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Old 31st May 2014, 08:38 AM   #1
alibear is offline alibear  United Kingdom
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Default Zobel network

Hi everyone, I am building an amplifier that has an output of 30watts RMS at 8 Ohms. The circuit gives values of 10ohms and 0.1uF for the zobel network at the output, what I do not know is the required wattage of the resistor.
Is there a way of determining this value, or any suggestions.
Thanks
Alan
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Old 31st May 2014, 08:48 AM   #2
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I suspect 2W will be large enough. Some go for 5W in large amplifiers. The circuit only presents a load at HF to help keep the amplifier stable.
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Old 31st May 2014, 08:50 AM   #3
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[QUOTE=alibear;3946049] what I do not know is the required wattage of the resistor.
/QUOTE]

2W.
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Old 31st May 2014, 09:07 AM   #4
AndrewT is online now AndrewT  Scotland
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The Zobel helps with keeping the amplifier stable.
That means (that in a competently designed amplifier) there is never oscillation due to instability.

The only sustained voltage at the output should be the audio signal.
In general this will have very limited sustained voltage above 20kHz.
For audio files the resistor can be very low power capability.
Install a 250mW and feel it's temperature when playing LOUD.

BUT !!!!!!!
We test at HF and with squarewave there is a lot of HF.
This sustained HF testing voltage at the Zobel is many decades more power than music or audio reproduction.

A short term test using 50kHz is likely to severely overheat a small resistor and due to esr in the capacitor also overheat the capacitor.

A 20kHz squarewave has enormous HF energy.
A 5W resistor and a highish esr capacitor can be damaged during testing.
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Last edited by AndrewT; 31st May 2014 at 09:09 AM.
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Old 31st May 2014, 10:02 AM   #5
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The amplifier, if designed properly, will have HF roll off at above 22kHZ as higher frequencies is just wasted energy. If one works out the 1/2Pii F C on 100n and 10R that is the centre frequency of maximum dissipation. (159kHz, I think I got my decimal in the right place!) and that will present a 10R load on the amplifier. The wattage is limited to the maximum DC power supply rail/s.
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Old 31st May 2014, 10:21 AM   #6
DF96 is offline DF96  England
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100nF and 10R form a high-pass filter (as seen by the resistor) with a corner frequency of 159kHz. From around twice that frequency upwards the amplifer will see around 10R as a load. At lower frequencies it will see mainly capacitance.

Other things being equal (which they may not be) I suspect the combination of a first-order HF rolloff in the amp and a first-order LF rolloff in the Zobel means that the maximum resistor dissipation will occur at the geometric mean of the two frequencies. For 22kHz and 159kHz this is 59kHz. I can't be bothered to work out the exact values, but here the amp will see something like 80-90R resistive in parallel with about 30ohms of capacitive reactance.
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Old 31st May 2014, 10:29 AM   #7
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So in essence as DF96 states, the power dissipated is not a lot, so a 2Watt resistor is more than enough and the power choice is purely cosmetic.
The DF96 is a lovely miniature pentode valve built circa 1953 for RF amplification in battery powered radio receivers as it has very low heater current requirements. I have one NOS left for the next repair.
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Old 31st May 2014, 10:38 AM   #8
AndrewT is online now AndrewT  Scotland
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Quote:
Originally Posted by JonSnell Electronic View Post
The amplifier, if designed properly, will have HF roll off at above 22kHZ as higher frequencies is just wasted energy.
?
do you mean the F-3dB of the single pole roll off should be at 22kHz?

Or

do you mean that at 22kHz, the roll off should be set such that the maximum attenuation of the 22kHz signal is less than 0.0001db?
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Old 31st May 2014, 05:00 PM   #9
alibear is offline alibear  United Kingdom
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Hi, many thanks to all you very knowledgable people. The DIY Audio forum is a wonderful source of information. I am learning slowly, but maybe in the future I may be in a position to give answers instead of asking so many questions.
Thanks again
Alan
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