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cepheid
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Hey everyone. I'm wondering if I'm analyzing the attached opamp circuit correctly. This is not homework, but rather, actual work. In the circuit diagram, V_{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 opamp (not shown). Here is what I came up with:
My understanding is that since an opamp 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 noninverting 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 opamp 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:
Next, I solved for V_{out}:
Now, here's the part that's puzzling me. The purpose of connecting the DAC chip's output to an external opamp 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 opamp 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_{ref}. It seems like I'm missing a factor of 1/2 somewhere. What did I do wrong?
My understanding is that since an opamp 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 noninverting 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 opamp 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 opamp 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 opamp 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_{ref}. It seems like I'm missing a factor of 1/2 somewhere. What did I do wrong?
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