Rosemary Ainslie Quantum Magazine Circuit COP > 17 Claims

[TinselKoala]

Exactly. And suppose you use the same battery and the same "element resistor" value, and you compare the Ainslie pulsed circuit, making the load rise to, say, 20 C over ambient.... to the straight DC with potentiometer circuit making the same temperature rise _in the load_.  What do you suppose you will find? Which battery will run down sooner?

Let's say you have a 10 ohm element resistor and a 24 volt power supply and you are using a 10 percent HI duty cycle. The mosfet circuit has a total On state resistance of, say, 2 ohms, so you have 12 ohms total resistance. The average current in the load is going to be I = (V/R) x Duty Cycle or (24/12)x 0.10 = 200 mA and this will produce a known power dissipation in the load of I²R or 0.2 x 0.2 x 10 = 0.4 Watt. The circuit itself with a resistance of 2 ohms will dissipate  0.2x0.2x2 = 0.08 Watt and the battery will be delivering 0.48 Watt average throughout the trial.

For the DC "control" you have the same element resistor and the same 24 volt power supply and you want the same 200 mA average current in the load for the same 0.4 W average power dissipation in the load.... so you need a _total_ resistance  R = V/I = 24/0.2 = 120 Ohms. Ten of these will of course be in the load element resistor in the oil. The other 110 will be in the potentiometer.... and this potentiometer will be dissipating I²R = 0.2 x 0.2 x 110 = 4.4 Watts! The total power drawn from the battery will be 24 V x 200 mA = 4.8 Watts! Which battery will run out sooner, and why? Ainslie magic... or Ainslie misdirection?

ETA: MarkE has pointed out that I should have used RMS current here, so I've corrected the calculation below.

[MarkE]

Exactly. And suppose you use the same battery and the same "element resistor" value, and you compare the Ainslie pulsed circuit, making the load rise to, say, 20 C over ambient.... to the straight DC with potentiometer circuit making the same temperature rise _in the load_.  What do you suppose you will find? Which battery will run down sooner?

Let's say you have a 10 ohm element resistor and a 24 volt power supply and you are using a 10 percent HI duty cycle. The mosfet circuit has a total On state resistance of, say, 2 ohms, so you have 12 ohms total resistance. The average current in the load is going to be I = (V/R) x Duty Cycle or (24/12)x 0.10 = 200 mA and this will produce a known power dissipation in the load of I²R or 0.2 x 0.2 x 10 = 0.4 Watt. The circuit itself with a resistance of 2 ohms will dissipate  0.2x0.2x2 = 0.08 Watt and the battery will be delivering 0.48 Watt average throughout the trial.

For the DC "control" you have the same element resistor and the same 24 volt power supply and you want the same 200 mA average current in the load for the same 0.4 W average power dissipation in the load.... so you need a _total_ resistance  R = V/I = 24/0.2 = 120 Ohms. Ten of these will of course be in the load element resistor in the oil. The other 110 will be in the potentiometer.... and this potentiometer will be dissipating I²R = 0.2 x 0.2 x 110 = 4.4 Watts! The total power drawn from the battery will be 24 V x 200 mA = 4.8 Watts! Which battery will run out sooner, and why? Ainslie magic... or Ainslie misdirection?

Not quite but the measurement problem you point out is there.  We need to match rms currents for heating, not average currents.
24V supply
12 Ohm heater + MOSFET
2A peak
3.7% duty cycle
I_rms = 2A*(0.037)^0.5 = 385mA_rms  (632mA_rms @ 10% duty cycle)
P_HEATER = 385mA²*10 Ohms = 1.48W  (4W @ 10% duty cycle)

Now, to get 1.48W in the heater using a pot, we need that same 385mA as a steady state current.  From a 24V power supply:

R_TOTAL = 24V/385mA =  62.4 Ohms, of which 52.4 Ohms will end up in the potentiometer.  Since the current is the same in the potentiometer as in the heater resistor, the heater resistor will only put out 10/62.4W/W of the battery drain, yielding an efficiency of 16%.  84% of the power from the battery:  7.8W will be dissipated in the potentiometer.  So even with a relatively poor switching circuit, the switched heater circuit will take less power from the battery to obtain the same heater output.  This is why when we care about efficiency we use switching regulators instead of linear regulators.

If we had used the 10% duty cycle then things would not have been quite so bad percentage wise but worse power wise:

24V/632mArms = 38 Ohms total, and the loss is 28/38 Ohms or 74% instead of 84% but the power loss increases to 11.2W.

This is over and above any battery effect that they get by pulsing their load.  From what little we know of the experiment, it appears ill conceived and incapable of producing useful data.  They need someone competent to help them figure out a valid experiment protocol.  After her Tasmanian devil routine, we know that Ms. Ainslie can no longer ask Steve for advice.  Maybe Poynt99 is willing to try and straighten them out.

[Tseak]

The experiment as described so far is deeply flawed. 

Actually not. The experiment is essentially not described. Having said that I have no idea how they will keep the heat dissipation of both elements the same without other extraneous heat losses. I presume that they are planning to use an external reostat in which case the results will be meaningless. Lets wait and see. With luck it should be done by the end of the year.

[TinselKoala]

Not quite but the measurement problem you point out is there.  We need to match rms currents for heating, not average currents.
(snip)

Yes, that's right, I stand corrected, thanks. Where I was multiplying the peak current by the duty cycle, to get average current, I should have been multiplying by the square root of the duty cycle to obtain RMS current.
This would have given me the RMS current for the power dissipation. The chain of reasoning, however, as you have shown, is basically right: using a potentiometer to control the power in the load resistor simply moves a large part of the battery's supplied power out of the load; the total power drawn from the battery is mostly dissipated outside the load and won't be counted if load heating is the criterion.

So, correcting my first try:

Let's say you have a 10 ohm element resistor and a 24 volt power supply and you are using a 10 percent HI duty cycle. The mosfet circuit has a total On state resistance of, say, 2 ohms, so you have 12 ohms total resistance. The RMS current in the load is going to be I = (V/R) x (Duty Cycle)^0.5 or (24/12)x (0.10)^0.5 = 632 mA and this will produce a known power dissipation in the load of I_rms²R or 0.632 x 0.632 x 10 = 4 Watts. Alternatively, P_ave = (I_peak²R) x dutycycle, or ((2 x 2) x 10) x 0.10 = 4 Watts. The circuit itself with a resistance of 2 ohms will dissipate  0.632 x 0.632 x 2 = 0.8 Watt and the battery will be delivering 4.8 Watts average throughout the trial. ( P = I_rms x V_rms, or 0.623 x (24 x (0.10)^0.5) = 4.8, or P = (V_peak²/R) x (duty cycle) = (576/12) x 0.10 = 4.8 , or even V_rms²/R = 7.59²/12 = 4.8.  )

For the DC "control" you have the same element resistor and the same 24 volt power supply and you want the same 632 mA RMS current in the load for the same 4 W average power dissipation in the load.... so you need a _total_ resistance  R = V/I_rms = 24/0.632 = 38 Ohms. Ten of these will of course be in the load element resistor in the oil. The other 28 will be in the potentiometer (and the mosfet) .... and this potentiometer and mosfet will be dissipating I_rms²R = 0.632 x 0.632 x 28 = 11.2 Watts! The average power drawn from the battery will be 24 V x 632 mA = around 15 Watts! Which battery will run out sooner, and why?

(Did I do it right this time?      )

[MarkE]

Yes, that's right, I stand corrected, thanks. Where I was multiplying the peak current by the duty cycle, to get average current, I should have been multiplying by the square root of the duty cycle to obtain RMS current.
This would have given me the RMS current for the power dissipation. The chain of reasoning, however, as you have shown, is basically right: using a potentiometer to control the power in the load resistor simply moves a large part of the battery's supplied power out of the load; the total power drawn from the battery is mostly dissipated outside the load and won't be counted if load heating is the criterion.

So, correcting my first try:

Let's say you have a 10 ohm element resistor and a 24 volt power supply and you are using a 10 percent HI duty cycle. The mosfet circuit has a total On state resistance of, say, 2 ohms, so you have 12 ohms total resistance. The RMS current in the load is going to be I = (V/R) x (Duty Cycle)^0.5 or (24/12)x (0.10)^0.5 = 632 mA and this will produce a known power dissipation in the load of I_rms²R or 0.632 x 0.632 x 10 = 4 Watts. Alternatively, P_ave = (I_peak²R) x dutycycle, or ((2 x 2) x 10) x 0.10 = 4 Watts. The circuit itself with a resistance of 2 ohms will dissipate  0.632 x 0.632 x 2 = 0.8 Watt and the battery will be delivering 4.8 Watts (24V*2A*10% DC) average throughout the trial. ( P = I_rms x V_rms, or 0.623 x (24 x (0.10)^0.5) = 4.8, or P = (V_peak²/R) x (duty cycle) = (576/12) x 0.10 = 4.8 , or even V_rms²/R = 7.59²/12 = 4.8.  )

For the DC "control" you have the same element resistor and the same 24 volt power supply and you want the same 632 mA RMS current in the load for the same 4 W average power dissipation in the load.... so you need a _total_ resistance  R = V/I_rms = 24/0.632 = 38 Ohms. Ten of these will of course be in the load element resistor in the oil. The other 28 will be in the potentiometer (and the mosfet) .... and this potentiometer and mosfet will be dissipating I_rms²R = 0.632 x 0.632 x 28 = 11.2 Watts! The average power drawn from the battery will be 24 V x 632 mA = around 15 Watts! Which battery will run out sooner, and why?

(Did I do it right this time?      )

Yes, you got it right.  It is called the difference between a PWM switching regulator in the first case and a linear regulator in the second case.