Power supply for bench testing

[ultimate157 5,000+ posts]

What power supplies do you guys use to bench test amps? I'm looking to pick up the equipment to bench my own amps, however have been unable to find a 12vdc supply capable of putting out the necessary current. (~300 amps)

Click to read more...

 

[JimJ 5,000+ posts]

300 amps is a lot of current to expect out of an AC supply.

I have a 70A continuous Astron linear supply and it's a pain in the ass to move anywhere...

 

[ultimate157 5,000+ posts]

How are you guys bench testing amps that will pull 300A (DC) under full load?

 

[brokeitagain 10+ year member]

batteries, lots of batteries, and a kick ass charger

 

[Haunz 10+ year member]

what are you using it for ? power output testing ?

To test functionality (especially after repairs) you don't want or need 300amps of current..

 

[ultimate157 5,000+ posts]

what are you using it for ? power output testing ?


To test functionality (especially after repairs) you don't want or need 300amps of current..

Find out the true RMS output of the amp.

 

[audioarsonal 10+ year member]

why does everyone insist on trying to find the RMS power of there amp?Clamping the amp does not give you a true RMS reading if that is what you are doing.

Besides the power supply for 300amps would be so god awful big and expensive that you would need a fork lift just to move it!Besides if you can't produce the current you need to acheive RMS in the vehicle then what good would it do to know?

 

[lilmaniac2 5,000+ posts]

battery bank is much more practicle, i was gonna build one b/c i could get a ton of nsb100fts for cheap, just dont have the space for it

 

[JimJ 5,000+ posts]

why does everyone insist on trying to find the RMS power of there amp?Clamping the amp does not give you a true RMS reading if that is what you are doing.

Do you mean because the phase angle of the power is unknown?

 

[helotaxi 5,000+ posts]

Do you mean because the phase angle of the power is unknown?

...which is why you use a purely resistive precision load...

The current is still induced by the voltage and it's the current that does the work. Phase is actually not relavant.

 

[Immacomputer 10+ year member]

I would be just as concerned at finding a bunch of purely resistive loads and a way to keep them cool while testing. My boss had two paint cans that are resistive loads but they're only like 15 watts or something small like that.

 

[JimJ 5,000+ posts]

...which is why you use a purely resistive precision load...
The current is still induced by the voltage and it's the current that does the work. Phase is actually not relavant.

I was thinking back to Fryman's thread on SSA, I was thinking that was the same "logic" he was following //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[JimJ 5,000+ posts]

I would be just as concerned at finding a bunch of purely resistive loads and a way to keep them cool while testing. My boss had two paint cans that are resistive loads but they're only like 15 watts or something small like that.

I've built a couple 50-ohm dummy loads for RF using gallon paint cans, filled with mineral oil they're good for 2kW or so, at least for half a minute or so...I'd assume you could do the same with a low impedance load for audio.

 

[helotaxi 5,000+ posts]

I was thinking back to Fryman's thread on SSA, I was thinking that was the same "logic" he was following //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

But in typical Fryman fashion he only knew enough to not know exactly what he was talking about. HS physics FTL. Just enough knowledge to be dangerous. He assumed that since there is a phase difference between voltage and current with a reactive load, that you needed to account for that difference when figuring power applied. The problem with that logic is that if it were true, it would break Ohm's law because V would not equal I*R at any point in time. Actual power applied at any instant is not important because as I sad before, it's the current that does the work. The voltage induces the current IAW Ohm's law. That peak voltage and peak current don't occur at the same time is irrelevant. Welcome to reactive loads. Welcome to driver phase response.