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High voltage low current.

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Simon Chaddock29/12/2010 13:51:35
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There seems to be a trend for ever higher discharge currents.
20C used to be considered good now 35C is not uncommon.
 
All well and good but resistance losses are proportional to the current squared.. So 200W output at 10V needs 20A whereas at 50V it would only need 4A. but the resistance losses (and the heat produced) would only be 1/25th (4x4)/(20x20).
 
As an example nine 500mAh cells in series (33.3V) would have the same total capacity as a 1500mAh 3s, could produce the same output power and would weigh about the same but would only need a 10A ESC.
 
An interesting thought..
 

Steve W-O29/12/2010 14:06:43
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but the resistance losses (and the heat produced) would only be 1/25th (4x4)/(20x20).
 
 
Heat produced where?
Chris Bott - Moderator29/12/2010 14:17:36
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Simon this is a definite choice to be made when choosing a setup.
You are right, losses are less with higher voltage setups. This is why the electricity companies distribute power at such high voltages. 
 
Our problem is that high voltage speed controllers are more expensive, and it's pretty hard to find a 50V 5A controller for example.
 
With higher powered setups say 1KW, I always try to go for higher voltages and lower currents. 
Simon Chaddock29/12/2010 16:11:11
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Steve W-O
Does not the battery, ESC and motor heat up? It all represents wasted battery energy.
 
Chis
As you say difficult to find a high voltage low current ESC because nobody makes them - yet!
  
Peter Beeney29/12/2010 17:38:52
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   Simon,
            With the greatest respect and I certainly might well have got my wires slightly crossed here, but would your little equation stand another coat of looking at? Starting with a spot of Ohm’s Law? And that’s not, as one of my erstwhile colleagues of yesteryear used to think, was simply when the clock got around to quarter to five in the afternoon! Although that’s perhaps the most satisfactory version!

   I’m afraid I’m floundering here, I need a spot of help to decipher some of this.

   PB   
                  
Steve W-O29/12/2010 18:18:21
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Posted by Simon Chaddock on 29/12/2010 16:11:11:
Steve W-O
Does not the battery, ESC and motor heat up? It all represents wasted battery energy.
 
Chis
As you say difficult to find a high voltage low current ESC because nobody makes them - yet!
  
 
 
 
Yes they do, one of the reasons people go for higher C batteries, not so much for the ability do supply a higher current, but because they have a lower internal resistance, so less loss and less heat.
 
Motors heat up as a result of power, not volts or amps, and efficiency, ie the lower the efficiency under certain operating conditions, the more power is dissipated as heat.
 
Power companies transfer power at higher voltages as it reduces the line losses, anjd reduces the size of conductor, and so costs of the lines. If the line was short and fat, it wouldn't make much difference.
 
Using a higher voltage in models means you can extract more power from a battery of the same C rating, this is the main reason for doing so. If you want 500W, you won't get it from a 2000MaH 11.1V 10C battery. If that was a 25C battery you would. (or could). If you used a 30C battery, it would heat up less, while still supplying the same power.
Vecchio Austriaco29/12/2010 19:01:21
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Before we think about low current high voltage ESCs we need to rewire our motors. It is the number of windings times the current which is responsible for the magnetic force. If we just change to a higher voltage it will only create higher currents as long as the impedance of the motor stays constant. 
 
So not only higher voltage but also a motor with more windings (of a smaller diameter) is necessary to have really the benefit of lower losses.
 
We shouldn't go above 60 volts - otherwise we create new problems (electric shock which can be dangerous to people starts here...)
VA  
Peter Beeney30/12/2010 14:41:30
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   Simon,
            Regarding your little statement about resistance, I think I would be tempted to say that the resistance losses, or in other words, the power expended, are proportional to the current squared times the resistance. I think the assumption that increasing the voltage reduces the losses has been made before, there was a lengthy thread about Ohm’s Law which greatly discussed the same subject quite a while back; and I’m not sure that it really cracked it, even then!

   So in your little circuit you would need to know the resistance. From Ohm’s Law, R = V/I, 10V/20A = 0.5Ω. As you say, 200W would be expended in this circuit.

   If we now apply 50V to this circuit, which you appear to have done, without making any other changes, then the current will rise to 100A. The power then expended in this circuit would then be 5,000W. If we simply double the voltage in a given circuit the power is squared. Here we’ve raised the voltage by a factor of 5, from 10 to 50, so the power has therefore risen by a factor of 25, 5 × 5. This might go some way to explaining why, when we increase the voltage on our model’s motor, sometimes the so-called ‘magic’ smoke comes out. Not so much magic, perhaps, just simply a case of getting the sums wrong.

   If we want to restrict this circuit to 200W then we must adjust the other side of the equation also, and raise the resistance to 12.5Ω. Thus the resistance losses are then exactly the same as the 0.5Ω resistor.

   Maybe to make this clearer, we can look closer at Ohm’s Law.

   P (power) = V × A……(1) But from O. L., we also know that A = V/R. Substituting for A in (1),

   V × V/R = V²/R Likewise, of course, A = I × R, thus I² × R.

   So, we now have three ways of calculating power, V × A, V²/R and I² × R.

   If we apply any of these to the two little circuits, then the answer is 200 watts.

   This is just the theory, there may be some practical advantages to raising the voltage; as with everything else, it will be a compromise. However, I’m not convinced that reducing the losses is one such advantage of raising the voltage, but as always, I stand to be corrected.

   Hope this is of some interest, but it will probably result in activating most people’s off switch!

   PB
Chris Bott - Moderator30/12/2010 15:18:46
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Posted by Peter Beeney on 30/12/2010 14:41:30:

    
   If we want to restrict this circuit to 200W then we must adjust the other side of the equation also, and raise the resistance to 12.5Ω. Thus the resistance losses are then exactly the same as the 0.5Ω resistor.

  
But we wouldn't include resistance to reduce the power, we'd use a different (low Kv) motor and prop it accordingly. 
Hamish30/12/2010 15:32:49
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Steve, can you explain this  please
 
"Motors heat up as a result of power, not volts or amps, and efficiency, ie the lower the efficiency under certain operating conditions, the more power is dissipated as heat."

Edited By Hamish McNab on 30/12/2010 15:34:10

Steve W-O30/12/2010 15:42:51
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Posted by Hamish McNab on 30/12/2010 15:32:49:
Steve, can you explain this  please
 
"Motors heat up as a result of power, not volts or amps, and efficiency, ie the lower the efficiency under certain operating conditions, the more power is dissipated as heat."

Edited By Hamish McNab on 30/12/2010 15:34:10

 
 
You need to take into account the voltage and the current, not just the voltage or current alone.
 
Motors only convert a portion of the power applied to them into turning the load, this portion is affected by motor design, power applied, load aplied, and even the temperature of the motor. The portion that is not used to turn the load is dissipated as heat.
A motor may operate more efficeintly at a particular throttle setting, depending on load etc in the actual application.
Peter Beeney30/12/2010 16:23:50
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   Chris,
           Exactly so, I was just trying to keep it simple. But the impedance of the motor would still be 12.5 ohms. In this particular circuit anyway. The point I was trying to make was that to maintain equilibrium, if we make changes on one side of an equation, we have to do the same on the other.
   One very small point when using a high voltage installation might be concerning short circuits. With a supply of 50 volts, and a short-circuit fault impedance of 0.1 ohm, that could result in an instantaneous fault current of 500 amps. That could cause a fire, so I reckon that any one that does use high voltage set-ups has a fairly robust procedure for insuring that the wrong two wires do not get attached together!

   Or, like the gas turbine pilots, a fire extinguisher to hand……….

   PB
              
Brian Parker30/12/2010 16:30:21
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Heat is the enemy.
 Ohms law is dependent on constant temperature. The Current flowing is proportional to the source/applied voltage and inversely proportional to resistance.
 Resistance changes with temperature. Under load the battery, motor, ESC and wiring will all be at varying temperatures.
 Maximum power transfer (efficiency) will only occur when the load resistance is equal to the source (battery) resistance. Source resistance limits the total current and will drop the voltage at the battery terminals.
 Load voltage across battery terminals = Source voltage - Current drawn x source resistance.
birdy30/12/2010 16:42:07
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My 2ps worth is that one of the big problems is the battery weight; each cell needs to be packaged individually, and with a high number of cells you'll end up with the the pros being outweighed by the negatives. 
Tony Smith 730/12/2010 16:45:52
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Posted by Chris Bott on 30/12/2010 15:18:46:
 
But we wouldn't include resistance to reduce the power, we'd use a different (low Kv) motor and prop it accordingly. 

Isn't this the critical factor?  I mean even at normal voltages (say for 3S) if you divided the voltage by the DC resistance of the motor you'd burn it up in an instant.   Its the back emf generated by the motor running that makes it all work together, so it can't really be assessed by just looking at static resistance figures.

Steve W-O30/12/2010 16:48:07
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Posted by Brian Parker on 30/12/2010 16:30:21:
Heat is the enemy.
 Ohms law is dependent on constant temperature.
 
 
Is it?
 
Ohms law remains the same regardless of temperature.
 
Resistance may change with temperature, but put that changed temperature into the equation, and Ohms law remains the same.
 
You will get different results when changing the value of E, I or R, whatever the cause for the change. But the law remains the same.
Steve W-O30/12/2010 16:52:05
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Posted by Tony Smith 7 on 30/12/2010 16:45:52:
Posted by Chris Bott on 30/12/2010 15:18:46:
 
But we wouldn't include resistance to reduce the power, we'd use a different (low Kv) motor and prop it accordingly. 

Isn't this the critical factor?  I mean even at normal voltages (say for 3S) if you divided the voltage by the DC resistance of the motor you'd burn it up in an instant.   Its the back emf generated by the motor running that makes it all work together, so it can't really be assessed by just looking at static resistance figures.


 
 
 
If you look at just the DC resistance of the motor, maybe, I don't know about back emf making it work together, rather that most of the power applied goes into turning the motor, not burning out the wires.
 
 
 
Chris Bott - Moderator30/12/2010 16:56:08
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For 200W our choice is to buy components for low volts high current or high volts low current. For a comparison, we'd need to know the characteristics of both.
 
Another way to look at it would be to use 2 lipos both suitable for 100W setups. We could use them in series or parallel. As long as we could source suitable motors and ESC's for our chosen method.
 
With the high current setup, we would at the very least need thicker wire and better (bulkier) connectors. 
 

 
Steve W-O30/12/2010 17:11:11
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Posted by Chris Bott on 30/12/2010 16:56:08:
 
With the high current setup, we would at the very least need thicker wire and better (bulkier) connectors. 
 

 
 
 
This is closer to the realities, and is probably the most relevant comment.
 
The difference this would make in practice is a fraction of an ounce, probably much less than using a bigger ESC than needed, as many of us do (certainly me)
Brian Parker30/12/2010 18:20:47
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Steve,
The Ohms law calculation is only true for a constant temperature. Change the temperature and you need to re-calculate due to the change in resistance. The formula is simply an algebraic form of the law and assumes a constant temperature.

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