[Boatanchors] 120 - 240vac and output
Gary Schafer
garyschafer at comcast.net
Sun Nov 28 21:52:43 EST 2004
Hi Ron,
Yes I agree with what you are saying. but we have to keep in mind that
in both our examples that the load resistance had to be changed
(amplifier retuned) in order to keep the current the same with the lower
plate voltage. Sometimes that may not be possible, the amp may not be
able to draw the same current at a lower plate voltage.
So we could also call the transformer a voltage device as the voltage
transfers by the turns ratio as well as the current.
The bigger problem is that we do not know what the resistance of the
wire carrying the power in the house is. This whole thing was intended
to be a simple example of how to get a rough idea of what the difference
in losses are.
With what you are saying, "power lost is equal to the square of the
current times resistance", could be measured by measuring the AC current
and voltage on the power line. Set the amplifier to draw say 10 and
measure the line voltage at the amp again. You could then calculate
what the line resistance was and use I squared R to find the power lost.
That would still give use about the same estimate as the other method as
we don't really know what the amp will do with more or less plate
voltage. Just an estimate.
I guess that with the 120 / 240 volt example we should be saying
multiply the 2X difference in current times the 2X transformer step up
ratio difference to be correct.
In the case of the 2 times 2 it equals 4. The same as 2 plus 2 equals 4.
But in your example with 4 times the current and 4 times the transformer
ratio difference only multiplying the two differences give the correct
answer. 4 times 4 equals 16.
But the 4X with the 120 verses 240 volt system is still correct for what
was being discussed if 2X 2X is multiplied. :>)
It gives the same answer as your example.
73
Gary K4FMX
Ron wrote:
> Hi Gary and all interested.
> If you are saying that you are looking for a rough estimate ....well ,
> that's what you calculated by using voltages. Almost every time I
> actually measured a line , I never get 120 or 240 so my point here is
> that it is futile to start working voltages.
> What I object to is that you and everyone else is tossing around 4X as a
> general answer where it is the square of the current or the inverse
> square of the voltage that is set by the transformer turns ratio
> differences.
> If the turns ratio of the transformer is 2 , then the answer is 4 ( the
> turns ratio sets the differences in current) but not because it is 4X
> but because it is 2 squared. Has nothing to do with a 2X drop in the
> line resistance and a 2X multiplication from turns ratio difference in
> the primary to secondary even though in your hypothetical example that
> both are true.
> If transformer turns ratio are different than 2:1 then is is the square
> of the difference as the transformer sets both the current and voltage
> ratios. Some transformers have several taps for low and high line
> conditions.
> I am not disagreeing with you statements other than I am saying that the
> differences track the square of the differences in current because the
> loss IS I squared R even if R is unknown. I am saying nothing more or
> nothing less.
> Isquared R and Vsquared divided by R both have units of power and that
> is what you are discussing ....power lost.. I just can see any way to
> discuss power loss without addressing the basic concept that power loss
> is I squared R or V squared divide R..
>
> Work the hypothetical example with a turns ratio difference of 4:1 to
> follow what I am trying to convey.
>
> We don't know the line resistance but we give one for calculation
> purpose of 0.5 ohms
> Condition 1 is 120VAC line drawing 20 AC amps. Pin =2,400 W
> Turns ratio is 25: 1 Racmains=0.5ohm
>
> Vin =120 V Vloss is 20A X 0.5=10 V V excite =Vin-Vloss= 110V
>
> Vsec = Vexcite X 25=2,750V Isec
> 0.8A Pout= 2,200W
> Ploss = 200W
>
> Condition 2 is 480VAC line drawing 5 AC amps. Pin =2,400 W
> Turns ratio is 6.25: 1 Turns ratio difference is 25/6.25=4:1
> Racmains=0.5ohm
>
> Vin =480 V Vloss is 5A X 0.5=2.5 V V excite =Vin-Vloss= 480V-2.5V
> =477.5V
>
> Vsec = Vexcite X 6.25=477.5 X
> 6.25=2,984.375V Isec 0.8A
> Pout= 2,984.375 X 0.8A = 2,387.5W
> Ploss = 2,400-2,387.5W = 12.5 W
> Notice that the currents transfer across the turns ratio without any
> change i.e. a transformer is a current device.
>
> Now ....using your methods... "The first 2X has to do with twice the
> voltage drop on the wire due to twice the current. IR if you must. But
> we don't know the resistance of the wire and we don't care. "
> In this example , the voltage drop is 4X due to 4 times the current.
>
> "The second 2X has to do with the multiplication factor of the
> transformer primary to secondary ratio differences."
> In this example the turns ratio difference is 4X.
>
> "The added voltage drop from supply line loss is ADDED to the already
> there voltage drop in the supply circuitry which includes transformer
> performance / resistance, capacitor values etc. Those are going to be
> the same on either 120 or 240 volts."
>
> When you add these ....you get 8X but clearly the answer is 200W /12.5
> watts and is 16X.
>
> Power loss is indeed I squared R and therefore power loss is always
> proportional to the differences in the currents squared.
> ---
> Ron
>
>
> Gary Schafer wrote:
>
>> I thought this thing was done! But sorry Ron you are totally wrong.
>> No one is / was talking about calculating power from resistance. The
>> 4X has nothing to do with calculating power. We are not doing I
>> squared R or even IV.
>>
>> The first 2X has to do with twice the voltage drop on the wire due to
>> twice the current. IR if you must. But we don't know the resistance of
>> the wire and we don't care.
>>
>> The second 2X has to do with the multiplication factor of the
>> transformer primary to secondary ratio differences.
>>
>> The added voltage drop from supply line loss is ADDED to the already
>> there voltage drop in the supply circuitry which includes transformer
>> performance / resistance, capacitor values etc. Those are going to be
>> the same on either 120 or 240 volts.
>>
>> What is being discussed here is ONLY the voltage drop contributed by
>> the size of the 120 or 240 volt AC supply line.
>>
>> Yes we know, that the amp may be able to run a little more power by
>> drawing extra current with less voltage drop over and above what the
>> voltage drop represents, which would change the primary current
>> values. But all the examples are using the same total power so the
>> current values are even multiples. If the amp is running a little more
>> power on 240 than on 120 that will make the first 2X a little less but
>> not by much.
>>
>> And there is no way to calculate what that extra power will be with a
>> particular amp anyway. You can use all the load lines you want to but
>> unless you have a load line for the power supply also you will not
>> know until actually trying it.
>>
>> The whole discussion was about a ball park idea of the advantage of a
>> 240 line over a 120 line.
>>
>> Please reread my last post and Vic's last post on this. They both
>> spell out exactly what is and is not being discussed.
>>
>> 73
>> Gary K4FMX
>>
>>
>> Ron wrote:
>>
>>> Vic,
>>> No one will disagree with your example but the point here is that the
>>> 4X you calculate is simply due to the fact that 2 Squared is 4. The
>>> specific answer to your specific example is 4. My fear is that your
>>> specific example will give everyone the impression that 4X is the
>>> general answer when it is not! The general answer is the square of
>>> the current .
>>>
>>> The only other example that I can think of is one 10 year old telling
>>> a 5 year old that green beans come from can in a grocery store. Well
>>> Yes ,,,they do .....but not really.
>>> I mean no dispespect with the above example but it was the only
>>> example I could think of where the answer is both right and wrong
>>> depending on how much detail you want. You are correct for your
>>> specific example but you example becomes really complicated when you
>>> start to go further and add the fact that the loading on the
>>> secondary is non- linear and then you value of 4 floats all over the
>>> place as the voltages and currents settle to the load line of the
>>> amplifier current voltage curve. That is specifically why you would
>>> work losses using currents and not voltages.
>>> None of this really matters on the signal strength that an amp will
>>> provide to a communications but if you want a real answer , that is
>>> what you must do . Other than that , your answer could simply be
>>> that the amp makes less power on 120 volts because of the voltage
>>> drop in the house wiring and green beans come in cans.
>>> ---
>>> Ron
>>>
>>>
>>>
>>> Vic Rosenthal wrote:
>>>
>>>> WA5CAB at cs.com wrote:
>>>>
>>>>> You don't have twice the voltage drop in the primary
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> God help me, I didn't SAY that! I said the following (and if I
>>>> don't get through this time, I'll give up):
>>>>
>>>> 1) Other things being equal, an amplifier will draw about twice as
>>>> much current on 120V than on 240V. Ignoring losses in the
>>>> transformer, VA out must equal VA in.
>>>>
>>>> 2) The IR drop on the resistance of the WIRING to the transformer
>>>> from the pole will therefore be twice as great. NOTHING to do with
>>>> how the transformer is wound, wired, etc.
>>>>
>>>> 3) Since the transformer is multiplying the voltage twice as much
>>>> when using 120V than 240V, any DROP is also multiplied twice as
>>>> much. Twice the drop times twice the multiplication = 4X greater
>>>> drop in HV due to wiring resistance. This drop is added to the
>>>> normal drop in the power supply due to secondary resistance, etc.
>>>>
>>>> Simplified example: the resistance of the house circuit is 1 ohm.
>>>> Amplifier draws 10 amps at 240V, 20 amps at 120V (full load). The
>>>> IR drop in the wiring is 10V at 240V and 20V at 120V. Suppose the
>>>> power supply puts out 3000V no load. Then it is multiplying the
>>>> voltage 12.5 times when operating on 2400V. In this case the voltage
>>>> drop of the power supply output due to the wiring resistance will be
>>>> 10V * 12.5 = 125 volts.
>>>>
>>>> Now suppose we use the same equipment on 120V. The voltage
>>>> multiplication is 25 times, so the output voltage will drop 20V * 25
>>>> = 500 volts, before even considering the drop due to secondary
>>>> resistance, etc.
>>>>
>>>
>>> _______________________________________________
>>> Boatanchors mailing list
>>> Boatanchors at mailman.qth.net
>>> http://mailman.qth.net/mailman/listinfo/boatanchors
>>> ** List Administrator - Duane Fischer, W8DBF/W9WZE ** ** For
>>> Assistance: dfischer at usol.com ** $$ For vintage radio info,
>>> see the HCI web site $$ http://www.w9wze.org
>>
>>
>>
>>
>> _______________________________________________
>> Boatanchors mailing list
>> Boatanchors at mailman.qth.net
>> http://mailman.qth.net/mailman/listinfo/boatanchors
>> ** List Administrator - Duane Fischer, W8DBF/W9WZE ** ** For
>> Assistance: dfischer at usol.com ** $$ For vintage radio info,
>> see the HCI web site $$ http://www.w9wze.org
>>
>
>
>
More information about the Boatanchors
mailing list