ESC Current Output Testing

[Xx Losungen xX]

Okay I'm done after this, it's getting silly.

Originally Posted by Alka

No, I don't think he is testing efficiency or even trying to. It looks like he is testing weather the BEC can handle the full load of a servo of a powerful servo without limiting current.

So testing which bec can handle a servo falling within its specs more efficiently.

Testing for efficiency, check.

Originally Posted by Alka

The current output can be fixed to a certain level by design and very simple circuitry.

Of course they exist, and they are simple. But now you're arguing over garbage. Didn't say they didn't exist nor that they weren't simple.

You can't limit voltage and current without dropping voltage, please find me one circuit that can, I'd love to find one that defies all known physics, I'd be a billionaire.

Originally Posted by Alka

To get efficiency we would need to know watts at the source vs watts used by the servo. The difference would be watts wasted BY the bec.

We would get watts with the VOLTAGE DROP, we already have the CURRENT.


Quit bloody arguing, this is merely just trolling at this point mate, enough.

[Alka]

I am not trolling here. I am respectfully giving my opinion.

You kept asking questions...

http://www.ti.com/lit/an/snva829/snva829.pdf
(this is a buck converter , so the output voltage will be lower than input if this is what you mean by dropping voltage.. however the voltage OUT from the device is fixed as is the current limit).

I believe I said that it appears the guy in the video is not testing efficiency at all.. and still stand by that

[Growler]

Originally Posted by Xx Losungen xX

We would get watts with the VOLTAGE DROP, we already have the CURRENT.

You do not have current measured at the source though. So you would still miss data for calculating efficiency.

[Xx Losungen xX]

Originally Posted by Growler

You do not have current measured at the source though. So you would still miss data for calculating efficiency.

Not for comparing efficiency, which is the scope of the video and the discussion.

That would certainly tell us what the vampire state of the entire esc is, but he is not testing the esc, he is comparing the efficiency of the BEC function across devices.

Set becs for same voltage.

Measure E at rest.

Measure E and I at FLA.

Confirm the load is doing the work it wants to be doing (watts).

Watts should be the same across all tested circuits, if not one has failed, if the load hasn't failed, the circuit has, it's out.

If watts are correct the one with the lowest voltage drop is the most efficient, it's a really simple test.

If we compared anything to EMF, it stands to reason that the castle stand alone bec would win, zero vampire state. Not what we want to test for.

[Xx Losungen xX]

Originally Posted by Alka

this is a buck converter , so the output voltage will be lower than input if this is what you mean by dropping voltage.. however the voltage OUT from the device is fixed as is the current limit)

Yes, it is.

It's not just what I mean, it's what is defined as voltage drop. It's simply what it sounds like, a drop in potential voltage between two points.

If I have two circuits, both with an EMF of 12v.

A load rated to draw 4amps at FLA with 6volts applied.

Both becs are set for 6volts and rated for a 5amp load.

Bec1 observes an FLA of 5amps.

Bec2 observes an FLA of 4amps.

Result for Bec1 - Bec is supplying less then 6 volts, confirm by observing voltage drop.

Result for Bec2 - Bec IS supplying 6 volts, confirm again.

Conclusion, Bec2 is dropping 6 volts and Bec1 is dropping MORE then 6 volts.

Which bec is more efficient for our load?

Also, you cannot limit voltage AND current without voltage drop, again, it goes against physics.

The white paper you posted states that in its data itself.

[Alka]

Originally Posted by Xx Losungen xX

Not for comparing efficiency, which is the scope of the video and the discussion.

That would certainly tell us what the vampire state of the entire esc is, but he is not testing the esc, he is comparing the efficiency of the BEC function across devices.

Set becs for same voltage.

Measure E at rest.

Measure E and I at FLA.

Confirm the load is doing the work it wants to be doing (watts).

Watts should be the same across all tested circuits, if not one has failed, if the load hasn't failed, the circuit has, it's out.

If watts are correct the one with the lowest voltage drop is the most efficient, it's a really simple test.

If we compared anything to EMF, it stands to reason that the castle stand alone bec would win, zero vampire state. Not what we want to test for.

please.. show us with some simple numbers then!!

[Alka]

Originally Posted by Xx Losungen xX

Yes, it is.

It's not just what I mean, it's what is defined as voltage drop. It's simply what it sounds like, a drop in potential voltage between two points.

If I have two circuits, both with an EMF of 12v.

A load rated to draw 4amps at FLA with 6volts applied.

Both becs are set for 6volts and rated for a 5amp load.

Bec1 observes an FLA of 5amps.

Bec2 observes an FLA of 4amps.

Result for Bec1 - Bec is supplying less then 6 volts, confirm by observing voltage drop.

Result for Bec2 - Bec IS supplying 6 volts, confirm again.

And why is BEC one supplying less than 6v when its under its rated current?
Also how are you measuring that BEC 1 is using 5 amps at full load unless you measure the current before the bec.

Still can't calculate efficiency without also measuring the current at the 12v source in your example.

bec 1 input: 12v, output 6v @ 5 amps;
bec 2 input 12v output 6v@ 5 amps;

If bec one is drawing 3 amps at 12 volts and bec2 is drawing 3.5 amps. Then you can tell that one is more efficient than the other. And the exact efficiency of the BEC at that load . 30watts / 36 watts = 83 percent efficient.. BEC 2 is 71 percent efficient (30/42)

[Growler]

Originally Posted by Xx Losungen xX

Yes, it is.

It's not just what I mean, it's what is defined as voltage drop. It's simply what it sounds like, a drop in potential voltage between two points.

If I have two circuits, both with an EMF of 12v.

A load rated to draw 4amps at FLA with 6volts applied.

Both becs are set for 6volts and rated for a 5amp load.

Bec1 observes an FLA of 5amps.

Bec2 observes an FLA of 4amps.

Result for Bec1 - Bec is supplying less then 6 volts, confirm by observing voltage drop.

Result for Bec2 - Bec IS supplying 6 volts, confirm again.

Conclusion, Bec2 is dropping 6 volts and Bec1 is dropping MORE then 6 volts.

Which bec is more efficient for our load?

Also, you cannot limit voltage AND current without voltage drop, again, it goes against physics.

The white paper you posted states that in its data itself.

This is exactly where you are wrong. If servo FLA at 6V is 4A then 5A reading means BEC supplies more than 6v, not less. So BEC 1 would drop less volts from 12V. R=6V : 4A =1,5 ohm. So U= I x R = 5A x 1,5 ohm = 7,5V as BEC 1 voltage.

[Alka]

Originally Posted by Xx Losungen xX

Yes, it is.

Both becs are set for 6volts and rated for a 5amp load.

This is exactly where you are wrong. If servo FLA at 6V is 4A then 5A reading means BEC supplies more than 6v, not less. So BEC 1 would drop less volts from 12V. R=6V : 4A =1,5 ohm. So U= I x R = 5A x 1,5 ohm = 7,5V as BEC 1 voltage.

Wait a minute here bud YOU said it was set for 6 volts !


So you have a bec that is set for 6 volts and put 4 amp load on it . Now it is putting out 7.5 volts. I would send the BEC back! And where do you measure this current, the 5 amps? not after the BEC because it will only show you the servo load.. If you measure before the BEC well then you are measuring the current at the 12v level.

[Xx Losungen xX]

Originally Posted by Growler

This is exactly where you are wrong. If servo FLA at 6V is 4A then 5A reading means BEC supplies more than 6v, not less. So BEC 1 would drop less volts from 12V. R=6V : 4A =1,5 ohm. So U= I x R = 5A x 1,5 ohm = 7,5V as BEC 1 voltage.

Lol no man, it's exactly where you are wrong.

Our load is designed to draw 4A at 6V, meaning it WILL draw 4A at 6V, can also state that it WILL want to do 24W of work.


Bec1 sees a load of 5A.

Result = 24W/5A = 4.8 volts

Bec2 sees a load of 4A.

Result = 24W/4A = 6 volts

This is not a purely resistive circuit, we have inductance and capacitance in this circuit as well, which all change based on voltage and current. We need another constant to use, such as the ability to do work (power).

Solving for R as you did, will not work, as you demonstrated.

[Xx Losungen xX]

Originally Posted by Alka

Wait a minute here bud YOU said it was set for 6 volts !


So you have a bec that is set for 6 volts and put 4 amp load on it . Now it is putting out 7.5 volts. I would send the BEC back! And where do you measure this current, the 5 amps? not after the BEC because it will only show you the servo load.. If you measure before the BEC well then you are measuring the current at the 12v level.

Actually we put a load on it rated to draw 4amps when supplied with 6 volts. But we know it's drawing 5amps, therefore its only getting 4.8 volts.

This isn't a resistor, it's more of a hair dryer then it is a toaster.

Can't use a purely resistive equation to solve for a circuit that is not purely resistive.

You two are confusing resistance, inductance and capacitance.

Yes, I'd send it back, the Bec is defective, set incorrectly or just really shitty to be blunt.

OR

The testing methodology is completely inaccurate.

AND

So is the conclusion that higher amps means a better bec.

Are you understanding now?

[Growler]

Originally Posted by Xx Losungen xX

Lol no man, it's exactly where you are wrong.

Our load is designed to draw 4A at 6V, meaning it WILL draw 4A at 6V, can also state that it WILL want to do 24W of work.


Bec1 sees a load of 5A.

Result = 24W/5A = 4.8 volts

Bec2 sees a load of 4A.

Result = 24W/4A = 6 volts

This is not a purely resistive circuit, we have inductance and capacitance in this circuit as well, which all change based on voltage and current. We need another constant to use, such as the ability to do work (power).

Solving for R as you did, will not work, as you demonstrated.

If you look at power formula that is P=UxI then people say that power equals amps times volts to find a result of that formula, however that is not what power is by definition, the formula shows how you can calculate it for DC circuit when you measure voltage and current. Meaning it is not a constant.

We are dealing with a low voltage DC circuit that has a stalled servo motor so the effects of inductance and capacitance are probably non existent for our purpose. Inductance and capacitance play much bigger role in AC circuits.

Resistance is the only thing we could call a constant here as it is a property of a conductor related to the type of material, dimensions and temperature etc. If you look at the Ohm`s law and say that the resistance R of an object is defined as the ratio of voltage U across it to current I through it then this is basically wrong as it is not the definition of resistance, that is how you can calculate it when you measure voltage and current.

[Growler]

Originally Posted by Xx Losungen xX

Actually we put a load on it rated to draw 4amps when supplied with 6 volts. But we know it's drawing 5amps, therefore its only getting 4.8 volts.

This isn't a resistor, it's more of a hair dryer then it is a toaster.

Can't use a purely resistive equation to solve for a circuit that is not purely resistive.

You two are confusing resistance, inductance and capacitance.

Yes, I'd send it back, the Bec is defective, set incorrectly or just really shitty to be blunt.

OR

The testing methodology is completely inaccurate.

AND

So is the conclusion that higher amps means a better bec.

Are you understanding now?

If a load draws 4A at 6V and you measure 5A then voltage is higher than 6V.

This is more like a stalled hair dryer thus more like a resistor.

Inductance and capacitance do not play a big role here because we are dealing with a DC current with flow in one direction. Even with a switching DC circuit you might get resistance when switching on and off, but it only opposes current flow when switching on and helps current flow when switching off so keeps flow in one direction. AC switches polarity so capacitance and inductance play a bigger role.

Testing methodology is simplified as it shows how much amps a servo pulls at whatever voltage BEC can actually hold. Using HobbyWing 1080 for that servo would make me nervous in this case as it is set to 7,4V but can deliver only 4,5A that is not much different from Axial AE-5 at 6V (4,4A). So this indicates that HW 1080 is over loaded as output voltage drops from 7,4V closer to 6V that AE-5 puts out. AE-5 BEC is rated to 3A so that is overloaded as well.

Higher amps do not necessarily mean a better BEC because it does not take into account how much it is over it`s rated power. Higher amps could mean you burn your BEC faster but not necessarily as you might not know how much safety margin you have at rated amps.

[Alka]

You hit the nail on the head growler.

At motor at stall just looks primarily like a resistor. If the BEC is within spec it will keep it's voltage. If more current is demanded than the BEC can put out.. well then a number of things can happen from chopping power to cutting out completely.

Efficiency can only be determined by knowing the loss in the BEC which you can only calculate if you measure the power drawn at the source and subtracting the power used by the final load. This is the same even if the load is not just resistive.