written 5.4 years ago by |
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.