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1. Rob Woudenberg
A BiTTof Flux tuning simulation
Results based on an AC model based on
M45 laminated steel

2. Rob Woudenberg
Flux tuning
04/12/2010Confidential
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 Basic FEMM model used is 5.5 x 5.5 x 3.3
cm:

3. Rob Woudenberg
Applied Input / output signals
 FEMM input parameters can only be current
 Primary current phase (red curve)is fixed
 Secondary current phase (blue curve) is changed from 0 to
359 degrees in steps of one degree to find unique situations
 Applied current values: 0.2 – 1.2 Amps
04/12/2010Confidential
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4. Rob Woudenberg
Cop vs phase findings (1)
04/12/2010Confidential
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0
1000
2000
3000
4000
5000
6000
184
188
192
196
200
204
208
212
216
220
224
228
232
236
240
244
248
252
256
260
264
268
272
276
280
284
288
292
296
300
304
308
312
316
320
324
328
332
336
340
344
348
352
356
PrimarytotalWatt
Primarypower
0.2
0.4
0.6
0.8
1
1.05
1.075
1.1
1.2

5. Rob Woudenberg
Cop vs phase (2)
04/12/2010Confidential
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 At low currents two COP peaks occur over the 180 – 359
range
 The lower the current the wider the COP peaks are apart
 At higher currents the two COP peaks merge at around 270
degrees
 Further fine tuning on current value will give one COP >>
100000 at 270 degrees @ I=1.075 A in this simulation≈
0
5 0
1 0 0
1 5 0
2 0 0
2 5 0
180
184
188
192
196
200
204
208
212
216
220
224
228
232
236
240
244
248
252
256
260
264
268
272
276
280
284
288
292
296
300
304
308
312
316
320
324
328
332
336
340
344
348
352
356
360
0 .2
0 .4
0 .6
0 .8
1
1 .0 5
1 .0 7 5
1 .1
1 .2

6. Rob Woudenberg
Power factor vs phase
04/12/2010Confidential
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 Power factor obtains it’s minimum around the 270 phase
difference area
 Power factor can be positive or negative
 PF = 0 270 degrees @ I=1.075 A in this simulation≈

7. Rob Woudenberg
Secondary Voltage vs current phase
 Load is aimed to be pure resistant.
 Phase between V and I of secondary coils is 0 degrees at
270 degrees @ I=1.075 A in this simulation
04/12/2010Confidential
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-100
-80
-60
-40
-20
0
20
40
60
80
100
184
188
192
196
200
204
208
212
216
220
224
228
232
236
240
244
248
252
256
260
264
268
272
276
280
284
288
292
296
300
304
308
312
316
320
324
328
332
336
340
344
348
352
356
0.2
0.4
0.6
0.8
1
1.05
1.075
1.1
1.2

8. Rob Woudenberg
Real Power overview (1)
 Real input power:
04/12/2010Confidential
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-4
-2
0
2
4
6
8
10
12
14
184
188
192
196
200
204
208
212
216
220
224
228
232
236
240
244
248
252
256
260
264
268
272
276
280
284
288
292
296
300
304
308
312
316
320
324
328
332
336
340
344
348
352
356
PrimarytotalWatt
Primarypower
0.2
0.4
0.6
0.8
1
1.05
1.075
1.1
1.2

9. Rob Woudenberg
ReaL Power overview (2)
 Real output power:
04/12/2010Confidential
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0
2
4
6
8
10
12
14
184
188
192
196
200
204
208
212
216
220
224
228
232
236
240
244
248
252
256
260
264
268
272
276
280
284
288
292
296
300
304
308
312
316
320
324
328
332
336
340
344
348
352
356
SecondarytotalWatt
Secondary power
0.2
0.4
0.6
0.8
1
1.05
1.075
1.1
1.2

10. Rob Woudenberg
General observations and conclusions
 Unique COP values appear at various input output current phase differences
 Some of these COP situations can never happen because of impossible load impedance causing
phase differences > 90 degrees
 At 270 degrees input / output phase difference COP > 1 situations occur where voltage
and current phase differences of the secondary coils are zero. This is corresponding a
pure resistive load
 The 270 degrees point can be explained as follows:
 Polarity of Iin and Iout definitions are 180 degrees different in the circuit model
 The additional 90 degrees is cause by the transition of signals from primary to secondary:
 Ip→p Vs Is ,current to flux to voltage to current.→ →
 The voltage to current occurs in an induction that causes 90 degrees phase difference
04/12/2010Confidential
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BiTTIin Iout