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In motorcycles broken stock voltage regulator is a very common issue. Stock solution is SCR based and commonly burns due to overheat/ SCR failure.

DIY regulators usually have the following design (mostly it's a copy-paste of original solution):

DIY voltage regulator

As moto alternator has no field winding, all the schematics are simply shorts windigs. From my point of view, it's quite ineffective. First, I thought about using IGBT/MOSFETs as rectifiers and PWM it's gates in order to regulate voltage as simple shunt regulator. But there are ideal rectifier controllers, like LT4320. Maybe there are ready to use IC controllers for ideal diodes that can do it?

Or is a SCR schematic is really enough? But change the design - use SCR as rectifier part. But it's not clear for me, what C terminal should be connected to. I think to Vout, but it's not shown

enter image description here

The main target is to get more current and increast ruggedness of current alternator. The question: is there a well known design of regulated rectifier with MOSFETs? Why it's not in use? It looks so simple. Shunt regulator inside rectifier itself.

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    \$\begingroup\$ They are not triacs - they are SCRs and what is your question? \$\endgroup\$
    – Andy aka
    Jan 13, 2017 at 9:51
  • \$\begingroup\$ Sorry, surely SCR. The question: is there a well known design of regulated rectifier with MOSFETs? Why it's not in use? It looks so dimple. Shunt regulator inside rectifier itself. \$\endgroup\$
    – pugnator
    Jan 13, 2017 at 13:55
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    \$\begingroup\$ Edit your question, don't add questions to the comments. \$\endgroup\$
    – winny
    Jan 13, 2017 at 15:04
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    \$\begingroup\$ Just a comment. The rationale to have a shunt regulator in motos is that the ignition will require more current at larger rpm, which goes in sync with the alternator power output. So there is a relatively constant load on the shunt regulator. \$\endgroup\$ Sep 11, 2017 at 10:57
  • \$\begingroup\$ The C terminal is connected to the switched power so the Zener diode current does not slowly drain the battery when the engine is off. \$\endgroup\$
    – KalleMP
    Sep 17, 2017 at 11:41

4 Answers 4

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Your proposed design is overly simplistic. Remember, the gate drive for each SCR is relative to its own cathode, so your common drive circuit won't work. It works in the shunt circuit because all of the cathodes are connected to a common point (ground).

Furthermore, when you put the SCRs in series with the alternator, you now want them to fire when the voltage is too low, rather than when it is too high.

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  • \$\begingroup\$ Yeah, there are error in the schematic, now I see. What about to use SCR/MOSFET as half bridge? \$\endgroup\$
    – pugnator
    Jan 13, 2017 at 15:41
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    \$\begingroup\$ What about it? Sure, it would be possible, but again, the trick is in generating the correct control signals. \$\endgroup\$
    – Dave Tweed
    Jan 13, 2017 at 15:44
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You just need to add 1n4007 to each gate of scr or it will blow up gates in this scheme. This schematic is working l have tried. C is connected to ignition key .

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    \$\begingroup\$ How are the diodes connected? In series with gate drive or across the gate and cathode? \$\endgroup\$
    – KalleMP
    Sep 17, 2017 at 11:38
  • \$\begingroup\$ Waiting for the answer. I'm still not giving up. I also want to try LTC3810 to build PWM based schematic \$\endgroup\$
    – pugnator
    Sep 18, 2017 at 10:09
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This is a proposal for a series circuit. At the beggining, when the thyristors still don't conduct, the voltage on capacitor C1 will rise above V batt a couple volts, enough to trigger them. You will lose a few degrees of the half cycle, not important I think. When V batt is high enough to make Z1 into conduction, tr1 will shunt the current trhough R1 and so the thyristors won't trigger. Only concern is: will the voltage across C1 rise dangerously high in this situation?

enter image description here

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  • \$\begingroup\$ I can use capacitor rated for 100+ volts. I'll try this schematics, thank you! \$\endgroup\$
    – pugnator
    Jun 28, 2018 at 13:35
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Resistor connected to positive of C1 (you could name resistors as well) needs to disperse a lot of power, as most of the time you need to loose 80V on it having enough current to trigger thyristors.

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