[vamfun]

Originally Posted by Al Skierkiewicz

Chris,
Vemf=Counter EMF=CEMF, which will be the same for either controller in the "off" condition or in coast, when connected to the same type motor running at the same speed.
CEMF will be less than Vbattery unless additional mechanical force is applied to the system or will be zero if the motor is not turning.
If we identify the four parts of the H bridge such that the two FET groups connected to positive battery terminal are labeled A & B, and the two FET groups connected to negative terminal are labeled C & D. Then for one motor direction group A & D are turned on and for the other motor direction B & C are turned on.
All FETs in both controllers have diodes from source to drain as part of their implementation/manufacture.
When any (or all) FET group in either controller is turned off, the motor inductance will make a voltage spike whose polarity is opposite of the applied voltage during the "on" phase of the PWM output and in the same polarity as CEMF.
The drop across the Victor diodes is 1.2 volts and the Jaguar is 1.25 volts.

OK so far so good.

If these conditions exist then two sets of diodes will conduct the voltage spike through the series circuit that includes the battery and all wiring to that branch. The current will continue to flow only until the voltage spike plus CEMF exceed the battery voltage and the voltage drops across the resistance of the wiring and the forward drop of the diodes.
If motor current is supplied by FET groups A & D then the diodes in B & C will conduct during the discharge

Here is where you lose me. If the motor current is being driven by A & D when A turns off D is still left "on" in the JAG. This provides a near short to ground on the C & D side of the motor and as I see it the spike can only flow through B diode and a short to ground . So only one diode and an "on" D FET in series with the motor need be considered in the circuit.

Signed Confused