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Old 23rd December 2012, 04:57 AM   #1
alttbm is offline alttbm  Canada
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Default Building a 400 watt amp from salvage

I have an older 400 watt stereo (it was a full featured stereo not an just an amp) so I already have my PSU built.

I'm looking to salvage what I can of the unit + Speakers and build a new stereo

Solid State of course, thats what came out (I think) thats whats going back in.

this is my first time doing any audio work, but I can handle electronics assemby.

whats this crap about ohm's? I'm completely new to this so maybe a little
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Old 23rd December 2012, 07:18 AM   #2
djk is offline djk
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An 'ohm is an Englishman's 'ouse.

What kind of V does this alleged 400W amplifier run on?

My guess would be 57V~70V, but that can make a big difference.
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Old 23rd December 2012, 11:20 AM   #3
JMFahey is offline JMFahey  Argentina
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Quote:
whats this crap about ohm's? I'm completely new to this so maybe a little [help]
Best help in this case is suggesting you study at least a little of Electronics and come back later.
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Old 23rd December 2012, 01:23 PM   #4
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yeap ....later much later
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Old 23rd December 2012, 01:28 PM   #5
gootee is offline gootee  United States
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To understand Ohms, you have to understand Volts and Amperes (Amps). (We will ignore capacitors and inductors, for now.)

Volts are relative so they are always between two points. Having some Volts between two points won't necessarily make anything else happen. It depends on whether the two points are connected to each other, by something conductive. If that is the case, then what happens depends on the Ohms between the two points. Amps will flow between the two points according to Amps = Volts divided by Ohms, which is called "Ohm's Law".

That can be "rearranged", for different situations: If you know that there is a known number of Amps flowing through some known number of Ohms, then you could figure out the Volts between the two points that are on opposite ends of the Ohms, with Volts = Amps times Ohms. That is also the same "Ohm's Law". Similarly, if you happened to know the Volts across some Ohms, and you knew the Amps running through those Ohms, then you could calculate the Ohms as Volts divided by Amps.

You will also want to understand power, which is Watts. Watts are always in terms of some load that is dissipating them (doing work and/or or making heat). The load is some number of Ohms (ignoring capacitance and inductance, still). Watts are calculated by multiplying Volts time Amps, where the Volts are measured between two points on opposite ends of the load's Ohms, i.e. the volts "across" the load times the Amps flowing through the load.

Since Ohm's Law gives us any one of Volts, Amps, or Ohms in terms of the other two, we only need two of those to calculate the Watts being dissipated in any Ohms load. So if you remember any algebra, you could take Ohm's Law, i.e. v = i * R (Volts = Amps x Ohms) and substitute it into Watts = v * i, giving Watts = (i*R) * i = i squared times R. Similarly, Watts also equals v squared divided by R.

Everything above is true for constant (unchanging) DC voltages and currents, like in a battery-powered circuit with only resistors or reistive loads. And it's all also true for AC circuits, with time-varying voltages and currents, with only resistances, as long as you're talking about (or calculating) the Volts, Amps, and Watts at any instant of time, for example at the peak levels for the Volts, Amps, and Watts waveforms (but anywhere else on the waveforms, too). But people often use "RMS" when talking about Volts, Amps, and Watts that have sinusoidal waveform shapes, because it's convenient and useful and is easy to calculate and understand.

If we use RMS levels for sine waves of Volts or Amps, then we can calculate as if we are talking about constant DC Volts and Amps.

You can think of the RMS value as the level that would produce the same amount of power as using a constant DC of that level, instead of a sinusoidal waveform. The RMS level for a sinusoid is just its peak value divided by the square root of two, which is about 1.414. Multiplying by 0.7071 gives the same result. i.e. The RMS value for a sine wave is just its peak level times 0.7071. (Note that for other shapes of waveforms, the RMS level would have to be calculated differently. So you will probably usually only "RMS" for sine-type waveforms.)

Example: If your amplifier's PEAK output voltage level was 50 Volts, which would be a sinusoidal waveform changing between +50 Volts and -50 Volts at some frequency, and you had a load of 8 Ohms, then the peak power dissipated by the load would be v squared divided by R, or 50 x 50 / 8 = 2500/8 = 312.5 Watts (Peak). But if we use the RMS value of the voltage, 50 x 0.7071, i.e. 35.355 Volts RMS, we get 156.25 Watts RMS, which is exactly half of the peak power level.

Depending on what number of Ohms your stereo's "400 Watts" figure was rated for, you can calculate the maximum peak and RMS output voltages that it would have to be able to produce. If we assume they meant 400 Watts RMS into 8 Ohms, then we would know that v^2 / 8 = 400, so v max would have to be the square root of 400 * 8, which is 56.57 Volts RMS, maximum, which would also be a peak voltage of 1.414 * 56.57, i.e. 80 Volts peak, maximum. From those voltages, we can also easily calculate the maximum RMS and peak currents, in Amps, which would be 56.57 / 8 = 7.07 Amps RMS, maximum, and 80 / 8 = 10 Amps peak, maximum.

Since the power supply is not perfect, it will have to initially be producing more than plus and minus 80 Volts, to get a usable +/- 80 Volt range. You can learn more about that at the following link:

http://www.zen22142.zen.co.uk/Design/dcpsu.htm

Cheers,

Tom

Last edited by gootee; 23rd December 2012 at 01:50 PM.
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Old 23rd December 2012, 01:50 PM   #6
DUG is offline DUG  Canada
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Study chipamp data sheets and schematics for amplifiers until you know what all of the parts are for and what is required of them.

Or buy a kit to be used in your chassis with physical and electrical compatibility.

IMHO

Sound like a very nice project.

I should dig up my old Technics amp/receiver and try to fix it the same way. (only 60W/ch though)
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Old 23rd December 2012, 02:41 PM   #7
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I think this guy had read the "400W" power draw spec on the rear of the amp and done what most young players do and think thats the speaker output power of the amp when in reality this might be a common 2x50W amp.
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Old 23rd December 2012, 02:58 PM   #8
DUG is offline DUG  Canada
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Make / model of 400W?
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Old 23rd December 2012, 05:12 PM   #9
djk is offline djk
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"this might be a common 2x50W amp. "

Could be only 40V or so.

I was hoping it would be more like a 57V supply so he could build a 2x100W amp.
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Old 23rd December 2012, 07:23 PM   #10
alttbm is offline alttbm  Canada
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I kno what an ohm is
I meant Why is It Relavent
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