discrete voltage regulator circuit

check out this voltage regulator:

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I used this circuit on a motorcycle I owned. I made a couple of changes: flipped all the polarities around in order to switch the low side of the field, and substituted a mosfet for the darlington. It worked pretty well. Now I have a couple of questions.

1) About cap C1 that rolls off the higher frequencies so T1 doesn't oscillate: How does one calculate the rolloff frequency? -- it's worth knowing for practical and theoretical reasons.

2) I think it would make sense to put a cap in series with R4, making the positive feedback transient and ensure a minimum PWM frequency in order to keep the headlight from flickering, a problem I had. The original circuit uses hysteresis, so the switching rate depends on external factors -- charging system, wiring harness, loads. f = 1/(2RC)? If so, then adding 0.01 uF in series with R4 at 120 k would ensure predictable switching at a bit over 400 Hz. Did I get this right?

Reply to
gearhead
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I don't think C1 is a good idea at all, it slows the switching speed greatly which increases the power dissipation, and was probably derived empirically with no follow on testing, with temperature especially. The T1-T2 combination bursts into oscillation at higher currents because T2 gain is higher there and the field winding acts like a gain peaking coil with a 90o phase advance, causing T1-T2 to steady state in the linear region in oscillation. This can be avoided by taking T2 and the field coil out of the feedback loop.

The AC-feedback will guarantee minimum on/off times and an upper limit on the frequency, not a lower limit.

Reply to
Fred Bloggs

Hmmm- he uses a 1N5408 which is only a 3A diode, seems a bit weak for a circuit banging around 5A currents at low frequency...

Reply to
Fred Bloggs

(snip)

The coil is a problem and there's nowehere else in the circuit to get feedback, so I guess one has to add another transitor in parallel with T2, and instead of connecting to the coil put a resistor in the collector. And take the feedback from there. I suppose that would work.

Reply to
gearhead

Something like this does just as well, there is no temperature compensation, which should be about -2mV/oC/cell on the Vbatt trip threshold, and you will need to add adjustment in there somewhere, no chance of oscillation here: View in a fixed-width font such as Courier.

. . . . . .-----[1k]---------. . | | . | .------|----------+-----. . | | | | | . | | |< ---------- . +-----[180]-+----| BC857 |RST V+ | . | | |\\ | | . | | | | 555 | . | | +----+---|THR OUT|---> to MOSFET . + o | | | | | . | [100] '---|TRIG DIS|---. . BATT | | | | | . -/ | | GND | | . - o 1N4740 ^ | ---------- | . | | [1.2k] | | . | | | | | . | | +--------------|--------' . | | | | . | | [1.2k] | . | | | | . '-----------+------+--------------+--------- . . .

Reply to
Fred Bloggs

-----T2 is connected to ground via R3, and T2 is conductive: current flows through the rotor winding L, and the magnetic field in the alternator increases, which in turn increases its output voltage. When the voltage is high enough to bring D1 in its conductive state, T1 will also become conductive and will connect the base of T2 against the positive line.---

I dont think the description is right. magnetic field goes up can increases voltage output? no! and the voltage increase can make D1 conduction? no. the correct description should be, current saturates gradually, which makes potential at DF reduce, this will cause T1 conduction, then positive beedback takes place. the oscillator occurs.

Reply to
readeraz

(snip)

=A0| =A0 =A0 |

----------

=A0 =A0 =A0 =A0 =A0|

=A0 555 =A0 =A0|

=A0OUT|---> to MOSFET

=A0 =A0 =A0 =A0|

=A0 DIS|---.

=A0 =A0 =A0 =A0 =A0| =A0 |

=A0 =A0GND =A0 | =A0 |

-- =A0 =A0|

=A0 | =A0 =A0 =A0 =A0|

=A0 =A0 =A0| =A0 =A0 =A0 =A0|

------'

=A0 =A0 =A0|

=A0 |

=A0 =A0 =A0|

How about

,----+-------------------, | | | | | ,----|--100K--, --- | | | | - | | |\\| .01 =3D --- \\ Vref--100---+--|+\\ | - / | >-------+---to mosfet | \\

Reply to
gearhead

The LinCMOS is kind of weak on output drive. Your bike does not have an ignition switch? How do you disable the field coil drive when it's parked?

Reply to
Fred Bloggs

If the alternator uses a centrifugal switch to throw the field coil in the circuit at RPM, then you can use this to supply power to your circuit, there will be no battery drain other than leakage when the bike is off. The comparator, CMP, works independently of your controller, to interrupt the field drive, very briefly and at low frequency, to maintain power as long as the field coil is in the circuit. When the engine is cut-off, the circuit shuts off. Drawback is the component count is getting ridiculous...Another thing is, you don't have enough hysteresis, you can do 100mV/cell, which at 6x cells comes to 600mV, that would be a switching scheme going from 2.2 to 2.3 volts/cell, a full charge maintenance. View in a fixed-width font such as Courier.

. . . BATT . | . | . .-------+-------------------------+----o(+) . | | | | . | [Rb1] | o . >| | | / csw . pnp |-----+---[Rb2]---------. | o . /| | | | . | | Dfb - L . | Cbp | ^ L . +------[180]-+-||-. | | L . | | | | | | . | | GND .------------+----o DF(-) . | | | | | . | -----+----------------------------------+----. . | | | | | | | | . ------------ | | | | | | . | BATT V+ | | | | - [100k] | . | | | | | v | | . | hysteretic | | pwr|- -| | | |\\| CMP . | controller |---------| 2N7000|-+---+----------------|-\\ . | | | MOS|> -. . | GND RST | | | | === | .-[100k]-+--|+/ | . ------------ -/ | | 0.1u [1M] | | |/| | . | | ^ 1N4740| | | | | [100k] | | . | | | | | | | | | | | . '-----|------+-------+----+---+---+-------------+----' | . | | | | . | GND | | . '--------------------------------+----------------' . . . . .

Reply to
Fred Bloggs

sfet

- Hide quoted text -

A rectifier prevents any flow from the battery back into the generator. Inside the generator, the high end of the field connects to the high end of the armature. The whole thing gets its start from remanent magnetism in the pole shoes, and builds up from the there. The battery doesn't have anything to do with it.

Reply to
gearhead

Hide quoted text -

Can you tap into the anode of that rectifier? And/or can you power your regulator off an ignition circuit?

Reply to
Fred Bloggs

fet

d?- Hide quoted text -

Yes, I can run a comparator (or whatever) from the anode side of the rectifier. But the bike has to be revving around a couple thousand rpm to get anything out of the generator -- at idle the voltage drops. Then the comparator inputs will sit at higher voltage than the comparator's supply, a big no-no. Powering off the ignition circuit would be impractical. People have these antique bikes and they want to take the guts out of the old relay box and pop in a circuit board. The bikes don't have a separate wire that runs from the ignition switch down to the relay box.

Actually one of the guys came up with a circuit and installed it in his bike already:

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Guess I ought to just leave well enough alone.

Reply to
gearhead

I just had a little bit of a Eureka moment about why that circuit oscillates. When I put that circuit together I tested it on a bench with a resistive load in the power mosfet drain to simmulate a field coil. When I used a 2 or 3 amp load, it oscillated. So inductance didn't cause it. But the top rail has some impedance, and when the load switches on it pulls the top rail down a fraction of a volt, which feeds back to the base of the small transistor that controls the power mosfet. As you increase the load, at some point the magnitude of this voltage droop negative feedback approches the magnitude of the positive feedback or exceeds it, and you get oscillation.

To solve the problem I could try

1) increasing the positive feedback, or 2) bypassing the negative feedback

but if I just hang a cap off the transistor base for bypass it will slow down the switching. So:

Vbatt | +---------------------------+----, | | | | 1K field | .01 | | diode ,--100k--||--|----+ string | | _| | | ,-------+-||_ | | / | | | | | +---1K--+--1K--+--| | | | | | 1K 10 =3D > | | uF | | | '-------+-----------+------------+ | gnd

The negative feedback has a time constant of 10 mS, much longer than the

1 mS time constant of the postive feedback. Should work?
Reply to
gearhead

for practical and theoretical reasons.

I just had a little bit of a Eureka moment about why this circuit oscillates. When I put that circuit together I tested it on a bench with a resistive load in the power mosfet drain to simulate a field coil. At light loads it didn't oscillate but when I used a 2 or 3 amp load, it oscillated. Resistive loads, so inductance didn't cause the oscillation. But the top rail has some impedance, and when the load switches on it pulls the top rail down a fraction of a volt, which feeds back to the base of the small transistor that controls the power mosfet. As you increase the load, at some point this increasing negative feedback approches the magnitude of the positive feedback or exceeds it, and you get oscillation.

To solve the problem I could try

1) increasing the positive feedback, or 2) bypassing the negative feedback

but if I just hang a cap off the transistor base for bypass it will slow down the switching. So:

Vbatt | +---------------------------+----, | | | | 1K field | .01 | | diode ,--100k--||--|----+ string | | _| | | ,-------+-||_ | | / | | | | | +---1K--+--1K--+--| | | | | | 1K 10 =3D > | | uF | | | '-------+-----------+------------+ | gnd

The negative feedback has a time constant of 10 mS, much longer than the

1 mS time constant of the postive feedback.

On the right track?

Reply to
gearhead

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