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Re: Compensation circuit when using active crossover [message #49197 is a reply to message #49196]

Mon, 27 March 2006 15:51

Wayne Parham
Messages: 18790
Registered: January 2001

Illuminati (33rd Degree)

The series resistor will be dependant on load impedance. Bypass cap is found using the capacitive reactance formula, which is Xc = 1/2πFC. I'd say go with R=100kΩ and C=400pF. Go up and down doubling or halving values 'till you get where you want to be.

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[Message index]

		Compensation circuit when using active crossover By: ccvogel on Sun, 26 March 2006 13:35
		Re: Compensation circuit when using active crossover By: Wayne Parham on Mon, 27 March 2006 08:32
		Re: Compensation circuit when using active crossover By: ccvogel on Mon, 27 March 2006 10:45
		Re: Compensation circuit when using active crossover By: Wayne Parham on Mon, 27 March 2006 11:26
		Re: Compensation circuit when using active crossover By: ccvogel on Mon, 27 March 2006 14:56
		Re: Compensation circuit when using active crossover By: Wayne Parham on Mon, 27 March 2006 15:51
		Thanks By: ccvogel on Mon, 27 March 2006 16:03
		Re: Compensation circuit when using active crossover By: Larry Acklin on Mon, 27 March 2006 09:50
		Re: Compensation circuit when using active crossover By: ccvogel on Mon, 27 March 2006 10:50

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