Hi
I am using LMH6517 followed by ADC16DV160 in my design. I want to simulate the AAF filter needed in between and would like to know the output impedance of LMH needed for the same. Can you provide me S-para file of LMH?
Regards,
Sarika
This thread has been locked.
If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
Hi
I am using LMH6517 followed by ADC16DV160 in my design. I want to simulate the AAF filter needed in between and would like to know the output impedance of LMH needed for the same. Can you provide me S-para file of LMH?
Regards,
Sarika
Hello Sarika,
I do not have the S paramater file at the moment. I will try to find it and post it soon.
Did you want to design a band pass filter? What order filter did you have in mind? Did you want a terminated filter? What is your target insertion loss for the filter and termination?
The input of the ADC16DV160 is a capacitive sample and hold. It is fairly high impedance. However, at the ADC input we recommend a resistive bias circuit to provide the ADC common mode voltage. We also recommend a parallel capacitor to reduce the impact of the sampling switching. This capacitance can be used to form the final stage of the filter.
For your frequency range the output of the LMH6517 can be considered to be ideal. If you design a doubly terminated filter you will need to use matching resistors and the matching resistors will dominate the amplifier S parameters.
If you want to reduce the loss between the amplifier and the ADC you can use an impedance transform filter. For example if the filter input impedance is 100 Ohms and the output impedance is 200 Ohms you will get 3dB less voltage loss than using a matched impedance filter.
We have found that simulated filters shift in bandwidth when built on a PCB. To compensate for this it is best to design the filter to have 5% to 10% higher frequency than desired.
Best Regards,
Loren
Hello Loren,
Thanks for the prompt reply.
I have designed a third order BPF with first element as shunt. The on board flatness is around 8dB with loss of around 14dB.
I have designed for max of 13dB loss. Right now, I am thinking if I can change the component values and achieve good flatness with low loss.
I have recently come across TI`s user guide snau079. I have tried to implement the filter (between LMH and ADC) in page 7 as it the design somewhat met my purpose, just that i need for a higher bandwidth (177-202MHz), or even higher considering the tolerance you suggested. Plus, the filter prototype is different.
I have made this schematic to replicate the suggested filter in page 7:
The got the response:
This does not match with the plot given in the user guide.
Can you suggest me where I am wrong?
I will share you my filter design soon.
Hello Sarika,
Can you send your board layout? The loss you are seeing looks like either a bad component (cold solder ?)or maybe something with the PCB layout. There should not be 12dB of filter loss.
I simulated your filter and it seems to have a peaked response at low frequencies. This would most likely get worse when put on a PCB.
I included another filter option in the TINA file that I attached. This filter is 100 Ohms in and 200 Ohms out for lower loss, but still flat response. It looks like you only use 10Ohm matching resistors on your filter. This gives lower loss, but has poor matching with most filter designs.
Regards,
Loren
Hello Sarika,
I don't have ADS. Can you import a PSpice netlist? I also attached an JPEG file.
Regards,
Loren
Hello Sarika,
The 200 Ohms for the filter termination is ideal to show filter performance. In your circuit the resistors will need to be slightly larger. In addition the value of C8 will need to be reduced from 82pF by the equivlaent of the ADC input capacitance. It will take some iteration to find the best resistor and capacitor values once you have built your board. I would recommend starting with a 76pF capacitor and two 125 Ohm resistors.
The reason that the resistors are required is that the real component of the ADC16DV160 input is very large (k ohms). This needs to be reduced to match the filter output stage. The reason that the final capacitor needs to be changed is that the ADC input is around 6pF capacitive. This capacitance will change the filter freuency response, but it can be cancelled out by reducing the capacitor in the filter. This only works if the final stage of the filter is a shunt stage, so I recommend that the filter design always have a shunt stage next to the ADC.
Regards,
Loren
Hello Loren
Thanks for the explanation. I will implement your suggestions and get back to you.
Meanwhile I tried filter simulations as per your previous mails.
Attaching two different schematics and their results:
The difference between these two schematics is the way i kept the terminations. In one I have used separate terminations (50 Ohm input and 100 Ohm output) for all the four ports and in the other I have used one termination for input (100 Ohm) and one for output (200 Ohm) differential pair.
As you can see, I got different results. Can you suggest me why?
I felt both the cases are one and the same and was shocked by different results :P
Hello Sarika,
The two termination methods are not exactly the same. With a differential termination there is no connection to the reference plane (typically ground). This means that while the filter is terminated with respect to the signal path the filter is floating with respect to ground.
When you terminate each leg of the filter separately you have an equivalent differential termination. There will also be an another termination component. The filter now has a tie to ground on each side. These two terminations, (in parallel) form a common mode termination.
In your simulations you are driving the filter with two separate sources. This means that the filter has both differential and common mode components in the drive signal.
I am pretty sure that if you drive the filter with a differential source the two termination options should show the same response.
Regards,
Loren