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!
May worth to read this AN,
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!
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.