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I'm working on modifying a board, while going through its schematic design I came across a section where a DAC output into the range of [0, +3.3V], which is then supposedly gained to +/-3.3V range, see extract below:

circuit

U24 is the an AD5662 DAC, with both VDD and Vref at +3.3V.

U25D is an OPA4170 opamp, with supply at +/-3.3V. It's in a non-inverting configuration. The transfer function is derived from:

$$V_{+}=\frac{1}{2}(V_{out}-3.3)$$, so $$V_{out}=2V_{+}+3.3$$

Two things that don't make sense to me:

  1. Input range of 0 to +3.3V would be amplified to [+3.3v, 9.9V] according to this transfer function, since the positive supply is only at +3.3V,that would result in Vout to be constantly at the rail.

  2. Since the input voltage is always positive, how can the output of the opamp ever be negative?

I would think that this is probably a mistake and that the left side of R61 should be connected to -3.3V, resulting in a transfer function $$V_{out}=2V_{+}-3.3$$, which would work as advertised.

However, the production board does indeed follow the schematic, and the opamp converts a strictly positive Vout (sinusoid with offset) from the DAC to a zero-centered signal. This wouldn't surprise me if the DAC output is capacitively coupled to the opamp. But R57 and C51 act as a low-pass filter, so I'm kinda stumped here.

Clearly this board's design works as advertised, but it doesn't make sense to me why. Can anyone point out where I'm making mistakes in my analysis?


Edit: Seems like I made a pretty silly mistake in my transfer function derivation, should be $$V_{+}=\frac{1}{2}(V_{out}-3.3)+3.3$$ to account for the voltage divider potential as being offset from 3.3 rather than ground.

Thanks all the for the help, I wish I could mark everyone's as solution.

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    \$\begingroup\$ Algebra. You made a mistake in your transfer function. \$\endgroup\$ – efox29 May 10 '16 at 1:39
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The input voltage for R61 is correct. The voltage at the V- pin will be the average of Vout and 3.3V because R60 and R61 are the same. In the case where V+ is at 0V, V- should also be driven to 0V. The output voltage that satisfies this requirement is -3.3V. The equation for V- is \begin{align} V_- = \frac 12*(V_{out}+3.3V) \end{align} The transfer function is then \begin{align} V_{out}=2V_+-3.3V \end{align} as you say it has to be.

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It's easier if you think of the circuit as an inverting configuration

schematic

simulate this circuit – Schematic created using CircuitLab

The voltage across R1 is 3.3 - V2, so the output is $$ V_O = 3.3 - 2(3.3-V2) = -3.3 + 2V2$$ Keep in mind that for this to be accurate, the 3.3 has to be very accurately matched to the 3.3 reference of the DAC. Any mismatch will result in a zero shift.

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I don't really know what else to put in this section. Foo bar.

$$ \frac {3.3V - V_+}{10k} = \frac {V_+ - V_{out}}{10k} $$

$$ 3.3V + V_{out} = 2V_+ $$

Negative feedback, so V+ = Vin

$$ Vout = 2_{Vin} - 3.3V $$

Vin = 0, Vout -3.3V Vin = 3.3, Vout = 3.3V

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The original design may have required only a single polarity output, so U25 is acting as a comparator. The behavior of this circuit is determined by R61. If connected to +3.3v, the op-amp would have put out -3.3v to keep its (-) input at zero volts, or if the (+) input went above zero the op-amp would put out +3.3v, so it would switch output voltages.

1) If R61 is connected to GND, the gain is 2, so that means the DAC will be twice as sensitive to input voltage, but also means that U250 will saturate quicker, but it is an amplifier with a gain of 2, not a comparator.

2) if R61 is omitted, then the gain is unity, acting as a buffer for the DAC, but it is a amplifier with a gain of 1, not a comparator.

3) If the op-amp power was +/- 15 volts, not much would change in terms of its behavior, only the output, because Vo - 2(15 - V2) = -15 + 2V2. V2 is only dependent on the DAC output, and is best defined as V2, or zero volts (for clarity).

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  • \$\begingroup\$ "... so U25 is acting as a comparator." You are confusing the circuit with a Schmitt trigger using positive feedback. This circuit is using negative feedback and behaves as a linear amplifier at all times. The other answers demonstrate this with the maths. \$\endgroup\$ – Transistor May 10 '16 at 19:31
  • \$\begingroup\$ No problem. I read "... or if the (+) input went above zero the op-amp would put out +3.3v, so it would switch output voltages." as a sudden switch (which I assumed you were thinking it was like a Schmitt trigger). In practice the op-amp output would approach 0 V as the ADC approached 3.3 / 2 = 1.65 V. Your points (1) and (2) are good (except it isn't the ADC that will saturate) and explain the circuit in a way the others don't. \$\endgroup\$ – Transistor May 10 '16 at 21:22
  • \$\begingroup\$ Point (3) assumes that the op-amp hits one of the supply rails and introduces the comparitor concept again. Work out what happens if the op-amp is powered from ±15 V. I think it will continue to behave as a linear amplifier rather than a compartor. Fix it up for a +1. \$\endgroup\$ – Transistor May 10 '16 at 21:23
  • \$\begingroup\$ I meant op-amp is powered from ±15 V and R61 tied to -3.3 V. Then the output becomes \$ V_0 = 2V_+ + 3.3 \$. At \$V_+\$ = 0, \$V_O\$ = +3.3 V. At \$V_+\$ = 3.3 V, \$V_O\$ = 9.9 V. The point being that it's still an amplifier and not a comparitor. Your (3) output could only reach -3.3 V if \$V_+\$ = -3.3 V (which it can't as the ADC isn't bipolar). \$\endgroup\$ – Transistor May 10 '16 at 22:13

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