Hey there.. this is my lab discussion.. which i'll attend it this coming monday.. however.. there are questions that i'm not sure the formula to calculate it.. i really hope u guys can varified or corrected my formula.. =) 1. The problem statement, all variables and given/known data Vin(pp) = 1.0V R1 = 10k Ω Rf = 33k Ω is it Vin = Vin(pp) / 2 ? Vout(pp) = -(Rf / R1) Vin(pp) What are the peak voltages across R1 and Rf? Calculate the peak currents of R1 and Rf 2. Relevant equations Current through R1 = Vin(pp)/R1 Current through Rf = Vout(pp)/Rf to find peak voltage, can i use voltage divider rule? VR1 = 10k/43k x Vin VRf = 33k/43k x Vin
Vout = -(Rf/R1) Vin This is correct. The junction of R1 and Rf (the -ve input) is a "Virtual earth". Because the open-loop gain of the op-amp is very high (more than a million), the -ve input voltage is always almost the same as the +ve input voltage. The +ve input is a 0V so you can assume that the -ve input is also at 0V. In fact it will differ by the "input offset voltage" but this is usually only a few mV or less. Therefore, the voltage across R1 is Vin(pp) and the voltage across Rf is Vout(pp). The currents are simply = V/R (ohms law).
I was not sure what you meant by this. If your input signal is a sine wave (with no DC offset - centered around 0V) then it has a positive peak value and a negative peak value that are equal and opposite and half the peak-to-peak value. This way of measuring the amplitude of a sine wave is not normally used. Usually the peak amplitude or the RMS amplitude is stated.
1. The problem statement, all variables and given/known data Vin(pp) = 1.0V R1 = 10k Ω Rf = 33k Ω is it Vin = Vin(pp) / 2 ? Vout(pp) = -(Rf / R1) Vin(pp) What are the peak voltages across R1 and Rf? Calculate the peak currents of R1 and Rf ********* Answers - For an inverting op amp - Vout = - Vin x (Rf/Rin) therefore Vout = -Vin x(33K/10k) = - Vin x 3.3 The input voltage stated here is 1 Vp-p. As one responder pointed out the type of signal was not indicated. It could be a sine wave, square wave or a pulse or actually any type of waveform. Let's assume it is a sine wave centered about ground (0 Volts). So the wave swings positive 0.5 volts and negative 0.5 volts. When the input swings positive 0.5 volts the output will swing negative (inverting amplifier). Vout = -3.3 x 0.5 = -1.65 volts. When the input swings negative 0.5 volts the output will swing positive. Vout = -3.3 x -0.5 = + 1.65 volts. The peak to peak output voltage is the difference between the positive and negative peaks, or - Vout(p-p) = 1.65 - (-1.65) - 3.3 volts. Normally AC signals are measured using average voltage or root mean square (rms). For a pure sine wave the average and rms voltage are the same. This is another discussion. If the signal in this example is a square wave of 1 v(p-p) from 0 v to 1 v then the output in this example would be 0 v when the input was at 0v and -3.3 V when the input was at 1 v. The peak voltages and currents for the square wave scenario - The junction of the resistors at the inverting input (-) of the ideal op amp is considered a virtual ground since the non-inverting (+) input is at 0 V (ground). So across Rin you see +1V - 0V = 1V. Across Rf you see 0V - (-3.3V) = 3.3V. The PEAK current for Rin is I(Rin) = V/R = 1V/10K = .0001 A or 0.1 mA. The current in Rf must be the same as none flows into the op-amp (in the ideal case). To prove it you can calculate this current. I(Rf) = 3.3V/33K = .0001 A = 0.1 mA I hope this clarifies things. Howard