Regulator failure

[amiklic1]

I have 4' Hugh Piggot's design wind generator for around three years now, and it's really good. I have plenty of power for small TV and few CFLs.

Today, after a few days of strong winds (10-15 m/s) I came to my mountain cabin, and I found out that the battery voltage is 16V (12V system), and that regulator is not working ( I didn't checking, just LEDs are nt working).

I use dump load calculated like this:

250W max power, 12V system voltage.

= 20A max current

From that, I get that resistance would be from 12V/10A= 1.2Ohm to 12V/20A=0.6Ohm

I se heater elements modified to get that resistance.

I am just using this regulator for a month or so, and before I used other one, but based on electromechanical relais.

Dump load (heaters) is connected in parallel to the battery, and is switched on when charging voltage gets higher that 14.4 V. The battery is still connected, just one large diode is preventing the battery from discharging thru the heaters.

With previous regulator, the windgen was slowing down visibly during the heaters switched-on period. Now, when I connected the new regulator, that's not happening. When I am looking into the voltage, it just gets lower a bit (from 14.4 to let's say 13.2 V) and that rises to 14.4, then gets lower....

Any help is appreciated, because I really don't know what's happening.

[AbyssUnderground]

Did you connect the regulator wrong some how? It seems the battery is being the providor of the power to the heaters rather than the generator. When it hits the cut-in voltage of 13.2 its turning the heaters off and charging the battery again, then 14.4 comes and it cuts off again.

[ghurd]

If the green LED is not on, the circuit is not connected to the batteries.

Sometimes a new switch can cause problems. Turn it off and on several times to see if the green LED comes on.

Check the 1N4001 diode near the switch.  It could have failed to an open condition.

Test the voltage across it. It should be about 0.6V.  If it shows battery voltage it is bad.

Be certain the circuit is connected to the battery with proper polarity.

It is proper if the green LED was on before it failed.

Other than that, all I can think of is a surge from a close lightning strike.

Good luck,

G-

[Flux]

That regulator is intended to work across the batteries, I don't see how you can have a diode to prevent the batteries discharging and still expect the regulator to work.

You could use a series diode and try the regulator on the input side of the diode but it may need modifying, I think it was intended to take its power and sensing from the battery.

Nearly all regulators work directly across the battery and to do that you must not have a diode. The snag with all these schemes is that a faulty regulator will run the battery flat.

Flux

[amiklic1]

OK, I looked a bit, and I saw that the old diode is remowed ( I suppose my sister's husband do so, because he was playing around with this setup).

Also, I guess that mosfets are preventing the current to go from the battery to the heaters.

None of the LEDs is lighting, and voltage rises up to more than 15.8. Maybe 555 chip is burned.

I would also like to ask what to use instead of 555 (for some other aplications) when one wants to go higher than 15-16 Volts?

[Nando]

I am not familiar with the device, do you have the schematic and if so can you send it to me, to see what type of circuit it is and how to replace the 555 or at least how to isolate it and have higher voltage capabilities ?.

Nando

[s4w2099]

cant you use a 12V zener and a resistor to power the 555? Then the MOSFETs should work up to 40Volts (the cheapest ones). You probably dont want to go over 20 volts anyways because most MOSFETs will not take a voltage greater than 20Vdc in the gate.

[ghurd]

The green LED should light, even if the 555 is burned.

Power must go from battery, through the switch, through the diode, the LED, then the resistor to ground.

If the green LED is Not lighted, the problem is almost certainly in that line.

For Nando,

http://www.homepower.com/files/shuntregulator18.pdf

[Flux]

Ghurd is correct. If the green light is not on, you have a failure with the switch, 1N4001 diode, 1k resistor or the LED or a connection.

If there is no power to the circuit there is nothing to turn the mosfet on.

For higher voltages you could add a 7812 regulator to feed the 12 power rail of the low power circuit. The 555 should be ok on a 12v supply when the thing is operating properly. It is far from being an elegant circuit but good enough if built properly.

First thing is to make sure you have nominal 12v dc on the cathode of 1N4001, without that nothing is going to work.

Flux

[Nando]

The solution is going to be limited to the LM723 maximum voltage break down.

Cut the line from LM723 Pin 11 to Q1 ( 2N2222) Collector resistor.

Place a Electrolytic capacitor, Positive to NE555 pin 8 line the Negative to ground, make the capacitor 10 microfarads breakdown voltage at least 25 volts.

Get a medium power NPN transistor like the TIP31C with Collector to LM723-pin 11, an Emitter to NE555-pin 8.

Take a resistor, 3.3 Kohms 1/2 watt from TIP31C-Collector to TIP31C-Base.

Take a 12 to 15 volts ZENER, 1/4 OK, 1/2 watt best and connect the lead with the Band to the TIP31C-Base and the otehr lead to ground.

Get a capacitor at least 1 microfarad 25 volts breakdown , positive to TIP31-Base and the Negative lead to ground.

You need to either add a resistor in series with the 3k that is in series with the trimpot, or replace with the proper value due to the increase in input voltage control range.

To calculate the NEW resistor, read the voltage at LM723-Pins 4 or 6 = Vz

Get the bias current = Vz / ( 500 + 3.3 K) = Iz, normally Vz ~ 6.4 volts

New voltage to control = Vn

(Vn-Vz)/Iz = Rn which replaces the upper 3 K ohms resistor.

Alternate Solution that is best.

Make the same circuit with a TIP31C but place it replacing the 1N4001 (or add it between the Cathode of the Diode and the upper power line feeding the LM723 and the NE555)

THEN DISCONNECT THE 3K resistor and replace it with the calculated resistor connecting the upper lead at the junction of the Switch and the

Anode of the 1N4001 diodes

This way the TIP31C will allow up to about 80 Volts ( you may need to place a heat sink on the TIP31C if the current goes higher than producing 1/2 watt TIP31C DISSIPATION.

NOW YOU NEED to change the POWER MOSFET with the proper breakdown ( at least 1.4 times the VN voltage) and proper current.

ALSO get a Capacitor at least 4.7 microfarads with a V breakdown of 1.4 times the VN and place from the UPPEAR Lead of the Dump resistor to Ground.

Good luck

Nando