Highpass RC circuit
- 1. Electronic Devices and Circuits UNIT – I SVEC19
- 2. The High-Pass RC Circuit 𝑋 𝐶 = 1 2𝜋𝑓𝐶 When f= 0; XC = ∞ (Capacitor is open circuited); Vo(t) = 0; then gain A= 𝑉𝑜(𝑡) 𝑉 𝑖(𝑡) = 0. When f increases, XC decreases, then output and gain increases. When f= ∞; XC = 0 (Capacitor is short circuited); M.Balaji, Department of ECE, SVEC 2 The circuit which transmits only high- frequency signals and attenuates or stops low frequency signals.
- 3. Sinusoidal input M.Balaji, Department of ECE, SVEC 3 Laplace transformed High- Pass RC circuit frequency response
- 4. 𝐴 = 𝑉𝑂 𝑠 𝑉𝑖 𝑠 = 𝑅 𝑅 + 1 𝐶𝑠 = 1 1 + 1 𝑅𝐶𝑠 Putting s= j 𝜔, 𝐴 = 1 1 − 𝑗 1 𝜔𝑅𝐶 = 1 1 − 𝑗 1 2𝜋𝑓𝑅𝐶 ∴ 𝐴 = 1 1 + 1 2𝜋𝑓𝑅𝐶 2 𝑎𝑛𝑑 𝜃 = −𝑡𝑎𝑛−1 1 2𝜋𝑓𝑅𝐶 M.Balaji, Department of ECE, SVEC 4
- 5. M.Balaji, Department of ECE, SVEC 5 𝐴𝑡 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑐𝑢𝑡𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓1, 𝐴 = 1 2 1 1 + 1 2𝜋𝑓1 𝑅𝐶 2 = 1 2 Squaring on both sides and equating the denominators, 1 2𝜋𝑓1 𝑅𝐶 = 1 ∴ 𝑓1 = 1 2𝜋𝑅𝐶
- 6. Step Voltage input A step signal is one which maintains the value zero for all times t < 0; and maintains the value V for all times t > 0. M.Balaji, Department of ECE, SVEC 6 t = 0 - t = 0 + t < 0 t > 0 t = 0 The transition between the two voltage levels takes place at t = 0. Vi = 0, immediately before t = 0 (to be referred to as time t= 0-) and Vi = V, immediately after t= 0 (to be referred to as time t= 0+).
- 7. Step input M.Balaji, Department of ECE, SVEC 7 Step input Step response for different time constants 𝑉0 𝑡 = 𝑉𝑓 − 𝑉𝑓 − 𝑉𝑖 𝑒− 𝑡 𝑅𝐶 V 𝑒− 𝑡 𝑅𝐶 When Vf = 0
- 8. Pulse input The pulse is equivalent to a positive step followed by a delayed negative step. M.Balaji, Department of ECE, SVEC 8 Pulse waveform Pulse waveform in terms of step
- 9. M.Balaji, Department of ECE, SVEC 9 RC >> tp
- 10. M.Balaji, Department of ECE, SVEC 10 RC comparable to tp
- 11. M.Balaji, Department of ECE, SVEC 11 RC << tp
- 12. M.Balaji, Department of ECE, SVEC 12 (a) RC >> tp (b) RC comparable to tp (c) RC << tp
- 13. Square wave input A square wave is a periodic waveform which maintains itself at one constant level V’ with respect to ground for a time T1, and then changes abruptly to another level V’’, and remains constant at that level for a time T2, and repeats itself at regular intervals of time T1+T2. A square wave may be treated as a series of positive and negative pulses. M.Balaji, Department of ECE, SVEC 13
- 14. M.Balaji, Department of ECE, SVEC 14 (a) Square wave input (b) When RC is arbitrarily large
- 15. M.Balaji, Department of ECE, SVEC 15 (c) RC > T
- 16. M.Balaji, Department of ECE, SVEC 16 (d) Output when RC comparable to T
- 17. M.Balaji, Department of ECE, SVEC 17 (e) Output when RC << T
- 18. Under steady state conditions, the capacitor charges and discharges to the same voltage levels in each cycle. So, the shape of the output waveform is fixed. 𝑓𝑜𝑟 0 < 𝑡 < 𝑇1, the output is given by 𝑣01 = 𝑉1 𝑒−𝑡/𝑅𝐶 At t=T1 ; 𝑣01 = 𝑉1 𝑒−𝑇1 /𝑅𝐶 𝑓𝑜𝑟 𝑇1 < 𝑡 < 𝑇1 + 𝑇2, the output is given by 𝑣02 = 𝑉2 𝑒−(𝑡−𝑇1 )/𝑅𝐶 At t=T1 +T2, 𝑣02 = 𝑉2 𝑒−𝑇2 /𝑅𝐶 Also 𝑉1 ′ − 𝑉2 = 𝑉 𝑎𝑛𝑑 𝑉1 − 𝑉2 ′ = 𝑉 M.Balaji, Department of ECE, SVEC 18
- 19. Expression for the percentage tilt The expression for the percentage tilt can be derived when the time constant RC of the circuit is very large compared to the period of the input waveform, i.e RC >> T. For a symmetrical square wave with zero average value 𝑉1 = −𝑉2, 𝑖. 𝑒 𝑉1 = 𝑉2 𝑉1 ′ = −𝑉2 ′ , 𝑖. 𝑒 𝑉1 ′ = 𝑉2 ′ 𝑇1 = −𝑇2 = 𝑇 2 M.Balaji, Department of ECE, SVEC 19
- 20. The output waveform for RC >> T is M.Balaji, Department of ECE, SVEC 20
- 21. 𝑉1 ′ = 𝑉1 𝑒− 𝑇 2𝑅𝐶 𝑎𝑛𝑑 𝑉2 ′ = 𝑉2 𝑒− 𝑇 2𝑅𝐶 𝑉1 − 𝑉2 ′ = 𝑉 𝑉1 − 𝑉2 𝑒− 𝑇 2𝑅𝐶 = 𝑉1 + 𝑉1 𝑒− 𝑇 2𝑅𝐶 = 𝑉 𝑉1 = 𝑉 1 + 𝑒− 𝑇 2𝑅𝐶 𝑉 = 𝑉1(1 + 𝑒− 𝑇 2𝑅𝐶) % 𝑇𝑖𝑙𝑡, 𝑃 = 𝑉1 − 𝑉1 ′ 𝑉 2 × 100 = 𝑉1 − 𝑉1 𝑒− 𝑇 2𝑅𝐶 𝑉1(1 + 𝑒− 𝑇 2𝑅𝐶) × 200% = 1 − 𝑒− 𝑇 2𝑅𝐶 1 + 𝑒− 𝑇 2𝑅𝐶 × 200% M.Balaji, Department of ECE, SVEC 21
- 22. When the time constant is very large, i.e, 𝑇 𝑅𝐶 ≪ 1 𝑃 = 1 − 1 + −𝑇 2𝑅𝐶 + ( −𝑇 2𝑅𝐶 )2 1 2! + ⋯ 1 + 1 + −𝑇 2𝑅𝐶 + ( −𝑇 2𝑅𝐶 )2 1 2! + ⋯ × 200% = 𝑇 2𝑅𝐶 2 × 200% = 𝑇 2𝑅𝐶 × 100% 𝜋𝑓1 𝑓 × 100% Where 𝑓1 = 1 2𝜋𝑅𝐶 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑐𝑢𝑡 − 𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑖𝑔ℎ − 𝑝𝑎𝑠𝑠 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 M.Balaji, Department of ECE, SVEC 22
- 23. Ramp input M.Balaji, Department of ECE, SVEC 23 When a high- pass RC circuit is excited with a ramp input, i.e 𝑣𝑖 𝑡 = 𝛼𝑡, 𝑤ℎ𝑒𝑟𝑒 𝛼 is the slope of the ramp then, 𝑉𝑖 𝑠 = 𝛼 𝑠2 From the Laplace transformed high-pass circuit, we get 𝑉0 𝑠 = 𝑉𝑖 𝑠 𝑅 𝑅 + 1 𝐶𝑠 = 𝛼 𝑠2 𝑅𝐶𝑠 1 + 𝑅𝐶𝑠 = 𝛼 𝑠(𝑠 + 1 𝑅𝐶 ) 𝛼𝑅𝐶 1 𝑠 − 1 𝑠 + 1 𝑅𝐶
- 24. Taking inverse Laplace transform on both sides, we get 𝑉0 𝑡 = 𝛼𝑅𝐶 1 − 𝑒− 𝑡 𝑅𝐶 For time t which are very small in comparison with RC, we have 𝑉0 𝑡 = 𝛼𝑅𝐶 1 − 1 + −𝑡 𝑅𝐶 + −𝑡 𝑅𝐶 2 1 2! + −𝑡 𝑅𝐶 3 1 3! + ⋯ = 𝛼𝑅𝐶 𝑡 𝑅𝐶 − 𝑡2 2(𝑅𝐶)2 + ⋯ = 𝛼𝑡 − 𝛼𝑡2 2𝑅𝐶 = 𝛼𝑡 1 − 𝑡 2𝑅𝐶 M.Balaji, Department of ECE, SVEC 24
- 25. M.Balaji, Department of ECE, SVEC 25 The above figures shows the response of the high-pass circuit for a ramp input when (a) RC >> T and (b) RC << T, where T is the duration of the ramp. For small values of T, the output signal falls away slightly from the input.
- 26. High-Pass RC circuit as a differentiator Sometimes, a square wave may need to be converted into sharp positive and negative spikes (pulses of short duration). By eliminating the positive spikes, we can generate a train of negative spikes and vice-versa. The pulses so generated may be used to trigger a multivibrator. If in a circuit, the output is a differential of the input signal, then the circuit is called a differentiator. M.Balaji, Department of ECE, SVEC 26
- 27. 𝜏 = 𝑅𝐶 ≪ 𝑇 then the circuit works as a differentiator 1/f ; here frequency must be small. At low frequencies, 𝑋𝑐 = 1 2𝜋𝑓𝐶 = 𝑙𝑎𝑟𝑔𝑒 𝑤ℎ𝑒𝑛 𝑐𝑜𝑚𝑝𝑎𝑟𝑒𝑑 𝑡𝑜 𝑅 The voltage drop across R is very small when compared to the drop across C. 𝑉𝑖 = 1 𝐶 𝑖(𝑡) 𝑑𝑡 + 𝑖 𝑡 𝑅 𝑉𝑖 = 1 𝐶 𝑖(𝑡) 𝑑𝑡 𝑉𝑖 = 1 𝑅𝐶 𝑉0 𝑑𝑡 (𝑖 𝑡 = 𝑉0 𝑅 ) Differentiating on both sides we get 𝑑𝑉𝑖 𝑑𝑡 = 1 𝑅𝐶 𝑉0 𝑉0 = 𝑅𝐶 𝑑𝑉𝑖 𝑑𝑡 Thus , the output is proportional to the derivative of the input. M.Balaji, Department of ECE, SVEC 27 V0 is small
- 28. Numerical Problem 1 A 1KHz symmetrical square wave of ±10𝑉 is applied to an RC circuit having 1ms time constant. Calculate and plot the output for the RC configuration as a High- Pass RC circuit. Solution: Given 𝑓 = 1𝐾𝐻𝑧, 𝑇 = 1𝑚𝑠 ∴ 𝑇𝑂𝑁 = 0.5𝑚𝑠 𝑎𝑛𝑑 𝑇𝑂𝐹𝐹 = 0.5𝑚𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐𝑎𝑙 𝑠𝑞𝑢𝑎𝑟𝑒 𝑤𝑎𝑣𝑒 𝜏 = 𝑅𝐶 = 1𝑚𝑠 Peak-to –peak amplitude VPP = 10- (-10)=20V Since RC is comparable to T, the capacitor charges and discharges exponentially. M.Balaji, Department of ECE, SVEC 28
- 29. High-Pass RC circuit M.Balaji, Department of ECE, SVEC 29
- 30. When the 1 KHz square wave shown by dotted line is applied to the RC high pass circuit. Under steady state conditions the output waveform will be as shown by the thick line. Since the input signal is a symmetrical square wave, we have 𝑉1 = −𝑉2 𝑎𝑛𝑑 𝑉1 ′ = −𝑉2 ′ 𝑉1 ′ = 𝑉1 𝑒− 𝑇 2𝑅𝐶 = 𝑉1 𝑒−0.5 1 = 0.6065 𝑉1 𝑉2 ′ = 𝑉2 𝑒− 𝑇 2𝑅𝐶 = 𝑉2 𝑒−0.5 1 = 0.6065 𝑉2 M.Balaji, Department of ECE, SVEC 30
- 31. 𝑉1 ′ − (−𝑉2) = 20 0.6065 𝑉1 + 𝑉1 = 20 ∴ 𝑉1 = 20 1.6065 = 12.449𝑉 𝑉2 = −𝑉1 = −12.449𝑉 𝑉1 ′ = 20 − 𝑉1 = 20 − 12.449 = 7.551𝑉 𝑉2 ′ = −𝑉1 ′ = −7.551 𝑉 M.Balaji, Department of ECE, SVEC 31
- 32. Numerical Problem 2 If a square wave of 5 KHz is applied to an RC high-pass circuit and the resultant waveform measured on a CRO was tilted from 15V to 10V, find out the lower 3- dB frequency of the high-pass circuit. Sol: f= 5 KHz 𝑉1= 15V 𝑉1 ′ = 10𝑉 𝑓1 = ? 𝑙𝑜𝑤𝑒𝑟 3 𝑑𝐵 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 M.Balaji, Department of ECE, SVEC 32
- 33. Peak-to-peak value of input 𝑉 = 𝑉1 + 𝑉2 = 15 + 15 = 30𝑉 𝑃 = 𝑉1 − 𝑉1 ′ 2 × 100 = 15 − 10 2 × 100 = 33.33% Also% tilt 𝑃 = 𝜋𝑓1 𝑓 × 100 33.33 = 3.14𝑓1 5 × 103 × 100 𝑓1 = 33.33 × 5 × 103 3.14 × 100 = 530.56𝐻𝑧 M.Balaji, Department of ECE, SVEC 33