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Re: Reverse attenuation and HF comp. networks, active X-over [message #36552 is a reply to message #36550]

Mon, 20 May 2002 22:34

Wayne Parham
Messages: 18789
Registered: January 2001

Illuminati (33rd Degree)

The formula is dB = 20 log V1/V2, so you'll need approximately 1/3rd value shunt resistance for a 10dB attenuator. The thing is, you're doing more than just padding, so a resistive voltage divider isn't all there is to it. The crossover provides the traditional voltage splitter function and it also provides some response shaping. The HF filter should be slightly underdamped. Traditional 6dB/octave augmentation is provided for mass rolloff compensation, and the two transfer functions form a response curve that's flat up to about 4kHz, with 6dB/octave augmentation above that. I'd model your circuit with Spice, to verify it provides a transfer function that looks like this:

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[Message index]

		Reverse attenuation and HF comp. networks, active X-over By: Chris R. on Mon, 20 May 2002 21:30
		Re: Reverse attenuation and HF comp. networks, active X-over By: Wayne Parham on Mon, 20 May 2002 22:34
		Re: Reverse attenuation and HF comp. networks, active X-over By: Chris R on Wed, 22 May 2002 21:17
		Re: Reverse attenuation and HF comp. networks, active X-over By: Wayne Parham on Wed, 22 May 2002 21:49
		Re: Reverse attenuation and HF comp. networks, active X-over By: Chris R. on Thu, 23 May 2002 16:50
		Re: Reverse attenuation and HF comp. networks, active X-over By: Wayne Parham on Thu, 23 May 2002 23:01

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