Designpages
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*But if we can change the windage volume, by making stator thicker, there is indeed a possibility to reduce '''Rgen/V<sup>2</sup>''' and increase the efficiency. | *But if we can change the windage volume, by making stator thicker, there is indeed a possibility to reduce '''Rgen/V<sup>2</sup>''' and increase the efficiency. | ||
--[[User:Ameet|Ameet]] 02:22, 15 December 2006 (EST) | --[[User:Ameet|Ameet]] 02:22, 15 December 2006 (EST) | ||
+ | |||
+ | ==Calculating efficiency of Tunecharger== | ||
+ | We use a pretty complicated formula to find efficiency of tunecharger. That is because, tunecharger is pretty complicated :-) | ||
+ | This page describes the formula & logic behind it. |
Current revision as of 05:18, 23 September 2007
Contents |
Study of UIUC circuit
UIUC design documents are here. Conclusions of Final report read that
The efficiency of the dc-dc converter circuit was unacceptably low. The main reason for the low efficiency was that the circuit could not be brought up to high power due to MOSFET heating and at low power the control circuit was taking a considerable percentage of the overall power. If output power were raised, the amount of power used by the control chip would stay approximately the same and the percentage of overall power that it consumed would be much less, therefore raising the efficiency. In an effort to get better efficiency or reduce MOSFET heating, flyback converter approach was attempted but we obtained similar unsatisfactory results
and in test data (appendix), around 6-12% efficiencies were reported. Converter seemed to work fine in all other respects. (protection, line regulation, ripple factor etc.)
Possible cause of inefficiency: I suspect inefficiency is caused due to all reasons below.
- Poor design & wrong testing: The design review document details that the circuit was designed to cater to output current of 8 to 10A. Output Voltage of 12 to 15V. From these specs, inductor is sized at 10microH. But in the actual test, load was 50ohm resistor, for input voltage of 48V, duty cycle of 25%, which should have given output voltage 48*0.25=12V. But actually, output voltage was 6V, and current drawn by load was 6/50=0.12A, this was way out of design specs of 8 to 10A. It is a usual design practice in buck converters (reference) to make sure that inductor current does not go to zero, or continuous charge mode is maintained even at 10% of nominal current. This dictates a higher inductor value, as it prevents a small load from discharging inductor quickly. This practice was not followed, leading to a very tight operation window for continuous charge operation. And the testing conditions were too severe/ far from nominal design. They should have used, 12V/10A=1.2 ohm load resistance.
- Discontinuous charge mode: As mentioned above, in above testing conditions, very probably converter was in discontinuous charge mode. (For explanation charging modes, refer Wikipedia)In this mode inductor discharges too quickly in off cycle, and inductor current goes to zero for partial cycle. This can be verified by calculating current at the limit of continuous & discontinuous modes from above link. For duty cycle of 25%, it is (48V * (1 - 0.25) * 0.25 * 1/(100kHz)) / 2*(10e-6 microF) = 4.5A > 0.11A
- High switching in Mosfet: As is evident in oscilloscope images during test, the mosfet gate voltage rings a lot around high value. This oscillations seem to be much higher than switching frequency, and might be increasing switching losses drastically, dissipated as heat, thus heating Mosfet. Cause of such ringing is still a mystery. Can you volunteer to solve it? In the figure on side, the channels are as follows: Ch1-Gate Voltage; Ch2-Source Voltage; Math-Gate-Source Voltage; Ch3-Output Voltage; and Ch4-Output Ripple Voltage. Even though the output was not the desired 12V but was instead 6 V, the output was regulated well, user adjustable (via turning pot), and converted.
--Ameet 02:22, 15 December 2006 (EST)
Systems analysis with ideal converter
Schematic shown alongwith shows simplistic view of the converter. A generator is connected to a battery through a converter. We want to study power losses outside the converter, and power flows in the system. Hence in this analysis, we assume 100% efficient converter. Analysis can be easily modified to cater to converter losses.
Let
- P=average power input to generator by pedaling.
- V=backemf generated by generator
- Rgen=winding resistance of generator
- I=current through generator side
- D=duty cycle of converter, settable by a potentiometer on converter.
- Vload=load voltage=(V-I*gen)*D
- Iload=load/battery current=I/D
Note that V*I=Vload*Iload indicating 100% efficient converter
- Vbat=electrochemical potential of the battery. Fixed for battery particular chemistry
- Rbat=internal resistance of battery. This is dynamic quantity and changes with state of charging and current, but for now, we assume it constant.
Pedal Generator model
Just like any prime mover, each bike rider has a unique Torque-Speed curve for fatigue free long ride, above which he cant operate. To achieve fastest speed, he usually operates 'on' this curve. Do lookup this reference for more info. This curve (with torque as Y and speed as X) has same shape as the curve labeled 'Gen' in adjoining Power curves figure. Fortunately for AC PM machines, Voltage is proportional to only speed (and not torque),and Current proportional to only Torque (not speed), leading to same shape in V-I curve for Pedal powered generator. Note that constant power curve 1,touches 'Gen' at P. Hence P is the highest power point on the curve, which usually occurs at fixed pedaling rate. (around 90rpm). Our aim is to operate at P, thus maximize power and maximize the efficiency. Note that this V & I are same as on defined above.
Coupling without converter
This is same as having D=1.
This implies I = Iload and V - I x Rgen = Vload = Vbat + I x Rbat
Curve 2, shows V,I curve for Ideal battery, which can supply any current at constant voltage Vbat. Due to resistance, this curve shifts to curve 3, as additional voltage drop I*Rbat will add to the V.
When above generator and load are joined, they operate at point Q, where the curves Gen and 3 intersect. Note that power at Q is much lesser than optimal point P. Hence without converter, we invariably end up capturing less power than we can otherwise.
Coupling with converter
When converter is present, we have a greater freedom to shift the load curve, so as to intersect it with maximal power point P. Lets see how this happens.
The battery curve 3, when seen on supply side, undergoes expansion in X direction by factor 1/D, and contraction in Y direction by factor D. e.g. note that curve 4, corresponding to D=0.5, when compared to 3, has double intercepts on X axis and half on Y axis. Similarly for curve 5 (D=0.25), X intercept is 4 times curve 3. As we vary D, these curves, touch a hyperbola in 4th quadrant, and thus pass through P for some optimal duty cycle D.
Since operating point is determined by intersection of curves, at optimal D, we operate at point P, and extract maximum power from the source. Since D can be electronically varied, it is easier to implement optimal power capture by changing potentiometer or in real time by algorithm like MPPT, to cater to different riders, changing charge and dynamic resistance of battery.
Thus it is always advisable to have a converter. In photovoltaic cells, such algorithms have led to 30% more power than ordinary coupling. In a way, converter is exact counterpart of gear system in bicycle, allowing us to operate at our maximal power point.
Effect of coil winding factor on efficiency
If P is the max power generated by the rider, assumed constant.
- Power Loss at generator side in windings is
= I2 x Rgen = P2 x Rgen/V2
Thus, % power loss at generator side = Power Loss/P = P x Rgen / V2
As P is assumed constant, lower the factor Rgen/V2, higher the efficiency.
Let us see if we can affect it by increasing number of turns per phase, and thus stepping up the generator voltage.
- Rgen=rho*L/A
where rho=resistivity, L=coil length and A=cross-sectional area.
If we increase turns by factor of N, length grows by factor of N, but if we fit the windage in the same volume as before, A will have to decrease by factor of N. Thus Rgen grows by factor of N/(1/N)=N2.
- V=k x n x f
where n= number of turns/phase, f=pedaling speed in rpm, which is constant for max Power point P. Thus if we increase n, by factor of N, V grows by N as well.
Hence Rgen/V2 grows by N2/(N)2 = 1
- In Other words, N does not affect efficiency.
- But if we can change the windage volume, by making stator thicker, there is indeed a possibility to reduce Rgen/V2 and increase the efficiency.
--Ameet 02:22, 15 December 2006 (EST)
Calculating efficiency of Tunecharger
We use a pretty complicated formula to find efficiency of tunecharger. That is because, tunecharger is pretty complicated :-) This page describes the formula & logic behind it.