CONVERTING WATTS TO AMPLIFIER FORMULA

[TonyV225]

Ok I know we have played with this topic a few times but its driving me crazy it seems a bit off to me well frankly it is off for example.

Watts divided by Voltage= Amperage

so you take 100 Watts devide by say 13.8 Volts = 7.2 Amps

500 Watts divided by 13.8 Volts = 36.2 Amps

1000 Watts divided by 13.8 Volts = 72.4 Amps (I really dont see this being enough amperage am I wrong here Ive tried this in different wattages and it seems like its a bit short on amperage from my experience. I am thinking this also has alot to do with efficiancy of amplifier ETC.

I do this formula then ad 50% at the end result and it seems a bit more like one could believe

example

500 Watts divided by 13.8 Amps = 36.2 + 50% = 54.3 Amps
This to me is more in the ballpark of the figure. Am I not doing this right or understanding it? Is it based totally on a efficiancy factor? I also did this and instead of 50% I used 30% and those answers looked close to manufactures ratings like Texas Star ETC.

Can someone help me understand this here I really cant explain it to my kids until I know I have it figured out for myself LOL! I get mixed answers on this around my area so what are your thoughts?

 

[W5LZ]

I think you may not be factoring in the efficiency of the amplifier. Which is something around 50% typically. That efficiency is the difference between the amperage required between the VA of the output power and the required amperage for the amplifier to produce that output power. The primary input current will always be about 150%(+/-) of the output power, depending on how efficient the amplifier is. There's no direct relation between primary input current and the current required for the output, at least no always the same relationship, which is due to peculiarities of construction, design, and probably the color of the cabinet (yeah, right, color of cabinet).
That 150% is a fairly good guess, ball-park anyway. Then, when you start figuring on the amount of current supplied by the primary source, you get to figure in that source's efficiency! ICS versus CCS for power supplies, meaning bigger than what you'd expect.
Does that help any?
- 'Doc


PS - That 150% thingy is just a marginal guess. 175% would be a much safer guess. Too much, more than likely, but certainly doesn't hurt.

 

[Wire Weasel]

Yeah what LZ said. Plus with a transceiver you have the power the rest of the radio is using to add into it.

I hate math. Math hates me. We're even.

 

[TonyV225]

I understand that if you have your radio hooked to the same power supply or source its consumption also has to be included in the figures but Im looking at just 1 thing drawing the simple way to start.

I personally have enough power supplies that each radio device has its own supply or supplies so each supply is dedicated to doing one job or feeding one thing I find this the safest way.

 

[HiDef]

If you are taking about power consumption verses R.F. power out it becomes a lot more difficult. You have peak power occuring at certain points and zero power at others.

200 watts of amp input power can yield 400 watts P.E.P. any day without serious distortion. Distort the waveform and peak power levels can approach that of pulse ratings where 200 watts input can give over 1000 watts peak out. This is what is done in radar. The peaks are very short and average power low.

You have to consider the waveform.

This is how one can squeeze a lot of power from a small amplification device. Yes, it can be done but it will be quite distorted.

 

[dudmuck]

500 Watts divided by 13.8 Amps = 36.2 + 50% = 54.3 Amps

You're almost there.
Think of 50% efficiency like this: half the DC input power is going as useful RF, the other half is going as heat. So if you have 500 watts RF output, then you need 1000 watts going in on the DC. In other words, for 50% efficiency you need to add 100% to get the total DC = 72.4Amps.

Its probably most useful to compare RF watts in RMS with DC watts.

 

[TonyV225]

acatually thats one of the formula adons I did when I first started messing with this and it came out to pretty close to what I actuallly
see here with things I run here and then tried it a few different ways aswell and thats when I brought this here and asked what you all think.

 

[Captain Kilowatt]

You're almost there.
Think of 50% efficiency like this: half the DC input power is going as useful RF, the other half is going as heat. So if you have 500 watts RF output, then you need 1000 watts going in on the DC. In other words, for 50% efficiency you need to add 100% to get the total DC = 72.4Amps.

Its probably most useful to compare RF watts in RMS with DC watts.

This is pretty much right on. For normal solidstate amps you can figure the efficiency as 50% for figuring out current draw for a given power output. There are variables such as devices used, bias, and circuit losses but for the basis of deytermining current draw figure 50%. Just be aware of what type of power output is being quoted,average (rarely) or peak (often).

 

[TonyV225]

Ok where I had the trouble was 500 divided by 13.8 = 36.2 with the 100% came out to 72.4 Amps Ok thjis is pretty damn close but now take 1000 watts divided by 13.8 = 72.4 + 100% = 144 Amps does this seem right in the real world? for 1000 watts? Im thinking so but I was always told with these solidstate amplifiers it was 120 Amps

 

[mackmobile43]

75 amps will power a 4/2sc2879 class c amp just fine and you should see 750 to 800 watts pep depending on your driver, with a few more amps say 85 or so you can power a class ab1 amp to the same output with fewer watts drive.

 

[TonyV225]

Mack Ive ran both of my 4/2879 with 40-50 in and yes I seen over 750+PEP that was on a 88 amp supply people say you cant drive these amplifiers I have with that much input but Ive done it for awhile now no problems just built a fan stand for the amps to sit on to keep the heat down and there flawless as far as operation goes.

there AB-1 Amplifiers and there not comp amps if Itold you what they are Ide have people telling me thats impossible they would burn up! Well Ive seen others do this so I knew they had no problems and I followed suit and also with no problems. I use the radio as a driver Ive ran my little amps on 10-12 meters without a problem although I dont drive them hard to keep the RF splatter down but these amps work better on 10 anyways as far as output goes with mimimal drive.