posted
i have like 40 20W resistors, if i hoke thies up in series, could i make a fesable resistor to bench test my amps???? and also, how can i test my amps wattage if the amp is already installed in my car???? i have a few multi metrs, an osiloscope(spelling????) and just about peice of electrical equipment avaible. the only thing i dont have is a clam meeter. any ideas and or help is much aprechiated. Thankx
Chris
-------------------- Founder of team NOIZE FACTORY Member of Team Gates Beliver of ABC boxes Clarion DXZ555MP Head Unit 1 HUGE ABC box Dissapointing 148.7 AC190 Mic
posted
That's perfect. The only thing you don't need is the clamp on meter.
Since you're using a set resistance, you'll be using v * v / r to calculate power.
Depending on the value of the resistors, you'll have to mix series/parallel wire them. You'll want to get as close as possible to 4 ohms or the load you're using.
You'll need the o-scope to check for clipping. As soon as you detect clipping on the scope, turn it down until it's completely gone. Measure the voltage and you're done. voltage * voltage / resistace = power.
40 20W resistors makes for 800W. It will handle up to 2000 for very short moments, no more than a few seconds.
What's the value of the resistor and what amp do you have?
Posts: 768 | From: Ottawa, On, Canada | Registered: May 2000
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posted
as per amps, i have about 30, not all mine, but i would like to see what they do. as for the ohm on the resistors, they varry from 1-1kw ohm. it will take me a while to figure out the right combination for the resistors, buh hey, iam here to learn Thankx for the info
Chris
-------------------- Founder of team NOIZE FACTORY Member of Team Gates Beliver of ABC boxes Clarion DXZ555MP Head Unit 1 HUGE ABC box Dissapointing 148.7 AC190 Mic
Let's say I had a pile of 4 and 8 ohm resistors. I would takwe pairs of 8 ohm resitors, parallel those to make some 4 ohm resistors.
You want all the resistors to have the same value. Even if it were 40 ohms a piece, 10 of those in parallel and you have 4 ohms. You just want each resistor to dissipate the same amount of heat and to simplify the wiring alot.
Posts: 768 | From: Ottawa, On, Canada | Registered: May 2000
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quote:Originally posted by emu: Since you're using a set resistance, you'll be using v * v / r to calculate power.
umm.....how does this formula work to calculate how much power it makes.....isn't Watt's Law P/A*V i know i am right about Watts Law..... but prove to me that your formula works....
posted
just get dummy loads. They come in both 8 and 4ohm @ 100 & 200w. http://www.partsexpress.com has em, easiest way to do it bro. 200w ones arent cheap though ~$30.
-------------------- Andrew Kemendo Team Pro Creations FAA Pilot #2718765 Oderint dum metuant
umm.....how does this formula work to calculate how much power it makes.....isn't Watt's Law P/A*V i know i am right about Watts Law..... but prove to me that your formula works....
Pleasem tell me you were kidding.
But anyway. Since you know ohm's law. You know that I = V / R You may have also realized that Power = I * V.
Using an old age secret called substitution. I can't tell you how it works, that's top secret.
Power = I * V I = V / R Power = (V / R) * V Power = V * V / R
Keep in mind that the context this formula is used assumes a set resistance. By elimating any frequency dependant variables, we do not need to treat current and voltage seperately.
Posts: 768 | From: Ottawa, On, Canada | Registered: May 2000
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quote:Originally posted by emu: Depending on the value of the resistors, you'll have to mix series/parallel wire them. You'll want to get as close as possible to 4 ohms or the load you're using.
Should one measure DCR, or the 2^n ohm rating? I saw some drivers ( think on parts express, mabye they were tweeters, can't recall) that advertised a DCR of 3.6 for a 4Ohm rating, or a DCR of 5.something for a Ohm rating of 8. Please don't smack me emu.
posted
If you're using a dummy load or the resistors you really don't have to worry about DC resistance and Impedance, they are the same thing in that case.
On the other hand, if you're dealing with actual speakers it's a bit more complicated, if you want to be accurate. The thing to do is hook your voltage probe up to the first channel of the O-scope. On the second channel, place the current probe. Now, display both on the screen at the same time, and you have to determine the phase of the current relative to the voltage. Once you have that phase, in degrees, use this formula:
P= V*I*cos(phase)
to get the real power delivered to the speaker. (V) & (I) are the magnitudes of the voltage and current. Hope this helps, Ash
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