Differential Amplifier - Page 1
Forum section: electronics / beginners

author: JacobEdward
author_rank: Regular Contributor
timestamp: April 08, 2015, 07:15:19 pm
content: I posted something similar to this a while back, but I am still confused.This is the book I'm using:http://www.allaboutcircuits.com/vol_3/chpt_8/9.htmlHow do you calculate the voltage drop from V1 over R1? If you were to remove the Op Amp from the circuit, it would be a simple matter of dividing V1 by the sum of R1+R2+R3 to get the current for the circuit, then taking the current value and multiplying it by R1 to get the voltage drop (with the op amp in the circuit, it would mean R2 was in parallel with the op amp, making it slightly more complicated but not terribly so)... but in this scenario, since the Op Amp is providing negative feedback, supposedly V1 should never reach R2, so why would R2 be included in the calculations for the V1 voltage drop?That then brings up a further confusion... if the Op Amp is providing negative feedback, thereby canceling out any voltage/current from V1 across R2, why wouldn't that negative feedback also split off into parallel lines, one going to the negative feedback, the other going to R1, which would mean V1 should actually be canceled by the Op Amp negative feedback before it actually reaches the Op Amp... This is all in the context of a DC circuit.Hopefully I explained that properly, but if not, here's a visual:

author: kezat
author_rank: Contributor
timestamp: April 08, 2015, 08:06:49 pm
content: Some points that may help you understand things a little more.The inputs to the op-amp are high impedance ie high resistance, you have almost no current flowing in to the opamp at this point, it is so small it can ignored for the basic understanding of this circuit. To give you a example your multimeter is also high impedance in the voltage range, do you expect a 9volt battery to drop in voltage if you measure it with your volt meter?The R3 "load" resister has no effect on the voltage divider of R1 and R2 or the circuit unless you remove the op-amp as you say or if the value is so low that the op-amp cant supply the current. (see below)Another point that might help is that the Opamp output is a Voltage source, you can think of it as a power supply that is adjusted by two inputs of the op-amp. If you add a load resistor to the output of this "power supply" the voltage will not change, again within reason.Lastly the voltage divider calculation for R1 and R2 is the voltage across it and not relative to ground. For example if V1 (Vin) is 5 volts and the output of the op-amp is 2 volts the Voltage divider calculations need to be done using 3 volts as the input V1 and of course the bottom of R2 as 0v.

author: TimFox
author_rank: Super Contributor
timestamp: April 08, 2015, 08:32:25 pm
content: Start with the non-inverting input ("+"). With no current flowing into the op-amp input, the voltage at that terminal is computed easily from V2 and the R4-R5 voltage divider. If the op-amp is within its output range, then the voltage at the inverting input ("-") must be equal to that at the + imput. Therefore, Vout will be that which gives the + input voltage for the voltage source V1 and the R1-R2 circuit. To calculate that, first compute the current through R1 (determined by V1 and the voltage at the two op-amp inputs). With no current into the op-amp, that current continues through R2, giving a voltage difference from the op-amp input to the output voltage Vout. Add that difference to the op-amp input voltage, paying careful attention to the direction of that current ("feedback"), to determine the output voltage Vout.That is the calculation routine. The logic is that Vout adjusts itself (within the limits determined by the power supply and the op-amp itself) to obtain the result in the arithmetic above.

author: jlmoon
author_rank: Unknown
timestamp: April 08, 2015, 08:50:54 pm
content: +1 TimFox. Explained very well.

author: dannyf
author_rank: Super Contributor
timestamp: April 09, 2015, 12:22:30 am
content: QuoteNo.The negative feedback works so that the output voltage will change to a point where the inverting input pin and the non-inverting input pin are at the same potential.The non-inverting pin sits at V+=R5/(R4+R5)*V2;The inverting pin sits at V-=V1-(V1-Vout)/(R1+R2) * R1.Set V+=V- and solve for Vout.