A certain op amp has a maximum output voltage range from -4 to +4 V. The output
ID: 2079104 • Letter: A
Question
A certain op amp has a maximum output voltage range from -4 to +4 V. The output can source or sink a maximum current of 10 mA. The slew-rate limit is SR = 5 V/mu s. This op-amp is used to create a noninverting amplifier with a feedback resistance of 100 k ohm a dc closed- loop gain of 10 and connected to a load R_L. Find the full-power bandwidth of the op-amp. For a sinusoidal input of 1 kHz and R_L = 1k ohm, find the maximum peak voltage possible without distortion. For a sinusoidal input of 1 kHz and R_L = 100 ohm, find the maximum peak voltage possible without distortion. For a sinusoidal input of 1 MHz and R_L = 1k ohm, find the maximum peak voltage possible without distortion. If R_L = 1k ohm and upsilon_s (t) = 0.5 sin(2 pi 10^6t), sketch the steady-state output waveform to scale against time.Explanation / Answer
1.we know that slew rate is rate of change of voltage of op amp so the the full power bandwidth will be equal to
dV/dt=w * V where w is angular frequency in radian which is given 2*pi*f . f is frequency in Hz. V is maximum peak voltage.
in above problem V is 4V and dV/dt=5*10^6V/s so fullpower bandwidth of opamp will be
=5*10^6/(8*pi)
2. we know that maximum current of sink is 10mA so the output voltage for 1kohms load is 10V which is not posible as opamp has max peak voltage is 4V and also full power bandwidth is higher than 1kHz frequency so maximum output voltage can be 4 V it means input voltage of 4/9V as non inverting opamp gain is 10V.
3. we know that maximum current of sink is 10mA so the output voltage for 100ohms load is 1V which is posible as opamp has max peak voltage is 4V and also full power bandwidth is higher than 1kHz frequency so maximum output voltage can be 1 V it means input voltage of 1/9V as non inverting opamp gain is 10V.
4. we know that maximum current of sink is 10mA so the output voltage for 1kohms load is 10V which is not posible as opamp has max peak voltage is 4V but full power bandwidth is less than 1kHz frequency so maximum output voltage cannot be 4 V so it shoud be =5*10^6/(2*pi*1*10^6) so input voltage will be =5*10^6/(2*pi*1*10^6) /10 as non inverting opamp gain is 10V.
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