10MHz function generator output stage design - Page 1
Forum section: electronics / projects

author: testian
author_rank: Regular Contributor
timestamp: February 02, 2017, 06:16:56 pm
content: Hi,i am currently working on an waveform generator based on an AD9834. And would like some feedback on my thoughts.The target frequency range is 0-10MHz and 0-10Vpp (sine, triangular with lower freq).Clock: 75Mhzdifferential output available (IOUT / IOUTB) 3mA , 600mV maxFollowing layout is planned:Signal -> Filter -> Amplifier (50 Ohm output)The filter will be a 7th order Butterworth with 18Mhz -3db point to suppress the noises without doping out too early. Does the 200 Ohm load resistor for the DDS-IC needs to be close to the source or the output stage? In my simulation i get strange results if i put it an the source, bit i have seen it in other designs.The output amplifier is a bit a problem for me. I have never worked with Op-AMPs before.Would you recommend to do my amplification with gain 17 at 10MHz (Gain Bandwidth Product good post:https://www.eevblog.com/forum/projects/ad9833-waveform-generator/msg344136/#msg344136) in one or two steps? What is the best way to include an offset control?My current thoughts are:FS Adj at AD9834 to control amplitude -> AD9834 ->Filter -> op amp 17 gain, offset control -> Output driver op amp

author: Kleinstein
author_rank: Super Contributor
timestamp: February 02, 2017, 08:00:09 pm
content: For the filter stage, there usually is a resistor before and after the filter. A butterworth filter is unusual (might be good for the triangle though). I am not so sure the filter will provide a super constant output impedance and thus is good for driving an precise attenuator.A gain of 17 might still work with a single current feedback OP. Otherwise the GBW would need to be rather high. It is often a good idea to have some gain (e.g. a factor of 2) in the output stage too. So this would be more like:DDS - filter - amplifer (e.g. x 6) - attenuators - amplifier (x3) - output.

author: testian
author_rank: Regular Contributor
timestamp: February 02, 2017, 10:10:15 pm
content: Amplitude controlled by AD9834FilterLM7171 amplifier (x6)LM7171 amplifier (x3)The LM7171 should be suitable due to to >600V/µs and the 200Mhz Bandwidth alternatively LM6181. Other recommendations (especially cheaper solutions) are recommended.A butterworth filter is used for its missing ripple.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 03, 2017, 10:22:44 am
content: The amplitude setting at the DDS chip is usually one for a small range, maybe 1:10 at most. So there should be another coarse amplitude setting maybe to switchable attenuator with something like 15 and 30 dB. At only 10 MHz it could be that CMOS switches are good enough. Depending on the gain chosen the first amplifier stage might use a cheaper lower voltage, lower slew rate OP. It is only the output stage that might needs a +-15 V or similar supply and a really high slew rate.If you use the amplifier in inverting configuration, one could add the offset at the non inverting input. This would give less RF current at the offset setting circuit.For a good rectangular signal one could alternatively send the signal from after the first amplifier to a comparator.

author: MarkF
author_rank: Super Contributor
timestamp: February 03, 2017, 11:09:05 am
content: Here is what I did. You would need to adjust the op-amp gains for your desired output levels.

author: testian
author_rank: Regular Contributor
timestamp: February 03, 2017, 10:16:00 pm
content: Thank you MarkF, but i don't understand your solution and i don't wanna just copy. Attached is my modified solution.An additional 5v rail is not a solution for me, to save a few cents on the opamp. For an additional attenuation an manual switchable resistor was added.Edit: Just noticed that the AC coupling will not work for low frequencies. The output of the AD9834 is from 0-650mV is it possible to remove the offset and get a symmetric signal?

author: MarkF
author_rank: Super Contributor
timestamp: February 03, 2017, 11:29:29 pm
content: The MCP4802 is a dual 8-bit DAC. VoutA sets the amplitude and VoutB sets the DC offset. The top op-amp is setup as a differential amplifier and the bottom op-amp is the gain for the DC offset. I choose the differenial configuration so the wavrform stays centered around zero as you change the waveform amplitude. The gains are set for a 2Vpp output with a +/-2V DC offset. The output filter starts to roll off at 2MHz. All the functions are set with the rotary encoder and it's built-in switch. The switch selects between 'input' mode and 'selection' mode. The sweep up and sweep down waveforms are done in software with a 500Hz interrupt sweeping the freq.I hope that helps you understand. Ask if you need more. I don't care if you just copy it but it would be nice if you understood the circuit.Edit. The software uses 99% of the PICs 32Mb memory.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 04, 2017, 10:54:34 am
content: Using the differential output is one way to avoid an offset. However it makes the reconstruction filter a little more difficult (more parts).Having the filter directly at the output is a little tricky, as this makes it harder to keep the 50 Ohms impedance at all frequencies. It is also a good idea to have at least some of the filtering before the first amplifier, do avoid possible distortion and EMI problems due to the unfiltered high frequency part (e.g. > 50 MHz) for the DAC.The attenuator looks like rather high in impedance, but might be OK at 10 MHz.

author: testian
author_rank: Regular Contributor
timestamp: February 18, 2017, 08:32:40 pm
content: Thank you for all your help, now i understand you circuit. Using the differential output makes things a lot easier and the signal increases from 0,6Vpp to 1,2Vpp. That reduces to overall gain to 9, but I will stick with my 2-Stage design and add the filter in between.

author: MarkF
author_rank: Super Contributor
timestamp: February 19, 2017, 06:18:19 am
content: I've been going over my design and you may be able to eliminate the 470 and 100 ohm resistors and drive your 3.9K resistor directly from the 2nd opamp. See the Vbias on the attached circuit on the left.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 19, 2017, 09:23:55 am
content: There should be at least a little filtering before the 1st amplifier - a capacitor between the outputs at least. This reduces the very high frequencies, that can upset even a fast OP.Adding the offset so early can be a problem: the offset will also be attenuated from the coarse gain switch.The early version with the 470 Ohms and 100 Ohms resistors at the offset adjustment would need slightly different resistors at the differential amplifier - the 100 Ohms in parallel to 470 are effectively in sereis to the 3.9 K. So that resistor should be something like 3820 Ohms instead.

author: pitagoras
author_rank: Regular Contributor
timestamp: May 06, 2017, 12:46:19 am
content: I raise this not-so-dead thread to ask a simple question, because I'm in a similar quest.If the output impedance is 50 Ohm and it to be 10Vpp on 50 Ohm this means the output is 20Vpp with a high impedance load.Thus in this case the desired gain is 34 rather than 17, am I right?

author: testian
author_rank: Regular Contributor
timestamp: May 06, 2017, 08:36:11 pm
content: Here is an interesting video which explains the reflections in transmission lines and shows what happens if a line is not properly terminated.Code:https://www.youtube.com/watch?v=PPgxFd97taYhttps://www.youtube.com/watch?v=g_jxh0Qe_FY