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Old 1st January 2013, 01:53 PM   #1
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Default creating 1 watt rms power from 5 watt chip amp?

is it possible to get 1 watt rms or even 5 watts rms?from a 5 watt chip amp?
how much capacitance should I need for the output? 2200uF? 10,000uF? I want the amp to nail the low frequencies down to 0.5hz as best as it can without distorting..
Is it possible? this is the chip i used
LA4225 datasheet(2/3 Pages) SANYO | Monolithic Linear IC Audio Output for TV application 5W Monaural Power Amplifier
and i have a 4.7uF capacitor for the input instead of 2.2uF's
and at the moment I have 6,200uF's for the output instead of 470uF's
It already nails the lows really nicely and sounds clean. very minimal static too.
if I put 10,000uF's how good would the low frequencies get then?
At first I only had 2,000uF's but adding the extra 4,200uF's added a lot more to it already..
Would there really be any use for having a 10,000uF cap in place for this chip?
Or is that just overkill?

Last edited by realflow100; 1st January 2013 at 02:15 PM.
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Old 1st January 2013, 02:26 PM   #2
6L6 is offline 6L6  United States
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What are you using this for?

Why do you need output at .5hz?

Is this being used as a reference of some sort, hence your requirement for a specific power?

Output cap is for blocking DC on speaker. 6,200uf is overkill.

Save your 10,000uf cap for the PSU or another project.

Last edited by 6L6; 1st January 2013 at 02:37 PM.
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Old 1st January 2013, 02:33 PM   #3
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I want to hit good bass notes and have overall super good clear sound quality for my speakers so they will hit with the best clearest sounding bass and hit the lowest frequency that they can..
The song bass I love you plays notes a little lower than 1hz and my amplifier can hit them all even though it cost absolutely nothing to put together this little chip amp since i salvaged the parts from an old TV motherboard that i found lying in a pile of scraps on the side of the road.
and the speakers actually show cone movement at less than 1hz easily
And the overal sound quality is amazing compared to something that costs probably 50 dollars. And If I had really good nice set of speakers it would sound amazing..
Right now I have the chip with heatsink all mounted nicely in a cardboard box.. With the power supply hidden in the box as well..And I have speaker terminals so I can easily attach or unattach almost any kind of speakers... Also with a headphone jack plug that I can just simply plug right into my computers headphone output... the sound quality is really good too even for a mono setup..
I want to make another clone of this amplifier so I can have a stereo amplifier.. one for the left channel. and one for the right channel..
And if possible I could brigde the amplifiers somehow.. but that's another thing..
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Old 1st January 2013, 02:36 PM   #4
6L6 is offline 6L6  United States
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You can't hear 1hz.

Anyway, it sounds like you are getting great performance from that chip. Time to start a new project, it sounds like you would be served best by some better speakers.

Happy New Year!
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Old 1st January 2013, 02:48 PM   #5
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I know i cant hear 1hz but i can see it lol
And I really want to buy some two 12 inch pioneer subwoofers at the flee market! only 50 dollars and comes with the box! woohoo
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Old 1st January 2013, 03:00 PM   #6
DUG is offline DUG  Canada
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Maybe it would be better to go for a direct coupled output design.

At 0.5Hz the cap can get ridiculously large.

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Old 1st January 2013, 03:08 PM   #7
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it wouldn't have as much power would it though right? and wouldn't that introduce ultrasonic pulses into the audio signal instead of normally just using a capacitor to block DC?
Oh yikes I just tried putting 19 volts to the amplifier chip for a few seconds
And Omg the 10 watt speaker bottomed out from a 5 watt IC chip amplifier
And there was no amplifier distortion.... only the bottoming out of the 10 watt speaker
Yikes I just got over 10 watts out of a 5 watt amplifier chip O.O that's a little scary..
But I hooked it back up to the 12 volt power supply instead so I think the chip is okay... It didn't get warm when I unplugged the power cord and checked the chip..
Maybe the massive heatsink helped save the chip?

Last edited by realflow100; 1st January 2013 at 03:38 PM.
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Old 1st January 2013, 04:07 PM   #8
gootee is offline gootee  United States
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You can easily figure out how much each frequency will be affected by each cap.

A capacitor in series with the signal, with a resistance to ground downstream from that, makes a passive RC high-pass filter.

If you know how a two-resistor voltage divider works, you can think of the capacitor and resistor as a frequency-dependent voltage divider. At lower frequencies, the apparent resistance of the cap is higher and a larger proportion of the available voltage is dropped across the cap, and less across the resistor. If the RC are at the input end, the voltage across the resistor to ground is what your amp's input sees. If at the output, the voltage across the speaker's "resistance" to ground is what you hear. Either way, it's lower at lower frequencies. How MUCH lower depends on the capacitance (and the resistance to ground).

A common resistive voltage divider, with series resistance Rs followed by a resistance to ground R, has

Vout = Vin x (R / (R + Rs))

i.e. The proportion that the R to ground is, of the total resistance, is the proportion of the input that gets to the output.

It's almost the same, for a series capacitor and a resistance to ground, but we need to know that the apparent resistance of a capacitance, to AC at some frequency, f, is 1/(2πfC).

So then the voltage-divider equation becomes:

Vout = Vin x (R / (R + (1/(2πfC))))

You could plug in values and use the equation in that form, or, we can re-arrange it to get:

Vout/Vin = 2πfRC / (2πfRC + 1)

Just remember to divide microfarads by 1000000, to get Farads to use in the equation. And use f in Hz and R in Ohms. π is about 3.14.

We could make the equation more useful, for your case, if we solved it for C, so you could pick a frequency and a Vout/Vin ratio and just calculate the needed capacitance.

I think this is right:

C = (Vout/Vin) / (2πfR x (1 - (Vout/Vin)))

If you wanted it solved for R, instead of C, you should be able to just SWAP the R and C, in that equation.

NOTE that if you make your input resistor larger, less capacitance will give you the same response. i.e. Increasing the resistance to ground gives exactly the same effect on the frequency response as increasing the capacitance.

But you shouldn't raise the input resistor's value TOO awfully high, because higher-value resistors generate more noise, and also pick up interference more easily, from the environment. (On the other hand, they're also easier for the source to drive.) A value of 10K to 20K or so should be fine, and maybe higher. I'd probably just try some much larger values and see if it has too much noise or not. Doubling the resistance is just like doubling the capacitance, as far as the frequency response is concerned. (And using 10x the resistance would give the same frequency response as using 10x the capacitance.)

If calculating with the R for a speaker, you can use 8 Ohms or 4 Ohms or whatever the rating is. But be aware that the speaker's actual impedance can change as the frequency changes.

Most people look at what's called "the minus 3 dB frequency", or "cutoff frequency", when talking about filters. That's just the frequency where the output is down to 3 dB less than the input. For voltages, that's a factor of 1/sqrt(2), or 0.7071, i.e. 70.71% of the input voltage makes it to the output. For power in Watts, -3 dB would be a decrease of 50%.

The -3dB frequency for a high-pass (or low-pass) passive RC filter is:

f(-3dB) = 1 / (2πRC)

Usually, for a high-pass filter, you'd want to make that frequency be at most 1/10th to 1/100th of the lowest frequency that you wanted to keep.

Last edited by gootee; 1st January 2013 at 04:37 PM.
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Old 1st January 2013, 04:33 PM   #9
sreten is offline sreten  United Kingdom
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No music or sound format has notes below 1Hz, period.

The lowest notes will be in the 0.1 channel of multi-channel
and hardly any of them will have any notes below about 16Hz.

You may think the modulation of cone excursion is a frequency,
the cone "appears" to move in and out at low frequency, but
that isn't producing low frequencies, its an optical illusion.

rgds, sreten.
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Old 1st January 2013, 05:02 PM   #10
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I can physically see the speaker moving at 1hz if I play a 1hz tone in audacity... 1hz is seriously slow.. i can clearly tell the difference between 1hz and 30hz
you can hear 30hz
but you cant hear 1hz its utterly impossible.. unless the entire earth shakes in an apocalpyse earthquake..
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