3
\$\begingroup\$

When looking at AD9850 DDS or other similar parts made by Analog Devices, it is obvious that the output signal amplitude changes with frequency. This can be shown as \$x= \pi \dfrac{f_{out}}{f_{clock}}\$ and amplitude changes by \$\dfrac{\sin x }{x}\$.

Using this formula, we should see an amplitude drop less that 20% between 1Hz to 50 MHz but this is not the case .

As I am driving this part, I see the amplitude is around 1 volt pk-pk at 10Hz and 10mV at 50 MHz. Am I missing something in these calculations?

enter image description here

\$\endgroup\$
14
  • 1
    \$\begingroup\$ I don't think that plot is showing what you think... It is telling you how to calculate the frequency and amplitude of extra spurious frequencies that will be generated when operating at certain fundamental and reference clock rates. It is not telling you how the amplitude of the fundamental varies as you change the fundamental frequency. \$\endgroup\$
    – Justin
    Oct 2, 2013 at 14:34
  • 2
    \$\begingroup\$ Yes, it is about aliased signals (images) and discusses about how to avoid them. But if we take the first part ( 0-100MHz),it is the fundamental frequency (0-50MHz) and the first image (50-fClock). \$\endgroup\$
    – Aug
    Oct 2, 2013 at 14:37
  • 2
    \$\begingroup\$ My guess is that you have too much capacitance (or a lower capacitance with a high series resistance) on the output. At 10Hz, the output has time to charge the capacitor before it changes polarity. At the higher frequency, the time constant exceeds the half-period and so the entire time is spent charging the capacitor, with the result that the voltage drops nearly to zero. \$\endgroup\$ Oct 2, 2013 at 14:38
  • 1
    \$\begingroup\$ @ScottWinder - while Aug's circuit may be making the situation worse, please understand that this is an inherent property of this type of DDS, irrespective of loading. \$\endgroup\$ Oct 2, 2013 at 14:48
  • 1
    \$\begingroup\$ Aug - is your scope probe 1x? Try switching to 10x as that will reduce the loading capacitance Scott is talking about. I don't think the step vs impulse rolloff predicts the degree of difference you are seeing. \$\endgroup\$ Oct 2, 2013 at 14:55

1 Answer 1

4
\$\begingroup\$

In addition to the normal attenuation at increasing frequencies (as @ChrisStratton and OP pointed out), there is the potential to unwittingly add a low-pass filter to the output by decoupling it. This is based on the fact that the trace (or wire) between the DDS output and the decoupling capacitor will have some resistance, which combines with the capacitor to form an RC filter. The cutoff frequency is defined as: $$ f_{c}=\frac{1}{2 \pi RC} $$ Taking an arbitrary trace resistance of \$60 m\Omega\$ (consistant with a ½ inch, 8 mil, 0.5 oz trace) with your 100nF capacitor, we get: $$ f_{c}=\frac{1}{2 \pi \cdot 0.06 \cdot 100n} \approx 26.5MHz $$ This will change based on the actual resistance, as well as the capacitance of the scope probe (as discussed by @ChrisStratton), but it shows that the capacitance could be having a very real effect on the output amplitude at higher frequencies.

\$\endgroup\$
1
  • \$\begingroup\$ I never noticed the large impact of copper wire resistance. I reduced the distance between output and user part + reduced coupling capacitor to 50pF and also used 10x on oscilloscope. The amplitude is now as expected in the formula. \$\endgroup\$
    – Aug
    Oct 2, 2013 at 21:01

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.