COP 17 Heater | Rosemary Ainslie

[witsend]

You take the voltage across the resistor / resistor Ohm to get the Amp and you multiply by the battery voltage seriously ? you kidding us right ?

you said 1.2V across the resistor , so 1.2/0.25 = 4.8A and you multiply the 4.8A by the batterie voltage is just a non sense, im sorry but i cant beleive you wrote that. Its Voltage across resistor MUTIPLIED BY the Amp (Vr/Ohm) = The number of watt.

EgmQC - I don't know what to say. How would you establish the energy delivered by the battery? Can you please explain your method perhaps?

[witsend]

Rosemary,

You keep referring to power as energy which is not the case. Energy is power * time (in seconds). The instantaneous energy levels in the re-charge are very small indeed compared with the total energy supplied to invoke the recharge on each switching cycle. This is what you will find out if you interpret the results of your future testing correctly.

Who will be writing the testing method and procedure and driving the storage scope now you have been loaned one?

Hoppy

I read through my post and see no inappropriate use of the word energy or power. If you've a problem kindly give me the reference.

[Hoppy]

EgmQC - I don't know what to say. How would you establish the energy delivered by the battery? Can you please explain your method perhaps?

Rosemary,

With respect, you really do need to properly understand the basics of EE before you launch into preparing valid test routines for your circuit.

Hoppy

[EgmQC]

EgmQC - I don't know what to say. How would you establish the energy delivered by the battery? Can you please explain your method perhaps?

I dont think i need , everybody here should know that already, Anyway , its exactly what i wrote in my previous post:

Voltage Across resistor = V
Voltage Across resistor / Ohm value = I
V*I = W

But we are talking about Watt here , not the energy, to get the energy number since its a pulsed circuit you need to take ALOT of sample at the shunt for 1 second , at the end, average the reading Sum(Vr*(Vr/Ohm)) and then / by Number of sample , that will give you the real energy used for 1 sec, so watt/sec or joule/sec, multiply by the time the battery got used and you will know exactly how much energy you used,. Ex: if the average for 1 sec is 0.25W , for 1 hour that will be (0.25*60)*60 = 900W used and that equale too 0.25W/Hour since the average equale a constant load extracted from the pulsed circuit.

[poynt99]

You take the voltage across the resistor / resistor Ohm to get the Amp and you multiply by the battery voltage seriously ? you kidding us right ?

you said 1.2V across the resistor , so 1.2/0.25 = 4.8A and you multiply the 4.8A by the batterie voltage is just a non sense, im sorry but i cant beleive you wrote that. Its Voltage across resistor MUTIPLIED BY the Amp (Vr/Ohm) = The number of watt.

Instantaneous power delivered by the battery is the instantaneous current from the battery times the battery voltage. The battery current is determined by the instantaneous shunt voltage and resistance as she stated.

If this is what Rosemary is referring to, then her calculation method is correct.

EgmQC you are talking about the instantaneous power in the shunt resistor, not the battery.

.99

[Hoppy]

I read through my post and see no inappropriate use of the word energy or power. If you've a problem kindly give me the reference.

What I mean but wrote badly is that IMO you appear to be confusing instantaneous power with energy in terms of charging the battery.

Hoppy

[witsend]

Thank you Poynt. That was getting tedious.

[Hoppy]

Instantaneous power delivered by the battery is the instantaneous current from the battery times the battery voltage. The battery current is determined by the instantaneous shunt voltage and resistance as she stated.

If this is what Rosemary is referring to, then her calculation method is correct.

EgmQC you are talking about the instantaneous power in the shunt resistor, not the battery.

.99

This was the original statement that Rosemary made which started the ball rolling and from there things appear to have got confused: -

TK - I'm hoping I can bend your mind around this problem.

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value). Then how do you work out the product of the energy available at that moment? In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

Hoppy

[poynt99]

This was the original statement that Rosemary made which started the ball rolling and from there things appear to have got confused: -

TK - I'm hoping I can bend your mind around this problem.

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value). Then how do you work out the product of the energy available at that moment? In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

Hoppy

I think it's safe to say that Rosemary is interchanging the terms "energy" and "power", but it is evident she is talking about "instantaneous power".

Having said that and accepted/converted the terminology in our heads, I see no problems in what she is saying. I think she is simply asking: what is the instantaneous power going back into the battery at that one particular IK spike.

.99

EDIT:

Of course her calculation as cited above does not indicate what power is going back into the battery. The 240W of instantaneous power cited is only what is dissipated in the shunt resistor at that instant. If we can assume (I wouldn't necessarily) that the shunt current at this time is also the same current in the coil, then the instantaneous power going back into the battery would be the shunt instantaneous current times the battery voltage, in this case -115.2W. Note to Rosemary: Don't take this number and go running to the bank with it just yet. Until this power number has an associated time attached to it, it is a meaningless value in terms of what effect this power has over an extended time period, or at least in comparison to all the other "powers" going in and out of the battery in one complete cycle. 50W for 1us is a much smaller power than 50W for 1ms over the same time period. The -115.2W spike I cited is most likely many magnitudes shorter in time compared to the 15.4us pulse of the forward input power from the battery.

In fact, if we do a very simple analysis, we can come up with an easy way to determine how long the kickback pulse would have to be in order to achieve total recharge of the battery--that is for an equal amount of energy going back in as is coming out.

I can tell you that for each forward pulse into your particular wire-wound resistor, about 662uJ of energy is going into it. That is 43W instantaneous power (measured) times 15.4us in time. In the case cited where the kickback spike was 115.2W back into the battery, this reverse power pulse would have to be 5.75us in length. That is -115.2W instantaneous power times 5.75us = -662uJ. Equal energy going back in as coming out.

[EgmQC]

Instantaneous power delivered by the battery is the instantaneous current from the battery times the battery voltage. The battery current is determined by the instantaneous shunt voltage and resistance as she stated.

If this is what Rosemary is referring to, then her calculation method is correct.

EgmQC you are talking about the instantaneous power in the shunt resistor, not the battery.

.99

Your right , im not talking about the same thing as Rosemary , i got a bit confuse realy sorry about that Rosemary, The point i wanted to make is about her mesurement at the RL resistor, Like Hoppy said, if you see 50V at RL, you dont multiply it by the amp the circuit use to extract the power, because there no energy coming at that instant from the battery , its like if you charge a capacitor and you disconnect it from the source and you step up the voltage , you will get the voltage you want but the Amp will not be at the same amplitude becasue its 2 source.

Her mesurement is right if the source is at 50v and connected but at 24v and disconnected its realy wrong.

i include the post i was refering to:

TK - I'm hoping I can bend your mind around this problem.

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value) . Then how do you work out the product of the energy available at that moment? In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

Now. If you do not do the integration simultaneously how is this advantage or gain made evident? And more alarmingly, if you simply do the product of both values and add it to the general loss - then your methodology is wrong.

If, as you say the LeCroy is doing the math - then I'm afraid I need to see it. I'm still concerned that it is not capable of doing that DC offset. To my primitive way of thinking this means that it cannot gauge zero in order to make a comparison as to where the waveform was found in relation to zero.

Please explain this. Sorry to impose.

You are right. Its's the best I can do to explain my concern. I use primative example and analogy. But it serves its purpose.

[Hoppy]

I think it's safe to say that Rosemary is interchanging the terms "energy" and "power", but it is evident she is talking about "instantaneous power".

Having said that and accepted/converted the terminology in our heads, I see no problems in what she is saying. I think she is simply asking: what is the instantaneous power going back into the battery at that one particular IK spike.

.99

Yes, I hope Rosemary can eventually assimilate this in respect to the real energy in that spike to get an idea of proportion between energy into the system and energy returned to the battery.

Hoppy

[Hoppy]

I think it's safe to say that Rosemary is interchanging the terms "energy" and "power", but it is evident she is talking about "instantaneous power".

Having said that and accepted/converted the terminology in our heads, I see no problems in what she is saying. I think she is simply asking: what is the instantaneous power going back into the battery at that one particular IK spike.

.99

EDIT:

Of course her calculation as cited above does not indicate what power (if any) is going back into the battery. The 240W of instantaneous power cited is only what is dissipated in the shunt resistor at that instant. If we can assume (I wouldn't necessarily) that the shunt current at this time is also the same current in the coil, then the instantaneous power going back into the battery would be the shunt instantaneous current times the battery voltage, in this case -57.6W.

This is why this statement from Rosemary was contested because she is talking about power back to the battery and 240W instantaneous would result in a mighty inacurate COP!

Hoppy

[poynt99]

This is why this statement from Rosemary was contested because she is talking about power back to the battery and 240W instantaneous would result in a mighty inacurate COP!

Hoppy

Indeed she overlooked that. But again it depends totally on Joules of energy. She is only high by a factor of about 2 times. i.e 240W vs. 115W instantaneous power, but the pulse width was not considered by her.

The major oversight is Joules coming out compared to Joules going back in. I have edited my post with more details illustrating with her example, the reverse pulse would have to be about 5.75us long for total battery recharge.

.99

[Hoppy]

Indeed she overlooked that. But again it depends totally on Joules of energy. She is only high by a factor of about 2 times. i.e 240W vs. 115W instantaneous power, but the pulse width was not considered by her.

The major oversight is Joules coming out compared to Joules going back in. I have edited my post with more details illustrating with her example, the reverse pulse would have to be about 5.75us long just for COP =1.

.99

I have been trying to emphasise to Rosemary, the importance of thinking in terms of energy when looking at COP. This system will likely run out at about COP 0.6 in electrical terms. I have found that most inductive flyback charging systems that I've built and tetsed come out in this order of COP.

Hoppy

[poynt99]

I have been trying to emphasise to Rosemary, the importance of thinking in terms of energy when looking at COP. This system will likely run out at about COP 0.6 in electrical terms. I have found that most inductive flyback charging systems that I've built and tetsed come out in this order of COP.

Hoppy

Indeed Hoppy.

.99 over and out.