Shunt Regulator PC board layout

[deloiter]

For those who are interested, here is a circuit board layout for Chris Greacen's Shunt Regulator (http://www.homepower.com/files/shuntregulator18.pdf). I recommend that you use sockets for the ICs and the transistor. The layout is sized for 1/6 or 1/4 watt resistors, small 100v ceramic caps and the 1K pot is a ceramic top-adjust type. Solder the 22v zener right to the FET. To make the project more user friendly you could panel-mount the pot, switch and the LEDs.  The main issue will be getting the trace image to print out correctly sized for the IC pins. Everything is available from digikey.com.

Some additional part number info:

 18v zener - 1N4746A (1 watt)

 22v zener - 1N5358B (5 watt)

 Higher power FET option: IRFB3207 (75v, 180a, much lower "on" resistance and still cheap at $5 US)

 LM723N

 NE555P

Component side:





Trace side:

[seanchan00]

Hi Deloiter,

Since you raise this subject again please help out here. I still can't get my shunt regulator to work properly. Nando gave this test procedure which I have done to debug but I think there is a light mistake in the test.

Nando's Test Procedure.

by Nando on Sun Oct 2nd, 2005 at 12:53:43 PM MST

(User Info)  

SeanChan:

Let's try the test procedure for this ckt.

Are you saying that the ballast (SHUNT) load has an indicator light across it ?.

If so, check to see if the Diode 22 V 5W is warm and connected wrongly ( backwards) the band should go to the Source and the ballast connection point, the opposite lead to ground.

Then if right, take a WIRE from the ground and the GATE of the MosFET to make sure that the GATE is at Zero volts, then measure the Source of the MosFET, it should read the Battery voltage,( Junction of MosFET source and Shunt load).

Measure the voltage from Source to ground, it should be Batt voltage if around 1 to 2 volts, then you may have the MosFET wired wrong, placing the Source to ground and the Drain to Bat voltage, turning the intrinsic MosFET diode on.

The MosFET PIN out for a TO220 case is, when facing the MosFET, left pin is GATE, Center PIN is Source and Right Pin is Drain.

Also Center PIN should connect to the Heat sink back metal if not isolated.

Make sure that the PIN3 NE555 LED diode is properly connected, it has a very small flatten edge near of the wires -- that is the Cathode and should be going to Ground ( Negative side of Battery)

Measure PIN 3 NE555 -- it should be below 0.3 volts

Measure PIN 2 NE555 -- It should be higher than 2/3 supply voltage ( bringing this pin momentarily below 1/3 Battery voltage the output ( PIN 3 NE555 ) should go high for about 1 second.

And if you keep PIN 2 shorted to ground the MOsFET should turn ON ( PIN 3 ON) and continues ON for 1 second after removing the PIN 2 short to ground.

DO NOT Measure PIN 6 & 7, BECAUSE the ckt has a 10 Meg resistor source and may bring the voltage down below the 2/3 bat voltage.

Regards

Nando

Flux said:

If you sit the fet on the bench with its plastic side up and the leads towards you, gate is on the left, source is on the right and the drain is in the middle.

Drain goes to your dump load, source goes to the circuit ground and the gate goes to the 100 ohm resistor to pin 3 of 555 and also to your led.

So if Flux is right, and he usually is, then Voltage from source to ground cannot be Batt Voltage since it is connected to ground. My first bug was due to reversed drain and source connection as I interpreted the schematic wrong.

I have done the test and my results are as follows:

Diode 22 V 5W = OK.

Ground to FET gate zero V correct.

Ground to Drain = Batt Voltage.

PIN 3 NE555 = 0.3 V correct.

PIN 2 NE555 =  0.2 V, not two thirds Batt Voltage.

PIN 2 shorted to ground mosfet didn't turn on the load!

Ground to collector of 2n2222A = 2/3 batt V.

Between 10 K resistor and PIN 2 NE555 0.2 V. Due to the resistor eating up the voltage?

My 22V Zener is soldered direct to the drain of the FET as you advised.

Where's my bug now. Have I shorted the NE 555 ?

Any better test procedure to follow?

SeanChan (:<)

[Flux]

Sorry you are still having trouble, I thought you got it working.

It seems as though Nando has called the source the drain and vice versa, otherwise his test procedure is correct.

You seem to have the fet bit of the circuit working but it looks as though you may have trouble with the 555.

There is one issue that may be worth checking at this stage. You say that pin 2 of 555 is at .2v and collector of transistor is at 2/3 battery volts. this is odd as you have two 10k resistors in series, one at ground and one at battery volts so the junction(transistor collector) should be at half battery volts. Check the value of the 10 k resistors.

With pin 9 of LM723 low (Q1) turned off, collector of Q1 and pin 2 of 555 should be high.You say it is low, check .1uf capacitor on pin 2( most unlikely to be a problem). Check the track round pin 2 for short, if no problem it is most likely a damaged 555.

If your 555 is in a holder, remove it and check that pin2 voltage comes up.

For the 555 to generate a pulse, pin 2 has to be high, then taken low as Nando suggested. If it is already low for some reason you will not get a pulse when you short it to ground.

You can check the LM723 stage by adjusting the 1k pot and make sure pin 9 will go high or low depending on the setting.

Any more problems dont be afraid to ask

[seanchan00]

Hi Flux,

Thanks for your guidance. You are right as always. Collector is at half battery voltage.

My readings are as follows:

Batt V = 14 V

Collector = 7 V

Pin 9 with 1 K pot all the way up to decrease Voltage shunted = 4 V

Pin 9 with 1 K pot all the way down to increase V shunted = 0.4 V

555 removed from holder - Pin 2 leg = 13 V ie up as you said so.

Pardon me but I am not sure how to check .1uf Capacitor on Pin 2 and how to check the track round Pin 2 for short.

The problem is pin 2 is always low. Does that mean the 555 is gone?

SeanChan.

[Flux]

Don't worry about checking the capacitor and pin 2 for shorts.

You have proved that they are ok by removing the 555 and seeing the volts come up.

Try a new 555 that seems likely to be the trouble. If that fails then the trouble is almost certainly something connected to a pin or between pins that shouldn't be.

I am not sure how you have constructed it, but when using strip board if you have it in your part of the world, it is easy to leave a thin whisker of track shorting between pins so have a good look with a magnifying glass.

Good luck

Flux

[Flux]

Just one more thing to check, have a look at the voltage on pin 5 of 555.  This should be at 2/3 of supply volts. if for some reason this is low it may affect pin2.

A short, wrong connection or faulty capacitor to pin 5 would hold the reference low and may hold pin 2 down. I am not certain it would reflect on pin 2 voltage but it would prevent operation.

Flux

[seanchan00]

Thanks Flux,

Will replace a new 555 when I get it. May have to take one from one of my desulphaters as I can't but them locally.

On a related thread, If I charge two 12 V batteries in parallel but use them separately for different loads do I need to protect each battery with a shunt regulator each ? My basic understanding is electricity will flow to the least resistance and when one battery is fully charged the resistance goes up so it will naturally go to the undercharged( more discharged by load ) battery until both are fully charged when the voltage will rise and get shunted by one shunt regulator attached to either battery.

If the above set up is workable, will it also work if I divide my 6 rectifier's DC output (from 3 AC Star input) into two positive and two negative and send one set to each battery? That way my two batteries will be effectively separated electrically when the wind is not blowing. Since each the three AC lines are rectified by two 35 amp rectifiers I can divide the power equally so each output DC line will get 3 phase AC feeding it. Can I get away with only one shunt regulator in second set up?

SeanChan.

[seanchan00]

Just Checked. Battery V 14 volts. Pin 5 = 9 volts, as close to two thirds as my crude meter will read.

Thanks again Flux.

[Ungrounded Lightning Rod]

If I charge two 12 V batteries in parallel but use them separately for different loads do I need to protect each battery with a shunt regulator each ?

No.

The voltage of a battery varies with state of charge.  This is what causes the plates to charge all-over rather than in spots, and why you can parallel batteries (provided they're of the same composition, which usually means you should buy them from the same production run and make sure they have the same charge/discharge history by leaving them connected rather than using them separately, or if you do use them separately try to equalize how far you discharge them and at what rate.)

Two paralleld batteries are like two bays in a single lake.  You only need one adjustable spillway/shunt controller to regulate the water/charge level in both.

My basic understanding is electricity will flow to the least resistance and when one battery is fully charged the resistance goes up...

Actually, with the state of charge the VOLTAGE goes up, and the flow is proportional to the voltage across the resistance divided by the resistance.  It doesn't "take the path of least resistance".  It divides proportionately to the resistance.

(Lighting does "take the path of least BREAKOVER voltage" which is usually misstated as taking the path of least resistance.  Ionizing air causes its resistance to drastically drop, and with a current-variable resistance the usual simplifications don't apply.)

... so it will naturally go to the undercharged( more discharged by load ) battery ...

Batteries have both resistance and charge voltage.  So the charging current divides between them.  But the resistance is very low, so most of the voltage across each battery is the charge voltage and very little is the resistive drop from charging current.  So if the batteries are not charged to an equal voltage the charging current splits VERY disporportionately - and if the charge-level difference is large enough the more fully-charged one will actually DISCHARGE into the undercharged one and help charge it.

(Imagine those two bays were connected by a narrow trench, with the input water feeding the middle of the trench.  If one bay is lower level than the other most of the water will go that way, and if they're different enough the more-full bay will empty into the less-full bay as well.)

... until both are fully charged when the voltage will rise and get shunted by one shunt regulator attached to either battery.

That part is right.  (If you dump water from either bay it will suck water out of the other through the connecting trench.)

By the way:  The bay analogy tells you why you want paralleled batteries to be the same manufacturer, lot, and service history:  The voltage for fully-charged varies slightly with battery chemistry, which varies with manufacturer, with manufacturing lot, and with service history.  A slightly over-full battery leaks power rapidly until it's just full.  (Then it leaks more slowly...)  So if you directly connect two batteries that have different voltage at full charge the one with the lower voltage will suck power from the other and keep it from being fully charged.  This tends to lead to both sulfation and deeper discharge (with resulting greater wear) on the higher-voltage battery.

If the above set up is workable, will it also work if I divide my 6 rectifier's DC output (from 3 AC Star input) into two positive and two negative and send one set to each battery? That way my two batteries will be effectively separated electrically when the wind is not blowing. ... Can I get away with only one shunt regulator in second set up?

No.  If you do this you'll need two shunt regulators, one for each battery.  There is voltage drop in the diodes, too, and they are one-way valves.  So even if the shunt regulator is sucking on one fully-charged battery and that's dragging down the voltage at the mill, there will still be some charging current into the other.  Since the shunt regulator can't discharge the second battery through the diodes you'll overcharge it.

If you have batteries that aren't matched, though, this might be a good way to go.

A downside to this is that you need bigger diodes.  When the batteries are unequally charged most of your current goes to the lower one.  That was OK when your diodes were tied in parallel - they split the load (assuming THEY were balanced...).  But with them hooked up separately each set takes the full charging load of its attached battery.  When essentially all the mill's output is going to just one of them (the typical case when they're used separately and not fully charged) that set of diodes gets essentially all the current.

[Ungrounded Lightning Rod]

Actually, with the state of charge the VOLTAGE goes up, and the flow is proportional to the voltage across the resistance divided by the resistance.  It doesn't "take the path of least resistance".  It divides proportionately to the resistance.

Oops.  That last part was confusing:

It divides so that the current in each resistance is proportional to the voltage across the resistance divided by the resistance.

When the two batteries are at different voltages the two internal resistances - which see the DIFFERENCE between the charger voltage and the battery voltage - see wildly different voltages.  Assuming the resistances are about equal the current will divide proportionally to the amount the battery is down below the charging voltage.

With low output from the mill that means the current goes almost entirely into the lower battery.  With higher output the more charged-battery gets a progressively larger share, but never an equal one.

[Ungrounded Lightning Rod]

When I first looked at the circuit I wondered:  "Where's the hysteresis?  This thing will suck down the battery when it turns on and will oscilate."

Then I looked deeper at both the circuit and the writeup:  All the necessary hysteresis is provided by the 555 (which also provides solid drive for the output transistor).  The 555 keeps the load on for a tad over a second after the battery voltage drops below the setpoint.  When the battery has just gotten to full charge it gets one-ish second discharge pulses.  If the charging rate continues the pulses get closer together, until they are comin often enough to match the charging rate - or they merge and the load stays on full time if the charge rate is higher than the load's draw.  Battery voltage is tightly controled, and the oscilation is GOOD.

Sweet!

[Nando]

Heck, one thousand "hecks" !!!

This is the third time this year where I exchange Source and Drain positions.

My most sincere apologies for such mistake

Regards

Nando

[seanchan00]

Thanks Ungrounded Lightning Rod,

That's a lot of clearing up in my understanding what options I have. I am asking all these because I try to recycle and not waste old batteries poisoning our environment. I have two set of unmatched Trojan T-105's. If I use them together it will soon result in both becoming as poor as my recycled one(Golf Club discard showing 3+ volt each 6 Volt cell which I have desulphated and used for 2 years).

From your advice my best bet will be to let the rectifier alone and charged the two batteries in parallel so the rectifier won't be overloaded. My charging of the batteries will be by two separate sets of fused positive and negative wires. The positive wires will be connected through a set of 6 diodes of 5 amp rating paralleled to make 30 amps rating for each positive wire. My theory is that the diodes will allow the wind power to charge the batteries but when the wind is not blowing the two battery sets will be kept apart by the diodes. The battery sets are then connected separately to their respective end use with a fuse in between. Then the last thing to get it working properly will be to have 2 shunt regulators, one for each battery.

For my shunt loads I intend to use old car workshop reject batteries(desulphated) to charge as back up to my system together with the usual car tail lights in parallel. Diodes will prevent these reserve batteries from discharging to the lights when the shunt is off.

I am taking the NE 555 from one of my desulphaters to replace the one in my non-functional shunt regulator and really hope that solves the problem. This option sure make my debugging that more urgent.

SeanChan(:>)

[seanchan00]

Hi Guys,

Finally debugged the Shunt Regulator. It was the NE 555 as Flux indicated. Swapped the NE 555 from the desulphater and when I tested the desulphater for 45 milliamps to confirm everything is ok it blew the multimeter fuse. Lucky I did'nt connect it just touch the probe to the battery, saw the needle fly past 250 milliamp and immediately let go. The regulator load now comes on when I shorted Pin 2 to ground as in Nando's test procedure and remains on for a second after.

Now I am wondering how to calibrate so I accurately shunt at 15 Volt during charging. Is waiting for the load to come on then take the voltage of the battery, then adjust the 1 K pot by trial and error the only way? Seems very laborious. Any short cuts?

I will be making the next shunt regulator next.

SeanChan (:>>).

[Flux]

I am pleased you have got it working.

Unless you have a smoothed variable power supply you will have to set the volts on a battery. If you have a battery charger you can use that.

Set the pot at maximum then when the battery is at the volts you want to control at, set the pot down until it starts to dump. 15v seems rather high but may be necessary for your reclaimed batteries.

Flux

[Nando]

You need a good DVM and a small calculator

Measure LM723 pin 4 or 6 as accurate as you can.

Measure the battery voltage as accurate as you can.

Set the pot to about half way and measure its voltage as accurate as you can.

Assume the shunt trip point, for your used batteries try not 15 volts BUT 14.6 volts.

Take the pin 4 voltage (let's say it is 6.95 volts)

Divide 6.95 / 14.6 = 0.4760273 ratio = Rcal

Take Rcal * Battery voltage (let's say it is 13 volts)

so 0.4760273 * 13 = 6.188356 Volts

Place the DVM to measure the Pot wiper voltage and adjust it to read 6.187 volts (not 6.188356)-- there you are, your system calibrated to Shunt Load at 14.6 Volts.

With a DVM with 3 decimal places, you may set the trip point to 0.1-0.2 % accuracy.

Regards

Nando