Hello (again)im still struggling to get my AD8333 to work properly.I try to demodulate Signals, similiar to the ones described in the Datasheet.
When i analyse the Outputs i measure on the one hand my RF Signal and on the other hand a ~108 MHz. These are the most dominant Frequencies.
I also get 4 times the expected frequency differenz (~40kHz). But i only need the frequency different. My clock is a single ended Sine wave with a frequency of 20 MHz and the RF Signal is a Sine wave too with a frequency of 5,01 MHz.
This Signal is provided to the AD833 by a AD8331 preAmp. The not used Pins are mostly floating except for the PX pins. For further information i attached the used Schematic.I tried different Values for the LFP but without any success. With so much frequencies it is hard to select the right frequency in my application so it would be great if someone has an idea what may help to receive a more accurate Signal.
In the configuration i use i would ecpect a 10 kHz Signalfor example.
Thanks for contacting us. There's not too much wrong with your schematic:
1. You really do need a differential drive for the 4LO, sine wave is OK. referring to fig. 64 in the data sheet, the input logic of the AD8333 4LO requires a LVDS level signal, the device shown is ideal for the application. I can't tell you exactly what happens when you try to drive the part with a single-ended signal but I do know it doesn't work.
2. The second fix is to drive the AD8021 used for an I-to-V converter with the negative output, not the positive. Driving with a positive output definitely won't work as the internal positive output circuitry is not the same as the negative.
Fixing these two items may give you what you're looking for but you should also change the filter caps in the I-to-V from 0.1uF to the 2.2nF shown in the data sheet fig. 64. The circuit shown there was used to show the waveforms shown in the data sheet, such as fig. 12.
Please advise, is this a commercial experiment or are you doing a very interesting university project?
Best of luck,
Hello,thanks for the fast reply. I will try the LVDS but i don't get No.2. I reconstructed this part of my schematic 1:1 like the Datasheet. Can you explain what you mean?I use the AD8333 for my Master Thesis. I want to measure bioimpedance via alternating elektromagnetic fields and need the AD8333 to demodulate this Signals.Kind regards
Re response #2,
Outputs: You're correct, I forgot the pin-naming convention for this part. Your output connections are correct.
Filtering: The filter needs to be effective to attenuate the 5MHz LO. Note in fig. 64 in the data sheet the RC filter values are 787 ohms and 2.2nF, so fc is 91kHz. If you use 787 ohms and a 0.1uF, your fc is 2kHz so you attenuate the 10kHz you want to use.