Inverting Differentiator

     The output of an inverting differentiator is derivative of its input. A sinusoidal input voltage is

connected to the op-amp with three different frequencies. Power supplies voltage of op-amp are ± 5v.

Figure 1: An Inverting Differentiator Op-amp

i) f = 1 KHz , amplitude = 200 mv , and offset = 0v

According to Volt/div of oscilloscope (measured voltage), Vin = 200 mv and Vout = 1000.25 mv

Vin = Asin(ωt) , Vout = Aω cos(ωt) , ω = 2πf  , Vout = -RC dVin / dt , Vout = -(RCAω)*cos(ωt)

If R = 1KΩ and C = 1μF so RC = 10E(-3). Vout = -(0.4π)*cos(ωt) , Vout = -1.257 *cos(ωt).

Output amplitude = -1257 mv.

percent error of output voltage = [(expected voltage - measured voltage) / expected voltage]*100%

percent error = [(1257 - 1000.25) / 1257]*100% = 22.43%

Figure 2: A Sinusoidal Input voltage of 1 KHz

ii) f = 2 KHz , amplitude = 200 mv , and offset = 0v

According to Volt/div of oscilloscope (measured voltage) , Vin = 200 mv and Vout = 2500 mv

Vin = Asin(ωt) , Vout = Aω cos(ωt) , ω = 2πf  , Vout = -RC dVin / dt , Vout = -(RCAω)*cos(ωt)

If R = 1KΩ and C = 1μF so RC = 10E(-3). Vout = -(0.8π)*cos(ωt) , Vout = -2.513 *cos(ωt).

Output amplitude = -2513 mv.

percent error of output voltage = [expected voltage - measured voltage| / expected voltage]*100%

percent error = [(2513 - 2500) / 2513]*100% = 0.52%

Figure 3: A Sinusoidal Input voltage of 2 KHz

iii) f = 500 Hz , amplitude = 200 mv , and offset = 0v

According to voltage div of oscilloscope (measured voltage), Vin = 200 mv and Vout = 500 mv

Vin = Asin(ωt) , Vout = Aω cos(ωt) , ω = 2πf  , Vout = -RC dVin / dt , Vout = -(RCAω)*cos(ωt)

If R = 1KΩ and C = 1μF so RC = 10E(-3). Vout = -(0.2π)*cos(ωt) , Vout = -0.6283 *cos(ωt).

Output amplitude = -628.3 mv.

percent error of output voltage = [expected voltage - measured voltage| / expected voltage]*100%

percent error = [(628.3 - 500) / 628.3]*100% = 20.42%

Figure 4: A Sinusoidal Input voltage of 500 Hz

According to the percent errors are calculated for three different frequencies, when f = 2 KHz percent

error is less than others low frequencies. In other words, percent error of output voltage decrease 

when frequencies increase.

Frequency (Hz)	Input Voltage (mv)	Expected Output Voltage (mv)	Measured Output Voltage (mv)	Percent Error%
500	200	500	628.3	20.42
1000	200	1257	1000.25	22.43
2000	200	2513	2500	0.52

Figure 5: A Schematic of an Inverting Differentiator Op-amp

When input voltage is 1 volt, the inverting differentiator op-amp will go to saturation state.

Vin = Asin(ωt) , Vout = Aω cos(ωt) , ω = 2πf  , Vout = -RC dVin / dt , Vout = -(RCAω)*cos(ωt)

If R = 1KΩ and C = 1μF so RC = 10E(-3). 

If f = 1 KHz. Vout = -(2π)*cos(ωt) , Vout = - 6.28 *cos(ωt).

Amplitude of output voltage is equal - 6.28 v  that is bigger than power supply voltage (± 5v).

Figure 6: An Inverting Differentiator Op-amp in Saturation Situation