boost / Buck boost

Remember the back EMF. The ultimate source being the vehicle's kinetic energy of course.

energy,

The back EMF of the PM motor

Nope, input is in the same direction (ie it will tend to accelerate the motor). The back EMF is the same polarity as the DC bus (+ to +).

I was reviewing my explanation and trying to develop a simulation in LTSpice. I realized I left out an important bin on the off cycle (sorry about that). I'm assuming a full H or 1/2H control and the flyback switch shorts the motor during the off cycle, there's where your current build up takes place.

Here's an LTSpice model if that's any use. Switches with anti-parallel diodes substitute for the usual power devices. The motor is modelled as an inductor with a voltage source for the back EMF.

• C:\Program Files\LTC\SwCADIII\regen.asc L1 N001 N004 .1m Rser=1m D1 N003 N001 D S1 N003 0 N002 0 mysw D2 0 N003 D V1 N001 0 50 Rser=1u V2 N002 0 PULSE(-10 10 0 0.01u 0.01u 0.025m .05m 80000) V3 N004 N003 SINE(25 25 1 0 0 0 10) Rser=1u S2 N001 N003 0 N002 mysw .model D D .lib C:\Program Files\LTC\SwCADIII\lib\cmp\standard.dio .tran 2 .model mysw sw(Ron=.000001 Roff=1Meg Vt=5 Vh=-0 Lser=0 Vser=0) .backanno .end

Sigh, I usually think in terms of the system's energy transfer rather than the switch details.

Robert

On 2007-05-27, Robert Adsett wrote: To: Robert Adsett Subject: Re: boost / Buck boost In-Reply-To: References:

X-Face: ?)Aw4rXwN5u0~$nqKj`xPz>xHCwgi^q+^?Ri*+R(&uv2=E1Q0Zk(>h!~o2ID@6{uf8s;a+M[5[U[QT7xFN%^gR"=tuJw%TXXR'Fp~W;(T"1(739R%m0Yyyv*gkGoPA.$b,D.w:z+> > - When the MOSFET is off the voltage will rise since there is now

the back EMF is less than the rail voltage, there's no way it's going to increase the current, unless the motor's shorted out instead of connected only to the bridge.

ah! yeah, that'd work, it'd be more efficient too, freewheeling the coils instead of feeding their energy back into the supply (when driving) and having the mosfets conduct instead of the protection diodes when braking.

Bye. Jasen