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 pwan 26th June 2010 11:44 PM

F5 - how to reduce bandwidth response

Hello,

The bandwidth response of the F5 probably too good for me. I'd like to scope it a bit narrower:

For the lower Freq, I plan to a coupling input C ( maybe 4.7 - 10UF) in the front ( before the R 1K).

Question is on the higher Freq

How to limit the bandwidth say, max 100-200KHz?

Thanks.

 bob91343 27th June 2010 05:04 AM

Put a simple RC low pass filter at the input of the amplifier. I assume you don't mind the amplifier having wide band response and just want to limit the signals it receives.

If you want to modify the amplifier itself you have a bit of a project ahead.

 bear 27th June 2010 06:21 AM

If you look at the most recent posts in the F5 thread, the discussion about parastics (oscillations) NP explains how to compensate to prevent those with a specific type of load, all that this is doing is lowering the bandwidth of the amplifier, you can do the same...

On the input you should figure out the RC time constant given the resistor to ground at the input... Unless you have DC coming out of something, there's no reason to put a cap there??

_-_-bear

 gootee 27th June 2010 07:30 AM

For a passive first-order RC low-pass filter for a single-ended amplifier input, you need a resistance R in series with the signal followed by a capacitance C to ground (except see the NOTE, below, for inverting opamp inputs).

The -3dB frequency of such a filter would be f(-3dB) = 1 / (2 * Pi * R * C), where Pi is about 3.14, R is in Ohms, C is in Farads, f is in Hertz, the slash (/) denotes division, and the asterisk (*) denotes multplication.

You should probably pick f(-3dB) to be at least 100 times the desired signal bandwidth, to minimize unintended loss of bandwidth.

After (downstream from) the filter, the amplitude of the signal will be half that of the input, at the -3dB frequency, and the amplitude will decrease by 20 dB (i.e. a factor of 100) for every decade (i.e. 10X increase) of frequency after that. You will probably need to include all upstream resistance (such as source impedance's resistance) in R. You can either specify R and calculate C, or, specify C and calculate R.

Solving the equation above for C gives C = 1 / (2 * Pi * R * f(-3dB) ) .

Solving for R gives R = 1 / (2 * Pi * C * f(-3dB) ) .

So, for example, if R is 150 Ohms and the desired f(-3dB) is 200 kHz, then C would be 1 / (2 * 3.14 * 150 * 200000) = 0.0053 uF (microFarads), which is the same as .0000000053 Farads, and 5.3 nF (nanoFarads), and 5300 pF (picoFarads). [After pico there's femto and atto but those won't ever appear for packaged components.]

At that point, you'd probably calculate what the -3dB frequency would be if you used the closest standard capacitor values, 4700 pF and 5600 pF. And if one of those wasn't acceptable, you could parallel smaller capacitances to sum the values, or, calculate the new R value needed to give the frequency you want with one of the standard values.

Sometimes, you will pick the type of capacitor you want to use, first, since the type (e.g. electrolytic, polyester film, polypropylene film, polystyrene, teflon, ceramic, silver/mica, et al) usually determines what range of values and physical sizes are available. And then you might calculate the R that would be needed for the upper and lower ends of the range of available standard C values for that type, to see if they are reasonable for your circuit application. Or, you might do it the other way around, if you already know what range of R values might be acceptable.

NOTE that for some amplifier inputs, such as the inverting input of an opamp, often, it is a bad idea to connect a capacitor from that input to ground, since it can cause instability of the system. In that case (and in every case, if you feel like it), you can usually just add 2R between the capacitor and the offended amplifier input, and change the original R to 2R, as long as the 2R resistance is large enough to prevent instability, which will be different for different amplifiers. This actually creates a bi-directional "T" (aka RCR) filter, which is usually "a good thing".

For a better discussion of amplifier input filters, see Chapter 7 (and others) of ADI - Analog Dialogue | Op Amp Applications Handbook .

Cheers,

Tom Gootee

 pwan 28th June 2010 05:18 PM

Big thanks, Bob, Bear & Tom for your help.

 gootee 29th June 2010 04:56 AM

I wrote, "You should probably pick f(-3dB) to be at least 100 times the desired signal bandwidth, to minimize unintended loss of bandwidth.".

That was incorrect.

It will normally be sufficient to set the f(-3dB) frequency of the low-pass filter to 10 times (or probably at most about 25 times) the desired signal bandwidth, not 100 times as I originally stated.

The trade-space is that we want the filter's cutoff frequency to be as low as possible, in order to better-attenuate any stray high frequency signals and noise, but we also want the cutoff frequency to be high-enough that it does not attenuate our signals of interest, so that it will not change the sound in any way, i.e. it will not appreciably change the existing bandwidth of the system.

If you do make the cutoff frequency 10X the desired bandwidth (i.e. 10 times the highest frequency for which you ideally wouldn't want to change the voltage's amplitude at all), then the filter's output voltage at the desired upper frequency/bandwidth limit will be attenuated (i.e. down) by about 0.0424 dB, which means that the peak voltage of a signal with a frequency at the upper edge of the desired bandwidth would be attenuated to around 99% of what it would have been without the filter in place. i.e. The Gain is around 0.99 at that frequency.

So, for example, if the filter's -3dB cutoff frequency is set to 220 kHz (220000 Hz, i.e. 10 times 22 kHz), then for signal frequency components up to 22 kHz the output voltage of the filter will be at least 99% of the input voltage.

Since we might prefer to set the gain, G, of the filter to some other number, at some particular frequency, f, I went ahead and solved the equation for the voltage gain across the output capacitor for the R * C product:

The voltage gain across the capacitor, at any frequency f, for any R and C, is:

G = 1 / sqrt [ 1 + ( 2 * Pi * f * R * C ) * ( 2 * Pi * f * R * C ) ]

And here it is solved for R * C, so it's easier to use:

R * C = [ 1 / (2 * Pi * f) ] * sqrt [ ( 1 / [ G * G ] ) - 1 ]

Now you can just substitute your G (a positive number < 1) and f (in Hertz) values into the right-hand side of that equation and calculate the result, and then either divide that by a desired R to calculate the required C, OR, divide it by a desired C to calculate the required R.

EXAMPLE:

Say that you want to design an RC low-pass filter that won't lower the amplitude of any signal components of up to 22 kHz by more than 0.1%. That would imply that the gain, G, would have to be at least 0.999 at 22 kHz.

Using the equation above, and a calculator, with G = 0.999 and f = 22000, we get RC = .000000324, for that example.

So now we can solve for either R or C, just by dividing by the other one:

C = .000000324 / R

and

R = .000000324 / C

You can use either one, depending on whether you prefer to pick R and calculate C or pick C and calculate R.

So, for example, if we already know that we have or want R = 100 Ohms, then C = .00000000324 Farads.

To get microfarads (millionths of a Farad), multiply by one million, i.e. remove six of the zeros, which gives .00324 uF, which is the same as 3.24 nF (nanoFarads) and 3240 pF (picoFarads).

The filter's -3dB frequency (where the voltage is down by 50%) will be:

f = 1 / ( 2 * Pi * R * C)

For the example, with R = 100 and C = .00324 uF (i.e. .00000000324 Farads), the -3dB frequency is about 491 kHz, which is around 22.3 times the 22kHz for which we wanted a gain of 0.999.

Note: When working with such small capacitance values and high frequencies, a "regular" calculator might run out of digits after the decimal point, or give fewer digits than needed. In that case, you can do the calculations with C in uF and f in MHz, and everything should still work out the same.

Cheers,

Tom

 Nelson Pass 30th June 2010 07:46 PM

Try 16nf across the 100 ohm feedback resistors (this means
two capacitors, one in parallel with each pair of 100 ohm which

Vary the capacitance to taste.

:cool:

 Beftus 30th June 2010 09:56 PM

How about a capacitor across R10?

 gootee 1st July 2010 01:32 AM

Quote:
 Originally Posted by Nelson Pass (http://www.diyaudio.com/forums/pass-labs/169382-f5-reduce-bandwidth-response-post2231579.html#post2231579) Try 16nf across the 100 ohm feedback resistors (this means two capacitors, one in parallel with each pair of 100 ohm which are already in parallel. Vary the capacitance to taste. :cool:
Yeah, sorry for blathering on for so long. I didn't have an F5 schematic and am not familiar with it. So I figured he might as well also try to keep RF out.

(And then, clearly, I got a little carried away <grin>.)

 pwan 1st July 2010 10:15 AM

Quote:
 Originally Posted by Nelson Pass (http://www.diyaudio.com/forums/pass-labs/169382-f5-reduce-bandwidth-response-post2231579.html#post2231579) Try 16nf across the 100 ohm feedback resistors (this means two capacitors, one in parallel with each pair of 100 ohm which are already in parallel. Vary the capacitance to taste. :cool:
Thanks Papa.

I mainly use CD player as source and seems the F5 gain is not sufficient. Just wonder how the F5 goes with the BOZ ??

PS: I came from the tube amp mania and this will be my first Class A solid state.

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