View attachment 1146Seems to me that the RR works harder, hence greater heat, when fully charging system loads. When shunting, it's only 1/2 wave rectification through the 3 SCR's. As compared to full wave rectification via the 6 diodes.... Meaning that if running low current LED's the RR is in shunt mode (less heat on RR) as compared to running high current loads (heated gear) when the RR is in full 'on' mode (full rectification, max current).
Nope. Not in that circuit anyway.
Half the power is going through each set of diodes, true. But that doesn't mean that no heat is generated when the other half is going through the SCRs. In fact, more heat is generated.
All things being equal, the voltage drop of an SCR is almost twice the voltage drop in an equivalent diode. A power diode is a dual-layer device with one P-N junction and a typical forward voltage drop of 0.7 to 1.2v whereas an SCR is a four layer device with 3 P-N junctions. The power in watts dissipated in such a device varies as the square of the current and is a product of the current and voltage drop. So if you double the voltage drop, you double the heat generated.
For the sake of simplicity, let's ignore minor factors such as lead inductance, capacitance, manufacturing tolerances, gate current, etc. and use some round numbers that don't need a calculator:
Suppose the alternator delivers 30 amps and the combined load is consuming 30 amps. Therefore each of the 6 diodes is passing 5 amps. Now let us suppose that the particular diode chosen dissipates 5 watts of heat at that current. Times 6 diodes, the regulator must therefore get rid of 30 watts of heat.
So now consider the case where the alternator delivers 30 amps, but the combined load is only 15 amps. 15 amps passes through the 3 diodes (as you pointed out) while the shunt passes the other 15 amps. The equivalent SCRs, with twice the voltage drop, would then dissipate 10 watts of heat per SCR at 5 amps. So we have 15 watts of heat from the diodes, plus 30 watts of heat from the SCR trio, for a total of 45 watts of heat generated inside the R/R! 50% less power consumed = 50% more heat to get rid of!
Now, there is a way to reduce the heat generated by shunting -- use power MOSFETs (or even IGBTs) instead of SCRs. The forward voltage drop of a typical MOSFET is typically a few tenths of a volt instead of multiple volts (although the resistance of a MOSFET increases with current). But at 5 amps, one could reasonably expect a MOSFET to generate a third of the heat compared to an SCR.
Another advantage of using MOSFETs is that they are available in both N-channel and P-channel designs. In the schematic above, the SCRs can only shunt up to half the power output -- what happens if the load is reduced to less than half by substituting LED bulbs? That other half can't be shunted, so voltage is forced to rise above the regulated set point -- possibly damaging electronics and boiling that expensive AGM battery dry. But it should be possible to use P-channel MOSFETs to shunt the remaining diodes and theoretically regulate the voltage even at near-0 load.
Just to be clear: I have no idea where the above schematic came from or if Yamaha used SCRs or MOSFETs in the FJR R/R.
Based on the next 2 posts, I had the concept right but the hotter/cooler cycles reversed.
