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Chip Amps Amplifiers based on integrated circuits 

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30th September 2009, 03:43 PM  #11 
diyAudio Member
Join Date: Jan 2009

All amps sounding music but not clean, If I mute the volume, some sound comes like VOU VOU VOU VOU VOU VOU ( Sorry I can not describe ) I have added RF filtter. It's not HUM. This sound does not come when single amp running. Should I purchase new transformer 18018 2Amp.(available to my supplyer ) 12012 transformer is not sufficient ? How big smoothing capacitor I should apply ?

30th September 2009, 04:04 PM  #12 
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Join Date: Dec 2008
Location: Burlington

I believe PB has it that you are asking too much of your transformer. (See my signature) the six amps are drawing too much current causing the bulb to light and the temperature to rise. The traffo isn't inherently faulty though because it works flawlessly with only one amp attached?
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30th September 2009, 04:09 PM  #13 
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Join Date: Jan 2009

Which transformer is applicable ?

30th September 2009, 05:06 PM  #14 
diyAudio Member

Your description sounds like a phenomenon called motorboating. That is a result of unwanted feedback, which leads to some kind of oscillation.
It is possible that you get that effect, when leads or traces are too close to each other. The speaker signal can e. g. couple into the input signal, and produce such an effect. Try to group the wires according to their function. E. g. power supply, speaker, line level. Then separate the groups from each other. Power supply wires should not run in parallel with line level signal wires. Speaker wires should not run in parallel with line level signal wires. The distance between speaker and power supply wires is not as critical, but it is always good to separate them as well. Review your amplifier with that in mind. If you have the PCBs very close to each other, the traces can also couple into each other through induction. Bring as much distance between the wire groups as you can, where they run in parallel. You need less distance, where the wires cross paths at right angles. The best solution is to take all line level signal wires to one side, all speaker wires to the opposite side and the power supply wires at a right angle to the others, i. e. upward, downward, towards the front or towards the back. You will need to look, how you can best do that in your amplifier case. When there is not enough space to keep the wires separate, you can try shielding them with metal plates in between. That is however not very effective at low frequencies, unless you can get your hands on the very expensive mumetal. Try grounded copper (e. g. untreated PCBs) or aluminium. The transformer should not degrade the sound at low listening levels. It could become an issue at high listening levels. A transformer should have a power rating at least as big as the combined output power of all amplifier channels.
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10th February 2013, 12:33 AM  #15 
diyAudio Member

hi
just wonder where can we buy transformer for LM3886 X4 it said that have power of 200 watts said like 30 volt but did not say the current going to be? 
10th February 2013, 12:34 AM  #16 
diyAudio Member

any information will be helpful
thanks 
10th February 2013, 01:16 AM  #17 
diyAudio Member
Join Date: Mar 2009
Location: Buenos Aires  Argentina

Watts is often used instead of the more precise VA (Volts x Amperes)
So what you need is: 200 "Watts"/30 Volts= 6.66A=7A 
10th February 2013, 02:04 AM  #18 
diyAudio Member

thanks so much

10th February 2013, 02:09 AM  #19 
diyAudio Member

jmfahey je
vivis en argentina? yo estaba en la pampa y ciudad de la paz que tal el mundo chico no? 
10th February 2013, 03:42 AM  #20 
diyAudio Member

If you want your final amplifier configuration to have good performance at the maximum rated power output, you would want to use a transformer with higher VA rating. Also good would be a higher output voltage, regulated down to what you need.
You could choose the minimum capacitance (per rail) to use, for unhindered bass response down to frequency f, with: C ≥ imax / ( πf∙(Δv  (ESR∙imax))) which was derived in the post at Power Supply Resevoir Size and gives the capacitance value, C, that would be required in order to supply the current for one halfcycle of a sine signal of frequency f (in Hz), with 0topeak amplitude imax Amperes, while causing the voltage across the capacitor to dip by no more than your choice of Δv Volts. imax = √(2 ∙ P_rated_rms / R_load) With the C equation above, you could first let ESR = 0, then calculate C, then find the ESR for that C at frequency f, then recalculate C with the ESR you found, and you might need to iterate like that until C does not change significantly. Or, using an approximate expression for ESR, ESR = 0.02 / (C x VR), where VR = Voltage Rating of capacitor, and substituting it into the first equation, you could use: C ≥ imax ((0.04πf / VR) + 1) / (πf∙Δv) [approximate, for electrolytic cap, only] If we assume that the charging pulses would always be sufficient to recharge the capacitors as much as needed, then the two equations above would become the two equations below: Capacitance Needed for a Sine Down to Bass Frequency = f, at Max Rated Power: (1) C ≥ ( (imax ∙ f ) / ( 2 ∙ fmains )) / ( π ∙ f ∙ (Δv  ( ESR ∙ imax ))) (2) C ≥ (( imax ∙ f ) / ( 2 ∙ fmains ))∙(( ( 0.04 ∙ π ∙ f ) / VR ) + 1) / ( π ∙ f ∙ Δv ) [approximate, for electrolytic cap, only] where VR = Voltage Rating of capacitor.  But, for it to be truly "bulletproof", for any signal shape, including DC at the maximum rated peak sine voltage and current, you would want to use: C ≥ (imax∙Δt ) / (Δv  (ESR∙imax)) which gives C as the capacitance that will allow only up to some desired maximum railvoltage dip Δv when a specified maximum rated current imax is pulled out of the cap for a specified time Δt. In this case, for a full wave bridge rectifier, we would use the time between charging pulses as Δt, i.e. Δt = 1/2f, where f is the AC mains frequency (50 Hz or 60 Hz), giving: Capacitance Needed for WorstCase Signal Shape, at Max Rated Power: (3) C ≥ ( 1 /( 2 ∙ f )) ∙ imax / (Δv  ( ESR ∙ imax )) Or, using the same approximation for ESR as was used before: (4) C ≥ ( imax / Δv ) ∙ ( ( 0.5 / f ) + ( 0.02 / VR )) (approximation, for electrolytic capacitors ONLY) where VR = Voltage Rating of capacitor. Note that if the ESR is not much lower than the desired Δv / imax (which is also the maximum impedance we want the load to see), then the required capacitance value would be excessive. Cheers, Tom Last edited by gootee; 10th February 2013 at 03:47 AM. 
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