Sekess, go back to a simpler equation. With a +/-20v output available a 16ohm load can only pull 20v/16ohms = 1.25A, so this 1.25A being the peak is multiplied times the peak voltage of 20V, coming out to 25W peak divided by 2 for 12.5 WattsRMS. The amp still has plenty of current, but has run out of voltage swing.
Sekess,
When you measure mains voltage (in US) you read 110v.
If you look it on a scope you will see a sinus from +155v to -155v. (310v peak to peak)
The efficient value of 110v is used for power calculations.
When you measure mains voltage (in US) you read 110v.
If you look it on a scope you will see a sinus from +155v to -155v. (310v peak to peak)
The efficient value of 110v is used for power calculations.
WRT the filter caps, why not simply bypass your electrolytic caps with a high quality film capacitor? That certainly will improve the esr and the HF reactance instantly...
_-_-bear
_-_-bear
Film caps has lower impedance at high frequencies.By the way it lowers esr.It is known that they have been used in power supplies
There is a famous (in audio circles) article from decades ago now in Audio Magazine (defunct) about Capacitors by Walt Jung and someone else, iirc. Find it online. Basic DIYer's reading!! 😀
_-_-bear
_-_-bear
Motor run caps are said to be the best for bypassing.
That would depend entirely upon the type of motor run cap, imo.
If it is a paper type, then not the best. Today there are polypropylene motor run caps, and they may be good depending on the exact spec of the particular cap. Most motor run applications are not concerned about how the cap performs at high frequencies, they are concerned about handling AC "ripple", which of course may be very high depending on the motor and voltages involved. Its probably safer to choose a cap intended for SMPS application because they generally are designed to handle rather high frequencies and substantial HF ripple as well!
Why do some polypropylene caps have fat metal tabs coming out the end, and not wires?


_-_-bear
As you are speaking about ESR, i have proceeded to an "ESR"google search and found this.
http://www.illinoiscapacitor.com/uploads/papers_application/A0A0A86888CE4C0591627D18FF09455A.pdf
It says ESR decreases with capacitance and frequency.
A big cap will have a much lower ESR than a small one so that the small cap ESR will be neglectable.
Or am i wrong.
http://www.illinoiscapacitor.com/uploads/papers_application/A0A0A86888CE4C0591627D18FF09455A.pdf
It says ESR decreases with capacitance and frequency.
A big cap will have a much lower ESR than a small one so that the small cap ESR will be neglectable.
Or am i wrong.
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Thanks for all your answers guys. I got it now. The voltage available number was screwing me up.
Thanks,
Steve
Thanks,
Steve
fat metal tabs
😕😕😕😕
Why do some polypropylene caps have fat metal tabs coming out the end, and not wires?![]()
😀
_-_-bear
😕😕😕😕
Hi,ok bobodioulasso,
So, if I'm following you correctly:
-- Let's say you've got +-24 volt rails. That would give you 48 volts total.
-- Let's take off 8 volts for various losses. I'm not sure if this is a good approximation or not. I just chose it so that we have an even 40 volts available at the rails.
-- So, if we have 40 volts as Vmax, we can now divide by 1.414. This would give you 28.29volts.
-- Now, using (V*V)/R as the calculation for Power, we get (28*28)/16 = 49 watts.............
you have a dual polarity power supply. This is exactly equivalent to two power supplies, one that is +24Vdc relative to zero volts/ground and the other one is -24Vdc relative to zero volts/ground.
You have a speaker connected to ground on one side and to the amp output rail on it's other side.
What is the maximum voltage that the positive supply can apply to the speaker?
What is the maximum voltage that the negative supply can apply to the speaker?
Why do some polypropylene caps have fat metal tabs coming out the end, and not wires?![]()
😀
fat metal tabs are shorter than wires, generally.😕😕😕😕
Short links = low inductance = lower impedance at HF.
That's why the two pin capacitors inserted into a PCB have relatively low inductance for their 10mm pin pitch.
fat metal tabs are shorter than wires, generally.
Short links = low inductance = lower impedance at HF.
That's why the two pin capacitors inserted into a PCB have relatively low inductance for their 10mm pin pitch.
Andrew,
Thanks for giving an answer.
However the question was not aimed at someone who is "highly knowledgeable about all matters electronic", rather for others to go and look, question and research. So, it would be better if you would restrain yourself from being didactic in many cases?
_-_-
PS. So, why not just short wires then?
PPS. I posit that I don't know squat.
Andrew ,
With all due respect to Bear , thanks for the clear and concise answers , 2 many times answers to questions are given with smoke and mirrors which leads to more confusion , where only those in the know can follow , the opposite of the actually intent to help i suppose.
regards,
With all due respect to Bear , thanks for the clear and concise answers , 2 many times answers to questions are given with smoke and mirrors which leads to more confusion , where only those in the know can follow , the opposite of the actually intent to help i suppose.
regards,
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there are times (many of them) when I too pose directed questions to the enquirer. I think I make it clear when I expect the enquirer to do the homework to come to an answer.
I do not play mind games with other Forum Members and I would respect more those that do play with me/us.
Clarity is important.
Just to go off topic.
Essay set in year8 (11/12years old).
Describe how you would build a model aeroplane.
I had previously read a book on precisely this subject.
I looked up my grammar text book and found out how to write a technical essay.
I submitted my essay and it came back with a zero. The English teacher refused to mark it. We never got the chance to discus his issue. I suspect he thought I had copied the book verbatim. He went on to become a university lecturer that I met again some 30 years later, when he retired and moved back to Edinburgh.
I do not play mind games with other Forum Members and I would respect more those that do play with me/us.
Clarity is important.
Just to go off topic.
Essay set in year8 (11/12years old).
Describe how you would build a model aeroplane.
I had previously read a book on precisely this subject.
I looked up my grammar text book and found out how to write a technical essay.
I submitted my essay and it came back with a zero. The English teacher refused to mark it. We never got the chance to discus his issue. I suspect he thought I had copied the book verbatim. He went on to become a university lecturer that I met again some 30 years later, when he retired and moved back to Edinburgh.
Had many teachers like that myself ... i guess it's an requisite when preparing for 'O' Levels... 🙂
Hey Andrew,
If for simplicity we assume no losses and that the positive and negative halves can output the full respective +24 and -24 volts, then the maximum voltage that the positive supply can apply to the speaker is +24 volts and the maximum that the negative can supply is -24 voilts (with respect to ground).
So theoretically, I would think that the driver could move -24 volts in the negative direction and +24 volts in the positive direction. This would give a total voltage swing of 48 volts from the bottom of the driver's movement (inward) to the top of the movement (outward).
This is what I had originally thought was the maximum theoretical voltage available to the driver (48 volts - minus losses and circuit limitations). But I was informed that - since the output is a sine wave - for calculating wattage, you use some fraction of the peak to peak voltage. I think it's the RMS voltage value. But, I'm not 100% sure that it's RMS. This is the part that had me somewhat confused.
Thanks,
Steve
If for simplicity we assume no losses and that the positive and negative halves can output the full respective +24 and -24 volts, then the maximum voltage that the positive supply can apply to the speaker is +24 volts and the maximum that the negative can supply is -24 voilts (with respect to ground).
So theoretically, I would think that the driver could move -24 volts in the negative direction and +24 volts in the positive direction. This would give a total voltage swing of 48 volts from the bottom of the driver's movement (inward) to the top of the movement (outward).
This is what I had originally thought was the maximum theoretical voltage available to the driver (48 volts - minus losses and circuit limitations). But I was informed that - since the output is a sine wave - for calculating wattage, you use some fraction of the peak to peak voltage. I think it's the RMS voltage value. But, I'm not 100% sure that it's RMS. This is the part that had me somewhat confused.
Thanks,
Steve
Hi,
you have a dual polarity power supply. This is exactly equivalent to two power supplies, one that is +24Vdc relative to zero volts/ground and the other one is -24Vdc relative to zero volts/ground.
You have a speaker connected to ground on one side and to the amp output rail on it's other side.
What is the maximum voltage that the positive supply can apply to the speaker?
What is the maximum voltage that the negative supply can apply to the speaker?
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