creating 1 watt rms power from 5 watt chip amp?

[lmaurits]

Originally Posted by realflow100

what's the maximum input capacitor that I could use for the little IC chip amp that I made?
Is 4.7uF already too much? Because it seems to work just fine for me... And the sound quality is spot on..
And the input capacitors maximum voltage rating. Should I use a higher voltage input capacitor? like 50 volt or more? or is 25 just fine?

There is no maximum, except that imposed by how much money you have. But there's no point in increasing it. It's part of a high-pass filter, just like your output capacitor, but the input impedance of your chip is so high that it won't affect your bass unless the cap is really small (like 0.001uF or something). 4.7uF is more than enough. You will not get more power if you increase it (just like the output cap). There is also no need to increase the voltage, since no significant voltage develops across an input capacitor anyway. You will not get more power if you increase it.

[lmaurits]

Gootee, thanks very much for taking the time to write up that long, detailed post. I'm going to read through it carefully and digest it all later tonight.

[gootee]

Originally Posted by lmaurits

First of all, I believe there is technically no such thing as "Watts RMS", or "RMS power", that's just confused marketing speak. Output power in Watts is computed using RMS voltage/current.

Secondly, you can certainly get 1 Watt out of a 5 Watt chip amp. A chip amp being rated at 5 Watts just means 5 Watts is the maximum it can put out before being damaged by overheating. The chip isnt always putting out 5 Watts no matter what. If you turn your music down quiet enough the chip will happily put out 0.1 Watts. The same is true of a 500 Watt amp or any other amp.

Earlier DUG suggested you get rid of the output capacitor and go direct coupled, to avoid filtering out the super low frequencies that you for some reason care about, and you said "it wouldn't have as much power would it though right". I don't believe that is true at all. The output capacitor does nothing to provide power, you won't get more power out of an amp by increasing the output cap and keeping the input levels and gain constant. The output cap just blocks DC, and filters low frequencies as a side-effect. In a design that doesn't need a DC blocking cap (like one with a dual power supply, or a virtual earth), there will be no filtering of low frequencies with no influence on the output power.

I am pretty sure putting 1 Hz through a loudspeaker is a Very Bad Idea, and will eventually kill a speaker for the same reasons that DC will.

Well, you are right. There is, technically, no such things as "Watts RMS".

But it seems to be a commonly-used shorthand for Watts that are calculated using Volts RMS.

According to Wikipedia, in the USA the term for that (i.e. Watts calculated from Vrms squared divided by R) should be "Continuous Power", which was specified in the Federal Trade Commision;s "1974 Amplifier Rule", so that advertising claims for audio amplifier "power" specs would all have to use the same type of measurement (Vrms is measured, using a specified load resistance, at a specified maximum THD value, which usually occurs just before the onset of clipping, and then CP rating = Vrms^2 / R).

However, for a purely resistive load, if you multiply Vrms x Irms (Volts RMS time Amps RMS, or, equivalently, calculate Vrms^2 / R or Irms^2 x R), you get the AVERAGE (aka "continuous average") power, in Watts, which, if a sine wave is used, also happens to be one-half of the PEAK SINE Power (also in Watts).

Anyway, for a sine wave and a purely-resistive load:

Pavg = (Vrms)^2 / R = (Irms)^2 x R = (Vpeak)^2 / (2R)

Actual "Peak Power", sometimes also called PMOP (Peak Music Output Power), does NOT have to refer to power while using a sine signal. So it can be many times higher than the Continuous Power rating (typically 5 or 6 times higher).

[sreten]

Hi,

Watts RMS is extremely well defined and not at all obscure.
The only issue being it assumes purely resistive loads.
Which speakers aren't, but that does not matter.

Best to assume speakers are as onerous a load (due to
reactance) as half the claimed impedance as resistance.

rgds, sreten.

[realflow100]

could I use a 30ohm resistor across the input to decrease gain and noise? I know it will put a small load across the computers output but it shouldn't be too much right?

[lmaurits]

Originally Posted by gootee

That's why I constantly harp on the benefits of tightly twisting together the input signal and signal ground wire pairs, and the AC input and transformer secondary wire pairs. Any space left between the AC wires makes them a tansmitting antenna for hum. And any space left between the input conductors makes them a hum-receiving antenna that's connected to a high-gain high-power amplifier! (Yikes!)

I feel like this is maybe a silly question, but why does current flow through the signal ground wire? At the amplifier end, signal ground is tied to power ground, right? So why would an electron travel all the way back across the signal ground wire to return to the ground of its "home device" when the amplifier's ground, at the same potential, is so very much closer? Do some flow through the signal wire and some to the amplifier's ground, in proportion to the resistance of those two paths?

[PetruV]

about the optical ilussion,it happened to me that the speakers started to move very much in and out while plaiyng music composed of mostly vocals at high levels,the voice is composed of 400hz with variyng amplitudes and the top of the peaks drive the speaker up and down slowly lookin like they ding 1-0.1 hz

[gootee]

Originally Posted by realflow100

could I use a 30ohm resistor across the input to decrease gain and noise? I know it will put a small load across the computers output but it shouldn't be too much right?

That would be an extremely low resistance, for an input (assuming you meant from input to ground). Then your DC blocking input capacitor, combined with the 30 Ohms to ground, would set the high-pass cutoff point at a fairly high frequency, losing all of the bass. f = 1 / (2 Pi R C) = 1 / (2 x 3.14 x 30 x .0000047) = 1129 Hz!

If you want less overall gain, you could use two resistors, to make a voltage divider. That way, you could select 'reasonable" R values but still make the gain of the divider be almost whatever you want. You would need one R in series with the input and one from the input to ground. If you call them Rs and Rg, they would make V_input_pin = Vin x (Rg / (Rg + Rs)). So, for example, to cut the signal in half, you could use two equal-valued resistors.

Your blocking capacitor would see Rs+Rg to ground, I think. So you can calculate the Rs+Rg needed to keep your cutoff frequncy down around 1 Hz or less. For 4.7 uF and 1 Hz, for example, Rtotal = 1 / (2 x 3.14 x 1 x .0000047) = about 33,900 = 33.9 kOhms, so anything above about 34k total would be fine. Then you would just need to find two R values that give the fractional gain you want, but still add up to at least 34k. (NOTE that if you were installing an actual volume control, which would vary the total resistance whenever the volume was changed, you would want it AHEAD of your capacitor. Otherwise, changing the volume would also change the cutoff frequency of the high-pass filter.)

[gootee]

Originally Posted by lmaurits

I feel like this is maybe a silly question, but why does current flow through the signal ground wire? At the amplifier end, signal ground is tied to power ground, right? So why would an electron travel all the way back across the signal ground wire to return to the ground of its "home device" when the amplifier's ground, at the same potential, is so very much closer? Do some flow through the signal wire and some to the amplifier's ground, in proportion to the resistance of those two paths?

As far as twisting the signal and signal ground wires together is concerned, it wouldn't matter if anything was flowing or not. We just want to twist them so they don't make such a good antenna.

But, regarding your question: There is not really such a thing as "ground" (or at least there doesn't need to be, for circuits). It's just a convenient concept. There are only loops, where current flows. Whenever the source tries to push electrons out of one of the two wires, it is also trying to pull them in through the other wire. If electrons do go out through one of the source's output wires, then they must also come in through the other wire, at exactly the same rate, unless there is some other path they could follow, to get all the way back to where the other wire connects to the source. But that would be a "ground loop", which we hope doesn't exist.

[jan.didden]

Originally Posted by lmaurits

I feel like this is maybe a silly question, but why does current flow through the signal ground wire? At the amplifier end, signal ground is tied to power ground, right? So why would an electron travel all the way back across the signal ground wire to return to the ground of its "home device" when the amplifier's ground, at the same potential, is so very much closer? Do some flow through the signal wire and some to the amplifier's ground, in proportion to the resistance of those two paths?

Current always flows in a loop; at any point, what flows out of it must also flow back into it.
If not, all electrons would accumulate at one point and we would run out of electrons, eventually

If you're really interested (which I doubt) you might Google Thevenin (which says all voltages in a loop sum to zero) or Norton (all currents in a node sum to zero).

jan