Part B: Component Questions - LM358 OP-Amp The LM358 op amp can be a single or d
ID: 2989105 • Letter: P
Question
Part B: Component Questions - LM358 OP-Amp
The LM358 op amp can be a single or dual supply op-amp. This mean that it can operate with just a single supply i.e. + 5 volt or with dual supplies i.e. (+ and -) 5 volt supplies. With a non-inverting configuration with dual supplies (+5v and -5v) being supplied to the LM358 Please answer the following questions.
1. What is the gain formula for a non inverting op amp?
2. With this configuration, is there a maximum output voltage peak to peak, if so what would it be?
3. This op amp is configured for a gain of 11. Input signal is an AC sine wave signal. What is the maximum AC peak to peak voltage input?
4. What happens to your output signal when the input signal is above the maximum peak to peak input voltage?
The LM358 op amp can be a single or dual supply op-amp. This mean that it can operate with just a single supply i.e. + 5 volt or with dual supplies i.e. (+ and -) 5 volt supplies. With a inverting configuration with dual supplies (+5v and -5v) being supplied to the LM358 Please answer the following questions.
1. What is the gain formula for an inverting op amp?
2. With this configuration, is there a maximum output voltage peak to peak, if so what would it be?
3. This op amp is configured for a gain of 10. Input signal is an AC sine wave signal. What is the maximum AC peak to peak voltage input?
4. What happens to your output signal when the input signal is above the maximum peak to peak input voltage?
The LM358 op amp can be a single or dual supply op-amp. This mean that it can operate with just a single supply i.e. + 5 volt or with dual supplies i.e. (+ and -) 5 volt supplies. With a non-inverting configuration with single supply (+5v) being supplied to the LM358 Please answer the following questions.
1. With this configuration, is there a maximum output voltage peak to peak, if so what would it be?
2. In the above configuration with a gain of 10 and the input signal is an AC sine wave at 450mv peak to peak, please draw in your notebook what the output signal will should look like.
3. Explain why the output signal looks the way it does.
Explanation / Answer
1. for the noninverting structure we can see that,
Vout*Rf/(Rf+Rg) =Vin (assuming ideal opamp)
so,
Vout/Vin = (1 + Rf/Rg)---------------ans.
2. the maximum peak to peak voltage accurs when it saturates or goes out of the linear region then
Vmax = +5 volt nd Vmin = -5 volt-----------------ans.
3.
if u draw the transfer curve u see that this is odd symmetric. it is a linear curve for (-Vin, +Vin) where the slope is defined by the gain Av.
for vin> Vin+ he out put voltage saturates to +5 V
similar;y if Vin <Vin- then the output voltage saturates to -5 Volt.
so, to calculate the max vin in the linear region we have,
Vin*10 = +/- 5
or,
Vin = +/- 0.5 V------------------------ans.
4. as stated above if the input is higher than the above shown Vin then the opamp saturate and goes to Vout = +5 Volts or-5 V respectively.
in part B:
1. here also the maximum voltage could be +5 V and 0 V------------ans.
2. as stated in the first section the output voltage can be defined as,
Vout = gain*Vin = 10*(0.45sin(wt))--------------volt.
but as u know the voltage variation can only be from 0 to 5 V.
so, for the positive half of the output the opamp will give full swing output but for the negative half opamp will saturte and give 0 output result.
so, the output will look like half wave rectified output with a positive half swing of 4.5 V-------------------ans,
3.
simply because we know that the output can not cross the defined biasde voltages. the opamps are having MOS or BJT transistor to act as a proper amplifier all the devices must lie in proper region. (saturated for MOS and linear for the BJT devices ). if the output voltage crosses the defined bias voltages many of these devices will get shutt off so the opamp will ot act as an amplifier and the output will saturets to its nearby voltage defined by the bias circuit which is 0 for the above circuit.