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Chip Amps Amplifiers based on integrated circuits

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Old 2nd June 2006, 06:06 PM   #1
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Default Check my math?

I've been reading up on Ohm's Law, and i made up some equasions here for an amp / psu i'd like to make, im not sure how exact the math has to be, so i figured i'd run it past you folks, maybe you can give me a hand

Quote:
========================
= From Wall Outlet =
========================
Input:
Voltage: 120 Volts
Power: 2400 Watts
Current: 20 Amps

========================
= PSU Output =
========================
R = Resistance (Ohms)
P = Power (Watts)
C = Current (Amps)
V = Voltage (Volts)

R = 12
V = 54 Volts
C = 4.5
P = 243 Watts

Output:
Power: 243 Watts
Current: 4.5
Volts: 54
========================
= Out From Amp =
========================
I = Current (Amps)
R = Resistance (Ohms)
V = Voltage (Volts)
P = Power(Watts)

P = 100 Watts

V = 50 Volts

R = ((V * V) / P)
R = ((50 * 50) / 100)
R = (2500 / 100)
R = 25 Ohms

I = V / R
I = 50 / 25
I = 2 Amps

Output:
Channels: 2
Power: 100 watts / Channel
Current: 2 Amps / Channel
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Old 2nd June 2006, 06:15 PM   #2
poobah is offline poobah  United States
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Looks good... I don't understand why you have R listed in your power supply calc's.

Use I for current in the future... not C. C is used for capacitance.

And don't expect to pull 20 Amps from a wall outlet. If you need that much power... change your equipment to 240 V... like a dryer or stove.


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Old 2nd June 2006, 06:17 PM   #3
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How many amps should i pull from the outlet? 4 - 5?
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Old 2nd June 2006, 06:24 PM   #4
poobah is offline poobah  United States
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It might help to know what you are doing?

If your power supply is putting out 243 Watts, it is probably requiring 243 / .95 = 256 Watts at its input. This assumes 95% efficiency... could be lower like 85 - 90%

If 256 Watts were going into your supply; what would the current be?

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Old 2nd June 2006, 06:32 PM   #5
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2.13 Amps?
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Old 2nd June 2006, 06:39 PM   #6
poobah is offline poobah  United States
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Yes!

It is important to understand that the wall socket doesn't just cram 20 Amps into anything you plug in.

The input to the power supply has a resistance (not really a resistor) that determines what anount of current flows. With what you know... what is the "effective input resistance" of your power supply?

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Old 2nd June 2006, 06:53 PM   #7
preiter is offline preiter  United States
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Home electrical circuit usually put out either 15 or 20 amps (if we are talking US here) . Unless you are building a 1500+ watt amplifier, however that's not what you should be worried about.

Try going backwards from your amp wattage to determine what size transformer you need.
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Old 2nd June 2006, 07:01 PM   #8
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Ok, using the following variables
P = 256 Watts (What im drawing from my outlet)
E = 120 Volts (What im drawing from my outlet)

Now that i have 2 Set variables, i can now find the other 2
I (Current)
R (Resistance)

R = (E * E) / P
R = 260.86 Ohms

I = V / R
I = 0.46 Amps

So my effective input resistance is 206.86 Ohms?
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Old 2nd June 2006, 07:02 PM   #9
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Quote:
Originally posted by preiter
Home electrical circuit usually put out either 15 or 20 amps (if we are talking US here) . Unless you are building a 1500+ watt amplifier, however that's not what you should be worried about.

Try going backwards from your amp wattage to determine what size transformer you need.
Great idea. Thanks!


--------------------
edit
--------------------
Correct me if im wrong, but i should have the output of the powersupply give more power (Watts) than what my Amp calls for right?

I know for computers, you should always have more wattage than your computer draws.

Thanks!
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Old 2nd June 2006, 07:06 PM   #10
poobah is offline poobah  United States
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Nope,

You are drawing 2.13 Amps... remember?

R = V / I = 120 / 2.13 = 56.3 Ohms
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