3
\$\begingroup\$

I have tried to construct this circuit compensate circuit byy Maxim in LT spice as to compensate the imperfect of DAC of my DDS function generator,for my case fs=125MHz,thus C=6.6pF, and the result as shown :frequency response of the circuit, how do I verify it have compensate the sinc response by looking this result? and then when I am trying to change the resistance value to change the gain,the result become like this :changing of resistance value, but the gain still zero..so anyone can explain why the result will look like this? and I am trying to figure the formula for that circuit by google,but it end up with nothing,so anyone can explain it by using formula?

result for H(f)

In order to achieve 20 VPP output for 0 to 5MHz,an non-inverting circuit have been designed as shown:opamp circuit for vpp=20V

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Too long to read, but two observations: 1 - 420 Ohm load on the output of the opamp at high frequencies may be too much of a load. Check the datasheet. 2 - Your circuit will have unity gain until about 10 MHz, after which it eventually goes up to a gain of 2.1. However, is your opamp still good at that frequency? To have gain set by feedback, you want roughly 10x more open loop gain, so you'd need a gain-bandwidth product of 100 MHz just to get to the first rolloff frequency. Sounds rather unlikely unless you are doing something very special with special parts. \$\endgroup\$ Mar 12, 2014 at 13:30

1 Answer 1

2
\$\begingroup\$

The gain is zero dB. That means unity gain. The dB is used to express relative signal and power strength. If the gain was 6dB, the amplifier would have a gain of 2. If the gain was -6dB, the gain would be 0.5.

At the higher frequencies the small capacitor's Reactance has reduced to the point where it does influence the gain by starting to take current away from the inverting input. The opamp action at these high frequencies is to raise its output level to compensate: -

enter image description here

As frequency rises even higher the op-amp's bandwidth limit is reached and gain falls away - note that on the right-hand side of your graphs the gain has gone below 0dB and this means the gain has fallen below unity (and will continue to fall).

If you use the formula: -

H(f) = \$\dfrac{Sin(\dfrac{\pi f}{F_S})}{\dfrac{\pi f}{F_S}}\$

For Fs = 125 MHz and f = 40 MHz, H(f) in decibels is 20 log\$_{10}\$(0.8398) = -1.52dB

Looking at your top graph, it seems the op-amp is over-compensating (about +2.5dB at 40 MHz) so try reducing R1 to 200 ohms or 180 ohms.

\$\endgroup\$
11
  • \$\begingroup\$ super like your explanation,thanks @andy aka for your guidance along the way....1 more question,I just found out my actual result,not really like a sinc response,so any possible explantion about it? the result is shown in above \$\endgroup\$
    – user37970
    Mar 11, 2014 at 14:06
  • \$\begingroup\$ I looked at the table you added to your question and it looks like your sampling rate is lower than 125MHz. Other than that I'm struggling to think of an answer. \$\endgroup\$
    – Andy aka
    Mar 11, 2014 at 14:12
  • \$\begingroup\$ I am using this module ebay.com/itm/…, do the DAC have it own sampling frequency? that why I am headache to squeeze out an answer.. \$\endgroup\$
    – user37970
    Mar 11, 2014 at 15:45
  • \$\begingroup\$ You need to know Fs - if not just keep messing with values until it looks flat. \$\endgroup\$
    – Andy aka
    Mar 11, 2014 at 16:18
  • \$\begingroup\$ according to data sheet, the Fs for AD9850 is 125MHz.and one more problem, I am going to amplify 1Vpp input to 20Vpp from 0 to 5MHz is it need cascade the ompamp in order to have such high gain? \$\endgroup\$
    – user37970
    Mar 12, 2014 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.