low pass filter

I

values

Is this the same problem - noise spikes on the digital signal?

The corner frequency for an RC filter occurs when the impedence of the C is the same as the R.

eg when

1/(2*pi*F*C) = R

Pick R or C and rearrange the equation to get the other

or cheat and use the link I posted.

thanks again for the link, It looks like I need 5R with 0.1UF, I was using

100R with 0.1UF, even so I couldn't even get a few hz though it, it seems to short out my signal generator. 100R with 0.005UF I will try shortly...

It is the same problem as I posted before, I've been on google all day trying to find a possible cure, the only thing which is closest is the

74LS14, Schmitt device with 0.8 Hysteresis , if that could be like 2V Hysteresis then that might work. A schmitt sure does help cut down the problem but far from a cure.

The problem is a just like it is here

Its hard to say how large the spike is, but I know its between 1V and 2V, it could be a rail to rail spike I know ive had that on the fall). Though its a bit to fast for my CRT scope to see, the spike seems to be 0.1US, its filtering that out so its not bounceing on and off when its not supposed to.

I've been playing around with duty cycle, now its a bit hard to explain but I will try, at say 50:50 ratio the spike is on the rise of the cycle, now if you adjust the duty either + or - about 10% then you can see the spike move onto the ground level, OR, onto the high level... When the spike is at

12V it really does not matter since the wave is going via some buffer's which are ON at about 2.5V so the spike there does not matter really.... On the LOW spike, the spike Can't be more than 2.5V else the problem wouldn't go away. But when you adjust the duty you can see the spike moving across the low level, then slowly moving up the rise and across the high level..

Like I say when the spike is on the low or high its not efecting the circuit, but when its inbetween (on the rise) its causing false turn on/off's.

I found a artice on adding Hysteresis to a comparitor, though the values I worked out I doubt would work without overdriving the comparitor, actually I am not sure if they even will work at 200khz... probably not....

Cheers, Chris

What is the impedance? What components are you using?

Tom

My value was a bit off (my bad) I was using 100nf with 100R, After some testing and using that page which the other chap posted, What actually works is 33R with 10nf, it starts cutting off about 100khz (you loose about a volt)..... If I use 100R with 10nf that still works though not as good, and the wave ends up looking like a triangle wave and not a square, pluging values for about 500khz filter seems to work out about best.

Chris

works

and

What sort of 100nF capacitor? The tollerance on some types is as high as

80%. If using 100R then the source impedence might also be a factor.

its a polyester 100nf, I think philips made them, the orange jobbies. Tried to find them in farnell but they might not to them anymore.

The best value seems to be 33R with 10nF, give me about 0.5V spikes then, A schmitt gate should clean up the rest of it...

Chris

--- Unfortunately, that's in the nature of the beast.

That is, if you use a simple RC lowpass, like this,:

Vin>--[R]--+---->Vout | [C] | GND>-------+---->GND

Then RC will form a frequency dependent voltage divider where Vout will be:

Vin * Xc Vout = ---------- (1) Z

and Z will be:

Z = sqrt (R² + Xc²)

so Vout will _always_ be less than Vin, no matter what.

In order to tie things down, the "corner" frequency of the filter is defined as that frequency which results in Vout being 3dB down from Vin, which is when

Vout ------ = 0.707 Vin

---

--- Not without you specifing the output impedance of the source and the input impedance of the load.

---

Unfortunately, it doesn't work that way.

Looking at the circuit again,

Vin>--[R1]--+---->Vout | [C1] | GND>--------+---->GND

If we arbitrarily choose 1V for Vin and 50 ohms for R1, and rearrange

Vin * Xc Vout = ---------- Z

to solve for Xc we'll have:

Vout * Z Xc = ------------- Vin

and since we don't yet have Z, we can get it with

Z = sqrt (R² + Xc²)

and, with f = 200kHz, that comes out to:

0.707V * 50R Xc = -------------- ~ 121 ohms. 1.0V- 0.707R

Now, since we're looking for a capacitor with a reactance of 121 ohms at 200kHz, to find the value of the capacitance we can write;

1 1 C = ---------- = ------------------------- ~ 6.6E-9F = 6.6nF 2pi f Xc 6.28 * 2E5Hz * 1.21E2Hz

So, here's what we've got with a 3dB loss, at 200kHz, through the filter:

1V>---[50R]--+---->0.707V | [6.6nF] | GND>---------+---->0V

To find out what the output voltage will be at 250kHz we'll need to first calculate the output reactance of the cap at 250kHz:

1 1 Xc = ---------- = ------------------------ ~ 96 ohms 2pi f C 6.28 * 2.5E5 * 6.6E-9F

and then plug that into (1):

Vin * Xc 1.0 * 96 Vout = ---------- = ---------- ~ 0.657V R + Xc 50 + 96

So, your 1V 200kHz signal going through the filter will result in an output voltage of 0.707V, and an errant 1V 250kHz signal going through the filter will come out at 0.657V, a difference of only 50mV, so you can see that "Pass everything 200kHz or lower and block everything 250 kHz or higher" is easier said than done.

-- John Fields Professional Circuit Designer