# LMF 100 - Mode 3 - design calculation

Thread Starter

#### atferrari

Joined Jan 6, 2004
3,930
LMF100 switched capacitors filter.

I intend to use it for an f0 =10 Hz thus fCLK to drive the filter (ratio 100:1) should be 10*100 =1000Hz but I have available only 120 Hz. The micro in charge cannot provide 1KHz what would make my life easier.

To use it in Mode 3 as a LPF where I can adjust the fCKLK / f0 ratio via resistors R2 and R4, I calculate:

Asuming R1 =18K and HOLP =1, R4 is also =18K.

To have a relationship 100:1, I calculate R2 as per the equation

fo = fCLK/100 * SQRT (R2/R4), obtaining

R2 = 1250000 ohms for a ratio of 1000/120 =8.33333...

My questions:

a) Is it actually appropriate to use R1 =18K? How to find out that?

b) R2 seems to be quite high (!) Is it convenient for this kind of circuit?

c) To make R2 smaller I should make R1 also smaller. Is that convenient as well? If yes, how much less?

Datasheet here:

http://www.google.com.ar/url?sa=t&rct=j&q=lmf100%20datasheet&source=web&cd=1&ved=0CCMQFjAA&url=http%3A%2F%2Fwww.swarthmore.edu%2FNatSci%2Fecheeve1%2FRef%2FDataSheet%2FLMF100.pdf&ei=oumqTojPCcLq0gGXt7GGDw&usg=AFQjCNGEdtviaBKR6vBPoKNYZc9PBKlkng

Help appreciated.

#### bertus

Joined Apr 5, 2008
20,465
Hello,

If you use a clock of "only" 12 times the wanted frequency, you will end up with a lot of distortion.
Here is a comparison of the 50 X and 100 X sampling modes from the datasheet:

imagine what will happen with the 12 X sampling.

You could make the clock frequency higher using a PLL with a divider like the 4046 with any counter.

Bertus

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Thread Starter

#### atferrari

Joined Jan 6, 2004
3,930
In fact, mode 3 is supposed to overcome not having the exact clok available.

The example of calculation given in the datasheet deals with exactly that.

In my case I still intend to use no more nor less than a 100:1 relationship.

#### crutschow

Joined Mar 14, 2008
24,749
Why did you select R1=18k?

The internal op amps can drive a load as low as 5K so you can have a feedback resistor as low as that. The corresponding input resistor can be as low as you are able to drive.

Thread Starter

#### atferrari

Joined Jan 6, 2004
3,930
Why did you select R1=18k?

The internal op amps can drive a load as low as 5K so you can have a feedback resistor as low as that. The corresponding input resistor can be as low as you are able to drive.
That was mi initial choice, close to the 20K selected in the example in the datasheet.

Finally I tested everything with 6K8 for R1 and R4 thus R2 = 742K

So far, it is not working. I am using +/- 5V supply with a clock from the micro (0 to 5V).

Somewhat disapointed, but looking for eventual mistakes.

Gracias Carl.

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