$\tau_1 = RC$$

When this emitter voltage $V_E = V_c$ reaches the peak point voltage $V_{p}=\eta V_{B B}+V_{D}$,the UJT turns ON and Capacitor C discharges through the resistance $R_1$.

The dischargeing time constant is given by,

$$\tau_2 = R_1C$$

Here $\tau_2$ is much smaller than $\tau_1$

When discharging voltage dropped to $V_v$, UJT turns OFF. The charging and discharging process of capacitor repeats for each time period T and is given by,

$$T=RC \operatorname{ln}\left(\frac{1}{1-\eta}\right)$$

$R_2$ is used for thermal stability of $V_p$, the value of $R_2$ can be calculated by using formula,

$$R_{2}=\frac{10^{4}}{\eta{V_{B B}}}$$

The maximum value of R is determined by

$$R_{\max }=\frac{V_{B B}-V_{P}}{I_{p}}=\frac{V_{B B}-\left(\eta V_{B B}+V_{D}\right)}{I_{P}}$$

The minimum value of R is determined by

$$R_{\min }=\frac{V_{B B}-V_{v}}{I_{v}}$$

$R_1$ can be calculated as,

$$R_{1}=\frac{V_{B B}}{\text l e a k a g e \text { curent }}$$

Waveform Explanation:

At t = 0

• Supply voltage $+\mathrm{V}_{\mathrm{BB}}$ is given the circuit then $\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{E}}=\mathrm{V}_{\mathrm{A}}=0$ Volts.
• By internal potential divider voltage at cathode is $\mathrm{V}_{\mathrm{K}}=+\mathrm{n} V_{\mathrm{BB}}$
• If $\mathrm{V}_{\mathrm{A}}\left(\mathrm{V}_{\mathrm{E}}\right)\lt\mathrm{V}_{\mathrm{K}},$ then internal PN junction (diode) is reverse biased.
• Therefore, it acts as open switch and UJT is OFF.

At $t\gt0$

• Current flows from $+\mathrm{V}_{\mathrm{BB}}$ to ground through resistor $\mathrm{R} 1$ and capacitor $\mathrm{C}$ . Thus capacitor starts charging exponentially as constant voltage source $+\mathrm{V}_{\mathrm{BB}}$ is applied.
• Hence in waveform from t $=0, t \gt 0, \mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{E}}$ increasing exponentially.
• Therefore no current flows through UJT.
• $\therefore \mathrm{V}_{\mathrm{R} 1}=\mathrm{I.R}_1=0$

$\therefore \mathrm{V}_{\mathrm{B} 1}=0$

• $V_{B 2}=V_{B B}-V_{R 2}$
• $\therefore V_{B 2}=V_{B B}$

At $t = t_1$

• Increasing voltage across capacitor is now become $\mathrm{V}_{\mathrm{C}}=\left(\mathrm{\eta} \mathrm{V}_{\mathrm{BB}}+0.7\right)$ Volts. Which is equal to anode voltage from equivalent circuit.

$\therefore V_{C}=V_{E}=V_{A}=\left(n V_{B B}+0.7\right)$ Volts.

• As $V_{A}\gt V_{K}$ internal PN junction (diode) is forward biased hence it acts as a closed switch, and UJT is ON i.e. UJT starts conducting.
• $\mathrm{V}_{\mathrm{B} 1}=\mathrm{V}_{\mathrm{R} 1}=$ maximum and $\mathrm{V}_{\mathrm{B} 2}=\mathrm{V}_{\mathrm{BB}}-\mathrm{V}_{\mathrm{R} 2}=0$

At $t\gt t_{1 :}$

• Charged capacitor finds path for discharging. Charged capacitor discharges exponentially through ON UJT $\&$ resistor R1. i.e. $V_{c}$ decreases.
• As $\mathrm{V}_{\mathrm{B} 1}=\mathrm{V}_{\mathrm{c}},$ therefore $\mathrm{V}_{\mathrm{B} 1}$ also starts decreasing exponentially.

At $t= t_2$:

• Decreasing capacitor voltage reached to a point where $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{k}}\left(\text { as } \mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{E}}=\mathrm{V}_{\mathrm{A}}\right)$
• $\therefore$ Again internal PN junction (diode) reverse biases.
• Therefore UJT is OFF $\&$ no current flows through it.
• Therefore all the waveforms continuously repeat themselves.