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Hi,

I am having a lot of trouble understanding this configuration. When i first saw it, it thought that the 2.5V reference voltage would force 1.25V on the negative terminal but after opening it up again i saw that R97 and R98 are not going to ground, therefore that assumption is not true.

Can anybody explain to me how this particular configuration would work?

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    \$\begingroup\$ Such a circuits are extremely easy to analyze. You should only know two things to get started: 1) The voltage on both OPAMP inputs is equal because of negative feedback. 2) The current does not flow into or from the OPAMP inputs because of it's "infinite" input impedance. From here you can calculate currents and voltages and express the output in terms of input.. \$\endgroup\$ – Eugene Sh. Jun 3 at 19:05
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    \$\begingroup\$ This is intended to double the DAC output 0~5 V to +/-5V out, 2.5Vin=0Vout. thus needs a bipolar supply. \$\endgroup\$ – Sunnyskyguy EE75 Jun 3 at 19:27
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One simple approach, if you are familiar with standard op amp circuits, is to use superposition.

First, assume that OUT-DAC1 is connected to ground but that REF-2V5 is still attached. What is the output of the op amp?

Second, assume that OUT-DAC1 is still attached but that REF-2V5 is connected to ground. What is the output of the op amp, as a function of OUT-DAC1?

Add the two results and you have your answer.

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The opamp is in a negative feedback configuration because the output is connected to the inverting input via R98.

That means the opamp will try to make the voltage difference between its inputs (+ and -) equal to zero.

The Vdac voltage directly feeds into the + input. So the opamp will try to make the voltage at the - input equal to Vadc.

If Vadc = 2.5 V then what does the opamp need to do to make - input also 2.5 V?

The voltage across R97 needs to be 0 V (2.5 V at both sides) so no current flows, that means there should also no current be flowing through R98 (because if there was a current, where would it go?). So the opamp's output is 2.5 V as well.

Now let's make Vadc = 2.0 V => That means V(R97) = -0.5 V and V(R98) = +0.5 V

So the output becomes Vadc - V(R98) = 2.0 V - +0.5 V = 1.5 V.

Now let's make Vadc = 3.0 V => That means V(R97) = +0.5 V and V(R98) = -0.5 V

So the output becomes Vadc - V(R98) = 3.0 V - -0.5 V = 3.5 V.

So the amplifier amplifies does: Vout = Vref + 2 * (Vadc - Vref)

If you still have trouble grasping this, then assume Vref = 0 V and figure out what Vout will be when Vadc = -0.5 V, 0 V and + 0.5 V.

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A theoretical approach would be:

Since the ideal op-amp input impedance is infinite, you can pretend the negative input (Vn) does not affect the rest of the circuit. So, the lower part of the circuit becomes a voltage divider and, since both resistors have the same value, the middle voltage becomes the average of both ends.

Vn = (Vo + Vref) / 2

By that same argument, you can also say that the positive input (Vp) is the same as Vdac (since no current flows through the resistor).

Vp = Vdac

Now, if you take the basic op-amp gain equation and isolate Vn, you get:

Vo = A * (Vdac - Vn)
Vo = A*Vdac - A*Vn
A*Vn = A*Vdac - Vo
Vn = Vdac - Vo/A

Then you can join both equations and work your way into isolating the output (Vo):

(Vo + Vref) / 2 = Vdac - Vo/A
Vo + Vref = 2*Vdac - 2*Vo/A
Vo + 2*Vo/A = 2*Vdac - Vref
(A*Vo + 2*Vo) / A = 2*Vdac - Vref
Vo * (A + 2) = (2*Vdac - Vref) * A
Vo = (2*Vdac - Vref) * A / (A + 2)

That last equation seems to be a little complicated, but since the ideal op-amp gain is really large (infinite), you may consider that A ~= A + 2 (for A >>>> 0), then it becomes very simple:

Vo = (2*Vdac - Vref) * A / A
Vo = 2*Vdac - Vref

That's it, you've got Vo as a function of Vdac and Vref and I believe it makes it easier to see what Vo does.

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I think, using superposition, the answer can be written down very quickly:

(1) V(DAC)=0: Vout1=Vref*(R98/R97)=-Vref

(2) Vref=0: Vout2=V(DAC)*(1+R98/R97)=2V(DAC).

(3) Vout=Vout1+Vout2=2V(DAC)-Vref=2V(DAC)-2.5V.

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