Image of the B output stage of the AD9834 DDS EV board

Image of the B output stage of the EVAL-CN0304 AD9834 DDS Evaluation Board

IOUT_B is a current source, not potential source, so the initial say half of the circuit I get. The resistor gives us a proportional voltage, where it is centred on the input range + put through a unity gain amplifier, so we do not load down the input + cause issues. What I don't get then is why it is put through this 51 ohm voltage divider and all these capacitors and inductors. It seems like this would create a bunch of nasty resonant peaks from the LC oscillators.

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    \$\begingroup\$ Elliptical lowpass filter. \$\endgroup\$ Sep 4 at 20:47

1 Answer 1


Your circuit is a low pass circuit with a very sharp cut-off at about 17.8 MHz. The sharp cut-off is due to the high order of the filter.

The output of the circuit probably goes to a 50\$\Omega\$ transmission line, which is itself terminated with a 50 \$\Omega\$ termination resistor. The 51 \$\Omega\$ resistors would then be for impedance matching, to prevent reflections back into the amplifier part of the circuit, or back into the transmission line.


simulate this circuit – Schematic created using CircuitLab

enter image description here

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    \$\begingroup\$ This confuses me further then, this is a DDS that can produce up to 37.5MHz signals, why would they choose values to give a cutoff so much lower than the output? \$\endgroup\$ Sep 5 at 11:15

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