Question 3.10: Design the op-amp Pierce oscillator in Figure 3.33 to oscill......
Design the op-amp Pierce oscillator in Figure 3.33 to oscillate at 10 kHz.
The op amp used is a general-purpose op amp, such as the 1458 or 741. General purpose op amps have typical slew rates of 0.5 V/μs and gain-bandwidth products (f_{T}) of 1 MHz. These parameters limit the frequency of oscillation. Of course, there are many special purpose op amps with high slew-rate values and high f_{T} values, which can be used to attain higher frequency of oscillation
Using supply voltage values of V^{+}=5{V} and V^{-}=-5V, the op amp output is limited to its saturation values (i.e., approximately ±5V). In this case the SR limits the frequency of oscillation to
f_{o}\lt {\frac{S R}{2\pi|v_{o}|}}={\frac{0.5 \times 10^{6}}{2\,\pi(5)}}=15.9\,\,\mathrm{kHz}
Hence, the oscillator should perform well at 10 kHz.
Letting L = 100 μH, the required total capacitance, from (3.68), is
\omega_{o}={\frac{1}{\sqrt{L C_{T}}}} (3.68)
C_{T}=\frac{1}{(2\pi f_{o})^{2}L}=\frac{1}{(2\pi10^{4})^{2}(100 \times 10^{-6})}=2.53~\mu\mathrm{F}
Letting C_{1}=10~\mu{F}, then
C_{2}={\frac{C_{1}C_{T}}{C_{1}-C_{T}}}={\frac{10 \times 10^{-6}(2.53 \times 10^{-6})}{10 \times 10^{-6}-2.53 \times 10^{-6}}}=3.39\ \mu\mathrm{F}
From (3.69), the required gain is
|A_{v o}|=\frac{R_{2}}{R_{1}}\geq\frac{C_{1}}{C_{2}} (3.69)
\left|A_{v o}\right|\geq{\frac{C_{1}}{C_{2}}}={\frac{10 \times 10^{-6}}{3.39 \times 10^{-6}}}=2.9
The gain condition can be satisfied with R_{2}=200\;{\mathrm{k\Omega~and~}}R_{1}=50\;{\mathrm{kΩ}} (i.e.,|A_{v o}{\vert}=4) .
The simulation is shown in Figure 3.36. The output is shown across C_{1}. The fundamental frequency of oscillation is 10.79 kHz. Although not shown, the waveform across {{C}}_{2} 2 has a peak value of approximately 5V.
A circuit that limits the amplitude of v_o in an op-amp tuned oscillator is shown in Figure 3.37. Other limiting circuits that can be used are those shown in Figures 1.18, 1.19, and 1.21.
