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I have been looking into high-voltage op-amp LTC6090-5, here it is its datasheet: http://cds.linear.com/docs/en/datasheet/6090fe.pdf

On first page of it, there is a reference circuit for using with DAC: enter image description here

Since my intention was to use this high-voltage op-amp with DAC, I am looking into understanding this reference schematic. My target circuit is very similar to this example.

Questions that I have in mind are:

  • why is there 10k resistor at DAC output? Is it used as current limiter? Why is current limiter needed if op-amp sinks very little current at its input?
  • Why is there a capacitor at the same + input?
  • at the feedback resistor(453k one), why is there a capacitor across it?
  • why is there extra connection from from DAC REF voltage(2.5V) to the voltage divider of negative voltage feedback of op-amp?

(I understand negative feedback connection at the op-amp)

Thanks everyone! :)

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  1. Why is there 10k resistor at DAC output? Is it used as current limiter? Why is current limiter needed if op-amp sinks very little current at its input?
  2. Why is there a capacitor at the same + input?

The 10k and 470p combination form a low-pass filter. Cut-off frequency is given by \$ f = \frac {1}{2 \pi R C} = \frac {1}{2 \pi 10k \cdot 470p} = 34~kHz \$. It will reduce any high frequency switching noise from the DAC.

  1. At the feedback resistor(453k one), why is there a capacitor across it?

This reduces the gain of the amplifier at high frequencies. Again this helps eliminate switching noise and amplifier stability.

  1. Why is there extra connection from from DAC REF voltage(2.5V) to the voltage divider of negative voltage feedback of op-amp?

We can see that the DAC is powered from +3 V and GND. This is therefore the maximum and minimum possible output voltages. (It will actually only give up to 2.5 V on the output as that is the Vref setting.) The amplifier, however, is powered from +70 V and -70 V which implies that we are expecting to drive it positive and negative with respect to ground.

Since the DAC can only give a positive output we can assume that 1.25 V (half of Vref) represents the mid-point or "zero" of the AC signal output. We therefore reference the amplifier to that point.

Removing the offset

We have one little problem left: we want 0 V out of the amplifier when ADC out is 1.25 V. To do this we need to raise the inverting input reference point by \$ \frac {1.25}{A} \$ where A is the gain. This is done by tweaking the ratio slightly with the potential divider ratio of 16k2 and 16k9. (These add to 33k3 which makes me think the values were worked out for convenience on a 3.33 V reference.) In any case it's 51.06% of 2.5 V. The amplifier will have to drive the output to less than 1.25 V to pull the inverting input down to match the non-inverting input. If the calculation has been done correctly the balance will occur when the output is at 0 V.

I'll leave it as an exercise to the reader to confirm the values are correct!

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  • \$\begingroup\$ See the add-on - "Removing the offset". \$\endgroup\$ – Transistor Jul 27 '16 at 21:11
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why is there 10k resistor at DAC output? Is it used as current limiter? Why is current limiter needed if op-amp sinks very little current at its input?

The 10 kohm and 470 pF form a low pass filter that removes DAC quantization noise beyond 34 kHz.

at the feedback resistor(453k one), why is there a capacitor across it?

The 10 pF and 453 k also form a low pass filter for exactly the same reason. Cut-off for this pair is about 35 kHz.

why is there extra connection from from DAC REF voltage(2.5V) to the voltage divider of negative voltage feedback of op-amp?

The extra connection turns a unipolar DAC output (only positive voltage) to a bipolar +/- 70V output.

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To answer your first and second questions, the 10k resistor and 470pF capacitor form a low pass filter to smooth out some of the noise from the DAC output providing a cleaner signal to the op amp.

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