EE 255 - Chapter 3b

# EE 255 - Chapter 3b - (Note T he above discussion assumes...

This preview shows pages 1–4. Sign up to view the full content.

75 ( Note: The above discussion assumes that transformer T 3 is never in either state long enough for it to saturate.) 3-8. Figure P3-3 shows a relaxation oscillator with the following parameters: R 1 = variable Ω = 1500 2 R 1.0 F C μ = V 100 DC = V BO 30 V V = 0.5 mA H I = (a) Sketch the voltages v t C ( ) , v t D ( ) , and v t o ( ) for this circuit. (b) If R 1 is currently set to 500 k Ω , calculate the period of this relaxation oscillator. S OLUTION (a) The voltages v C (t) , v D (t) and v o (t) are shown below. Note that v C (t) and v D (t) look the same during the rising portion of the cycle. After the PNPN Diode triggers, the voltage across the capacitor decays with time constant τ 2 = R 1 R 2 R 1 + R 2 C , while the voltage across the diode drops immediately.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
76 (b) When voltage is first applied to the circuit, the capacitor C charges with a time constant τ 1 = R 1 C = (500 k Ω )(1.00 μ F) = 0.50 s. The equation for the voltage on the capacitor as a function of time during the charging portion of the cycle is ( ) 1 t R C C v t A B e - = + where A and B are constants depending upon the initial conditions in the circuit. Since v C (0) = 0 V and v C ( ) = 100 V, it is possible to solve for A and B . A = v C ( ) = 100 V A + B = v C (0) = 0 V g159 B = -100 V Therefore, ( ) 0.50 100 100 V t C v t e - = - The time at which the capacitor will reach breakover voltage is found by setting v C (t) = V BO and solving for time t 1 :
77 1 100 V 30 V 0.50 ln 178 ms 100 V t - = - = Once the PNPN Diode fires, the capacitor discharges through the parallel combination of R 1 and R 2 , so the time constant for the discharge is ( )( ) ( ) 1 2 2 1 2 500 k 1.5 k 1.0 F 0.0015 s 500 k 1.5 k R R C R R τ μ Ω Ω = = = + Ω + Ω The equation for the voltage on the capacitor during the discharge portion of the cycle is ( ) 2 t C v t A B e - = + ( ) 2 BO t C v t V e - = The current through the PNPN diode is given by ( ) 2 BO 2 t D V i t e R - = If we ignore the continuing trickle of current from R 1 , the time at which i D (t) reaches I H is ( ) ( )( ) 2 2 2 BO 0.0005 A 1500 ln 0.0015 ln 5.5 ms 30 V H I R t R C V Ω = - = - = Therefore, the period of the relaxation oscillator is T = 178 ms + 5.5 ms = 183.5 ms, and the frequency of the relaxation oscillator is f = 1/ T = 5.45 Hz. 3-9. In the circuit in Figure P3-4, T 1 is an autotransformer with the tap exactly in the center of its winding. Explain the operation of this circuit. Assuming that the load is inductive, sketch the voltage and current applied to the load. What is the purpose of SCR 2 ? What is the purpose of D 2 ? (This chopper circuit arrangement is known as a Jones circuit .) S OLUTION First, assume that SCR 1 is triggered. When that happens, current will flow from the power supply through SCR 1 and the bottom portion of transformer T 1 to the load. At that time, a voltage will be applied to the bottom part of the transformer which is positive at the top of the winding with respect to the

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

EE 255 - Chapter 3b - (Note T he above discussion assumes...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online