- #1
- 5,199
- 38
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, VDAC is the output voltage of a DAC chip. VDAC and Vref 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:
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 RINV into the Vref 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 Vout:
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 RFB/RINV = 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 -Vref/2." That suggests that it converts the DAC output range from (0 to +Vref) to (-Vref/2 to +Vref/2). HOWEVER, if I plug RFB/RINV = 1 into the equation I derived for Vout, I get a gain of 2 and an offset of -Vref. It seems like I'm missing a factor of 1/2 somewhere. What did I do wrong?
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- = VDAC
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 RINV into the Vref 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 Vout:
[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 RFB/RINV = 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 -Vref/2." That suggests that it converts the DAC output range from (0 to +Vref) to (-Vref/2 to +Vref/2). HOWEVER, if I plug RFB/RINV = 1 into the equation I derived for Vout, I get a gain of 2 and an offset of -Vref. It seems like I'm missing a factor of 1/2 somewhere. What did I do wrong?
Attachments
Last edited: