Hi, I am using AD9833 to generate a sine wave whose frequency is from 0.1Hz to 10kHz, the amplitude is 20mV and DC offset is 2.5V. The output of AD9833 is from 38mV to 650mV. What should I do to change the DC offset and amplitude? Thanks!
The AD9833 HAS a current output DAC which outputs ~3mA p-p, but this is channelled through an internal 200R resistor giving ~600mV p-p voltage out. If you parallel another resistor between Vout and ground then you can attenuate the output voltage to anything in between. Use Ohm's law: 20mV / 3mA = 6.7R total resistance. Required parallel resistance: 1 / (1 / 6.7 - 1 / 200) = ~6.8R. This has the additional advantage of lowering the circuit impedance and noise. The DC offset may be produced by adding a DC blocking capacitor, say 10uF X5R ceramic followed by a 2 resistor potential divider. Voila!
Tips: Use a thin film parallel resistor to minimise 1/f noise - a potential problem at 0.1Hz.
Likewise, don't ever use 741s for anything requiring low noise/distortion!
I suspect you may require a reconstruction filter to iron out the quantization steps in the waveform.
Regards, hope you crack it!
May worth to read this AN,
Thanks for your reply. The file you give to me is about a current output chip. For AD9833, to change the output offset and amplitude, I think this circuit may work. Can you check it for me please?
Thanks. It should work.
Retrieving data ...