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Ardni

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Hi,
I am completing schematics for our design where I have a an analog signal coming in going through a differential driver with a low pass filter and then on to the ADC.

I am having a little trouble understanding the low pass filter.

I have attached a part of the schematic where the differential driver and the low pass filter can be seen.

I understand that this filter is a 3rd order capacitor input low pass butterworth filter (from looking here). My doubt lies in how the cutoff frequency is calculated. According to the datasheet for the differential driver the cut off frequency of the filter is 125MHz and from looking here an explanation of the formulas is provided. I do not understand the terms Cproto and Lproto or the denormailzed C and L values well. I was hoping to find a formula where I could set lets say the capacitor values and solve the equation for the inductor value.
Could someone please explain how I can achieve this. The plan is that I need to be able to alter the cut off frequency of the filter and would like to know what values are required?

Also as an aside, I understand that the 33 ohm resistor at the output of the differential driver does not affect the filter operation? but is there to match the impedence between the output of the differential driver and the input of the ADC, is this true to say? So it can be changed to match both impedances. This is something i have yet to do, as it does not give me the input impedence of the ADC in the datasheet.

Thank you for your help
 

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JimB

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I do not understand the terms Cproto and Lproto or the denormailzed C and L values well.
Standardised designs exist for filters, these designs are based on the filter having an impedance of 1ohm and a cut-off frequency of 1 radian/second, (1rad/s = 1/(2xPi) hertz).
Looking at the table of L and C values in the link, we see that for a 3rd order filter C1 = 1 Farad, L2 = 2henry, and C3 = 1 Farad, these are the prototype (normalised) values.

If we want to design a practical filter with an impedance of 50ohm, and a cut-off frequency of 10 Mhz, we calculate de-normalised values as:

L = (Lproto x 50)/(2 x Pi x 10,000,000) = 1.59uH

C = Cproto/(2 x Pi x 10,000,000 x 50) = 318pF

I was hoping to find a formula where I could set lets say the capacitor values and solve the equation for the inductor value.
Could someone please explain how I can achieve this. The plan is that I need to be able to alter the cut off frequency of the filter and would like to know what values are required?
When designing filters, it is usual to set the impedance and cut-of frequency and calculate the L and C values which are both dependant on impedance and cut-off frequency.

JimB
 

Ardni

New Member
Thank you very much for the reply Jim.
Things are a little clearer now but there are 2 things that I don´t understand properly.

From looking at the normailised values in the table, the value for both capacitors are 1 Farad. So does this imply that both caps must have the same value. In the filter I obtained from the datasheet of the differential driver one cap is 10pF and the other 30pF. So how with one formula for both capacitors can they both have distinct values?

The second thing I dont understand well is the impedance value. Does the impedance value elected affect the output impedance from the differential driver or the input impedance of the ADC? I cannot find info on either the o/p or i/p impedance for the 2 components.

I used the inductor equation from JimB to solve for the impedance and it gives me a value of 21.98 ohms, but I do not understand how this value must be interpreted. How does it relate to the 33ohm resistor before the filter?

I am a little lost with a few concepts so please forgive me if my questions seem basic.

Many thanks for the help.
 

JimB

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From looking at the normailised values in the table, the value for both capacitors are 1 Farad. So does this imply that both caps must have the same value.
Yes.
In many applications the filter is used in a part of the circuit where the impedance is well defined. In RF work an impedance of 50 ohms is common, which is why I used that value.

Sometimes a filter will also act as an impedance transformer, in these case the input and output capacitors will be of different values.
The info in the link you gave only covers filters where the input and output impedance are the same.


In the filter I obtained from the datasheet of the differential driver one cap is 10pF and the other 30pF. So how with one formula for both capacitors can they both have distinct values?
You need different formulae.

The second thing I dont understand well is the impedance value. Does the impedance value elected affect the output impedance from the differential driver or the input impedance of the ADC? I cannot find info on either the o/p or i/p impedance for the 2 components.
I suspect that the impedances will be on the datasheet somewhere, as a guess I think that the output impedance of the differential driver will be "low", and the input impedance of the ADC will be "high"

I used the inductor equation from JimB to solve for the impedance and it gives me a value of 21.98 ohms, but I do not understand how this value must be interpreted. How does it relate to the 33ohm resistor before the filter?
The LPF in your circuit is the same as that shown on page 25 of the datasheet for the ADA4937.
The text makes no mention of how to design the filter, I have a feeling that the values are fairly arbitrary and just designed to give the required cut-off frequency in a high impedance circuit.

JimB
 

Ardni

New Member
Thanks again Jim for the reply.

I have designed the input to the differentail amplifer to be 50 ohms to match the output impedance of the RF part which is also 50 ohms.

Must the output impedance of the differential driver and the ADC also be matched?
I find no mention of the input or output impedances in the datasheets for the ADA4937 and AD9640 ADC. It does say that the ADA4937 has a low output impedance as you suspected. If the ADC has a high input impedance, what does this mean I need to do with the circuitary in between the 2?

You mentioned that I need different formulae to those provided on the link, is there a name for this particular configuration where both caps have different values that perhaps I can google?

Yes the filter in my circuit was copied from the ADA4937 datasheet. In your opinion is it safe to use it as it is presented in the datasheet or must I alter it? Without knowing the exact input and output impedances and is there anything I can do?

Many thanks
 

JimB

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Most Helpful Member
I have designed the input to the differentail amplifer to be 50 ohms to match the output impedance of the RF part which is also 50 ohms.
Good idea.

Must the output impedance of the differential driver and the ADC also be matched?
I dont think so.

I find no mention of the input or output impedances in the datasheets for the ADA4937 and AD9640 ADC. It does say that the ADA4937 has a low output impedance as you suspected. If the ADC has a high input impedance, what does this mean I need to do with the circuitary in between the 2?
I think AD have done it for you in their datasheet.

You mentioned that I need different formulae to those provided on the link, is there a name for this particular configuration where both caps have different values that perhaps I can google?
Have a look at L-networks, which is effectively what you have now between the driver and ADC. (A balanced L-network, there probably is another name for it but I cant remember it just now).

Yes the filter in my circuit was copied from the ADA4937 datasheet. In your opinion is it safe to use it as it is presented in the datasheet or must I alter it? Without knowing the exact input and output impedances and is there anything I can do?
Why not! I think AD are a respectable company and would not put dud information in their datasheets.
I think you are over analysing this thing, it is a simple LPF to reduce the nyquist images above the sampling frequency.
If you had a lot of noise or unwanted signals above the sampling frequency you may want to improve the rejection of the filter, but this may be easier to do on your nice 50ohm input circuit.

JimB
 

Ardni

New Member
Thank you so much for the help Jim. You´re probably right that I´m over analyzing it, but it´s just when I didn´t have a good understanding of the circuit I was a bit concerned.
Anyway as you say, I´m sure the data provided in the AD datasheets is trustworthy.
 
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