Can you someone help me with number 5? I have done steps 1-4 already. I do not u
ID: 2293438 • Letter: C
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Can you someone help me with number 5? I have done steps 1-4 already. I do not undestand what break frequency is and how to calculate it. The highest decibal is 12.041 dB as you can see on the graph already. Please find f_break.
LT SPICE Lab 4 Sinusoidal Frequency Analysis 1. Wire the circuit of LT SPICE Lab 3 using LT Spice window. Use VAC voltage source (Vp- 0.4V) Inverting amplifier Gain in this circuit is G=Vo/Vin=-Ro/Ri=-10 Find frequency-response characteristics for this circuit. Right Click input voltage source and select function (none). Small signal analysis AC -> AC Amplitude 0.4V, AC Phase 0 => OK. 2. Click: SPICE Analyses -> AC Analysis => Type of Sweep =>Decade=>Number Pts=> 100; Start Freq=> 1kHz; End Freq=>10000 kHz=> OK. 3. Simulate> Run To see frequency response graph click: Plot Settings=> Add Traces => (select Op Amp output voltage) => OK 14. Print graph for report report, graph should have its name 5. Find and show on the graph the break frequency which is determined on the level Vo max/v2 or 0.707 Vo max, or on the level 3dB below of amplitude at 1kHz frequency.by using LT SPICE measurement instrument. LT SICH For G=-10, f breakExplanation / Answer
The cutoff frequency, corner frequency, or break frequency for an amplifier can be defined as frequency at which the power consumption of the amplifier becomes half (-3dB point) OR the gain becomes 0.707 times the it's dc value.
In the given frequency response, the gain (dc) is 12db, so the break frequency occurs to be at the point where gain becones 12-3 = 9dB.
In the given amplifier, due to negative feedback configuration of op-amp, the open-gain decreases by loopback factor and bandwidth increases by loopback factor. due to increase in bandwidth, the graph doesn't contain the whole response. So, to know the bandwidth, increase the sweeping frequency range to find the 9dB point.
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