In this problem you will develop a simple spice model for an op amp with a domin

Question

In this problem you will develop a simple spice model for an op amp with a dominant pole frequency response. The circuit model is shown below. If Re = 1 M ohm, and the pole is at f0 = 10 Hz. find CC. If Gm = 0.1 A V-1, write the voltage transfer function for this model in the form A(f) = A0/(1 + j[f /10Hz]) and find the value of A0. Simulate this model in spice or this transfer function in MATLAB and plot both the magnitude and phase as a function of frequency from 0.1 Hz to 10 kHz. The frequency response of an op amp is modeled with a dominant pole at f = f0. A(f) = A0/1 + j(f/f0) Find the unity-gain bandwidth fT. which is defined as the frequency for which the magnitude of the gain is zero, in terms of A0 and f0. Use Matlab to plot |A(f)|, in dB for 0.1 Hz le 10kHz for A0 = 3, 10, and 100 . Derive the closed-loop voltage transfer function of the noninverting amplifier circuit shown below if the op amp open-loop gain is A(f). Use MATLAB to plot |v0/vi| in dB for 0.1 Hz le 10 kHz for A0 = 100 and f0 = 10 Hz if R1 = 10K ohm and R2 = 100 K ohm. Plot A(f) on the same graph. What do you notice about the high-frequency behaviors? 15.9nF. (b) The low-frequency gain is 10s. fT = f0

Explanation / Answer

a) The pole frequency is given by fo = 1/ 2RcCC CC = 1 / (2)(3.14)(106)(10) CC = 0.0159*10-6 CC = 15.9nF
b) Gm = 0.1A/V
Applying KCL Gm vd =( vo / Rc )+ (- vo/jXC) Gm vd = vo [ (1/Rc )- ( 1/jXC)] vo / vd = Gm /  [ (1/Rc )- ( 1/jXC) ] vo / vd = Gm /  [ (1/Rc )- (j CC ) ] Av = GmRc / (1 - jCcRc ) Av = A/ ( 1 - j/c ) Av = A / ( 1 - jf / fc )
A = GmRc = (0.1)(106) A = 105
2.
A(f) = Ao / [ 1 + j(f/fo) ] |A(f)| = Ao / [ 1 + (f/fo)2 ] At f = fT, |A(f) |= 1 1 = Ao / [ 1 + (fT /fo)2 ] Squaring 1 = Ao2 / [ 1 + (fT /fo)2 ] [ 1 + (fT /fo)2 ] = Ao2 (fT /fo)2 = Ao2 - 1 (fT /fo) = (Ao2 - 1) fT = fo (Ao2 - 1)



CC = 15.9nF
b) Gm = 0.1A/V
Applying KCL Gm vd =( vo / Rc )+ (- vo/jXC) Gm vd = vo [ (1/Rc )- ( 1/jXC)] vo / vd = Gm /  [ (1/Rc )- ( 1/jXC) ] vo / vd = Gm /  [ (1/Rc )- (j CC ) ] Av = GmRc / (1 - jCcRc ) Av = A/ ( 1 - j/c ) Av = A / ( 1 - jf / fc )
A = GmRc = (0.1)(106) A = 105
2.
A(f) = Ao / [ 1 + j(f/fo) ] |A(f)| = Ao / [ 1 + (f/fo)2 ] At f = fT, |A(f) |= 1 1 = Ao / [ 1 + (fT /fo)2 ] Squaring 1 = Ao2 / [ 1 + (fT /fo)2 ] [ 1 + (fT /fo)2 ] = Ao2 (fT /fo)2 = Ao2 - 1 (fT /fo) = (Ao2 - 1) fT = fo (Ao2 - 1)




2.
A(f) = Ao / [ 1 + j(f/fo) ] |A(f)| = Ao / [ 1 + (f/fo)2 ] At f = fT, |A(f) |= 1 1 = Ao / [ 1 + (fT /fo)2 ] Squaring 1 = Ao2 / [ 1 + (fT /fo)2 ] [ 1 + (fT /fo)2 ] = Ao2 (fT /fo)2 = Ao2 - 1 (fT /fo) = (Ao2 - 1) fT = fo (Ao2 - 1)



(fT /fo)2 = Ao2 - 1 (fT /fo) = (Ao2 - 1) fT = fo (Ao2 - 1)

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