1610D Volt Reg Wells VR728 no joy

[WhiteRock]

<font color="blue"> Unless there is something else left out of the schematic diagram, the circuit will do nothing if there is no battery connected.</font>

Looks like, as alternator blue top goes positive and bottom blue goes negative, current flows through D1, through sw-a. The negative voltage on the cathode of D4 puts forward bias on Q1 saturating its collector. This also puts a negative voltage on the cathode of S2. R1 now puts forward bias on D6 leaving its anode at .7V causing Q3 to saturate via R2, thus firing the gate of S2. No?

I agree with no regulation, although there is probably nothing to regulate. The light don't care.

I don't think there is any need for current limiting. It will be determine by the inherent voltage differential output from the alternator windings. When the voltage on the battery come up to the proper level, Q4 will start triggering and turn off Q3 which stops firing the SCRs. I may be wrong....

 

[reb]

There are things left out of the schematic, without a doubt. As it is shown (with no battery connected), no current can flow through the circuit to trigger the SCR unless the SCR is already on (ignoring leakage currents). Note that the chassis ground connection does not return to the alternator. As you said, it appears the circuit will do nothing without a battery connected.

Current limiting of some form is required, whether it is internal to the alternator or external. The rating of the alternator is 5 amps, according to a earlier post. Ohm's law says the current will be equal to the voltage difference between the battery and the rectified alternator output divided by the resistance between them. The alternator output (according to other posts) will be in the neighborhood of 25 to 30 volts. Therefore, the resistance in the circuit must be in the neighborhood of a few ohms to maintain operation within the alternator's rating while charging a low battery. Without some resistance in the circuit, smoke would be let out of the alternator; and that is bad, because as we all know, smoke is what makes electrical things work. When you let it out, they quit working. /forums/images/graemlins/grin.gif

 

[Soundguy]

Depending on the internal winding resistance and wire size in the alternator.. it might run at full load indeffinately. ( well.. you know.. long periods anyway ). Since it is permanently biased.. and well.. low current anyway..

There's lots of silicon rated for shunt loads at full rating 'indefinately'.

Soundguy

 

[WhiteRock]

Perhaps my description was very poor. But, I do have to disagree unless you can be specific about my error.

I could not see any parts missing. I agree that there does appear to be parts missing, however if you do a careful time domain analysis, you will see that the different transistors and SCRs trigger to protect the circuit. If you can point out which junction is over biased, I will be glad to attempt to describe why I think it is not.

I don't recall saying the circuit does nothing without a battery. I said that I don't think it regulates, but I also don't think it is necessary that it does. High voltage is ok.

What you have is an alternator with out an excitable armature. So at a specified rpm, the wattage output is fixed by an internal magnet. The maximum voltage is also fixed by the number of wire turns in the alternator. While I know you measured a higher voltage on the oscilloscope, you did not indicate that you had a 5 amp load on it. I suspect that at 5 amps your output voltage will be approximately 14.5 - 15 V. With no battery, I suspect the voltage on the lights will be a little high if they are small wattage lights. So I think this system's design has an inherently current limit via the alternator output.

I agree with you on the ohms law, however with a batter installed, and it is low, the voltage difference between the battery and the alternator output causes the transistor to fire the SCRs. As the battery charges the voltage will rise across the zener and the transistor will stop firing the SCRs. At this point there is no current being consumed from the alternator.

About the absents of the of the SCR path; With no battery, the alternator still sources a voltage. The fact that the ground isn't connect has nothing to do with this part. When the alternator generates a voltage, one of the cathodes of the SCRs will always be biased negative due to the fullwave diode bridge. Not via leakage. The positive voltage leg of the alternator causes the PNP transistor Q3 to fire the SCR gate positively relative to the SCR cathode. This was tough for me to see at first. Because the transistors interact with each other in the different modes.

Oh well, just trying to help. I think this circuits works. I think it is a cheap design, but I think it works.

 

[reb]

This is a relatively simple circuit, but the analysis can get convoluted.

The most obvious missing component is a current limit resistor in the base of Q1. Without it, the full alternator voltage appears across D1, D4, and the emitter-base junction of Q1, all low impedance devices. That is guaranteed to let the smoke out of Q1 and/or D4.

Other missing parts include resistors across emitter-base junctions of Q1 and Q4 to compensate for component leakages. Also required is some resistance in the collector path of Q3, and probably from the gates to the cathodes of the SCRs.

</font><font color="blue" class="small">( When the alternator generates a voltage, one of the cathodes of the SCRs will always be biased negative due to the fullwave diode bridge. Not via leakage. The positive voltage leg of the alternator causes the PNP transistor Q3 to fire the SCR gate positively relative to the SCR cathode. )</font>

What bothers me most in reading the schematic is in trying to define the paths for the collector current of Q1 and the base current of Q3 before any SCR is triggered. Before an SCR is triggered, it appears as an open circuit, except for what should be negligible leakage current. I assumed the leakage current to be zero. If no current flows through the SCR, no current can flow through D6 and, consequently, the collector of Q1 or the base of Q3. Without base current, Q3 cannot trigger the SCR. There must be something left out of the schematic to provide a low current path across the SCRs.

As you said, this is a cheap design, but it works. That is what companies like to produce.

 

[reb]

</font><font color="blue" class="small">( Depending on the internal winding resistance and wire size in the alternator.. it might run at full load indeffinately. ( well.. you know.. long periods anyway ). Since it is permanently biased.. and well.. low current anyway.. )</font>

I agree. If the internal impedance of the alternator limits the current to 5 or 6 amps, that is only around 100 watts with a 12V battery connected. That is probably not enough to cook the wiring.

 

[WhiteRock]

<font color="blue">The most obvious missing component is a current limit resistor in the base of Q1. Without it, the full alternator voltage appears across D1, D4, and the emitter-base junction of Q1, all low impedance devices. That is guaranteed to let the smoke out of Q1 and/or D4.
</font>

I agree and had this wrong. And in most cases it would be Q1. The 3 junctions will take the full wrath of the alternator and Q1 will loose. So Q1 base current has to be limited and appears to missing in the schematic.

I thought I saw a switch that protected Q1, but I don't see it today and I don't think it was ever there. /forums/images/graemlins/blush.gif

 

[WhiteRock]

It now looks to me like if the battery is missing, that the lights would not work. Is this true?

 

[reb]

I need to think about the circuit some more. I think the circuit path I mentioned around the SCRs is necessary for the circuit to work in any case. With no battery the full rectified alternator output would be across the eb junction of Q4, Z1, and D6. Since the schematic shows no means to limit the resulting current to a safe level for those components, I would expect a failure. Connecting the headlights to the circuit would not help the situation. However, once the regulator faulure occurred, the lights would probably work (until they burned out) if the rectifier circuit was not affected by the failure.

 

[WhiteRock]

<font color="blue">Other missing parts include resistors across emitter-base junctions of Q1 and Q4 to compensate for component leakages. Also required is some resistance in the collector path of Q3, and probably from the gates to the cathodes of the SCRs. </font>

I do feel that Q4 is ok because..... as the voltage rises on the emitter, at some voltage the base will be clamp through the zener and saturate Q4 protecting its V-eb current. Then the current will flow from emitter to collector through R2. This also turns the charger off. I think this circuit is intended to be a comparator with high gain and leakage isn't an issue.

If this clamping action occurs at 14.4V (assuming Z1 12V), then R2 would have to dissipate approx 12 watts and would need to be around 5 ohms. But this only occurs when the battery is charged.

I haven't worked with SCRs in a long time, but I thought you could typically take the gate voltage to the anode and above. I don't think the gate presents a diode like junction like a transistor. It may be device specific...

I may have it wrong again......