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cepheid

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Hey everyone. I'm wondering if I'm analyzing the attached op-amp circuit correctly. This is not homework, but rather, actual work. In the circuit diagram, V

My understanding is that since an op-amp is a differential amplifier with a very large open loop gain, using it in this negative feedback configuration will cause differences between the inverting and non-inverting inputs to disappear (be "corrected") very quickly. Therfore, we can safely say that:

There is a "virtual short" between them. Since the + input is connected to the DAC output voltage, it follows that:

Now, assuming infinite input impedance (as in the ideal case), no current flows into the inverting input. Therefore, I have assumed that a current I flows from the op-amp output, through the feedback resistor, and then through resistor R

Next, I solved for V

Now, here's the part that's puzzling me. The purpose of connecting the DAC chip's output to an external op-amp in this configuration is to create a bipolar output from a unipolar one. Now, on the one hand, the DAC chip's datasheet claims that R

_{DAC}is the output voltage of a DAC chip. V_{DAC}and V_{ref}are both measured relative to analog ground (AGND), as are the + and - power supply voltages for the op-amp (not shown). Here is what I came up with:My understanding is that since an op-amp is a differential amplifier with a very large open loop gain, using it in this negative feedback configuration will cause differences between the inverting and non-inverting inputs to disappear (be "corrected") very quickly. Therfore, we can safely say that:

V

_{+}= V_{-}There is a "virtual short" between them. Since the + input is connected to the DAC output voltage, it follows that:

V

_{-}= V_{DAC}Now, assuming infinite input impedance (as in the ideal case), no current flows into the inverting input. Therefore, I have assumed that a current I flows from the op-amp output, through the feedback resistor, and then through resistor R

_{INV}into the V_{ref}node. (All of these nodes are actually pins of ICs, and I'm assuming that they can sink some current). Since the resistors are in series, I obtain expressions for the current I very simply from Ohm's law:[tex] I = \frac{V_{\textrm{out}} - V_{\textrm{DAC}}}{R_{\textrm{FB}}} = \frac{V_{\textrm{DAC}} - V_{\textrm{ref}}}{R_{\textrm{INV}}} [/tex]

Next, I solved for V

_{out}:[tex] \frac{V_{\textrm{out}}}{R_{\textrm{FB}}} = \frac{V_{\textrm{DAC}}}{R_{\textrm{FB}}} + \frac{V_{\textrm{DAC}}}{R_{\textrm{INV}}} - \frac{V_{\textrm{ref}}}{R_{\textrm{INV}}} [/tex]

[tex] V_{\textrm{out}} = V_{\textrm{DAC}}\left(1 + \frac{R_{\textrm{FB}}}{R_{\textrm{INV}}}\right)- V_{\textrm{ref}}\frac{R_{\textrm{FB}}}{R_{\textrm{INV}}} [/tex]

[tex] V_{\textrm{out}} = V_{\textrm{DAC}}\left(1 + \frac{R_{\textrm{FB}}}{R_{\textrm{INV}}}\right)- V_{\textrm{ref}}\frac{R_{\textrm{FB}}}{R_{\textrm{INV}}} [/tex]

Now, here's the part that's puzzling me. The purpose of connecting the DAC chip's output to an external op-amp in this configuration is to create a bipolar output from a unipolar one. Now, on the one hand, the DAC chip's datasheet claims that R

_{FB}/R_{INV}= 1 (these resistors are internal to the DAC chip). On the other hand, it claims that this exact external op-amp configuration results in "unity gain and and offset of -V_{ref}/2." That suggests that it converts the DAC output range from (0 to +V_{ref}) to (-V_{ref}/2 to +V_{ref}/2). HOWEVER, if I plug R_{FB}/R_{INV}= 1 into the equation I derived for V_{out}, I get a gain of**2**and an offset of**-V**. It seems like I'm missing a factor of 1/2 somewhere. What did I do wrong?_{ref}#### Attachments

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