# How to determine what resistor to use?

• posted

Hello,

Background: We need some method of discharging a DeWalt 18v battery. The folks at sci.chem. electrochem.battery came up with a circuit design. We used a Bosch automotive relay with a 330ohm resistor in series. Our goal is to get the relay to open up at 15v. As this is the low limit that the battery can safely be discharged to. However, our first attempt didn't work out right. The relay didn't open up at 15v. We finally disconnected the battery at 14.98v.

Question: Is there a way to determine what size resistor(s) we need to get the relay to open up at 15v?

None of us are electronic folks. A suggestion from the sci.chem.electrochem.battery folks was to measure the resistance across the relay. If we hooked up the meter correctly (we used the upside down horseshoe setting), it looks like it's 68.4. We also check the resistor and got 74.8.

TIA, Mark

• posted

Trial and error for DC relays is the only option, A 24 volt coil may close at 13 and open at 7, each brand and voltage rating will be different.

I have a circuit design at work, that I am currently prototyping using zener diodes and darlington transistors to make a 24 volt relay open and close at precisely 22.6 volts. I really wanted it a little higher, but I have to live with what other people make for components. Otherwise the circuit could get really complex.

I'll look at it m>Hello,

• posted

I think a better method would be to us a variable supply to determine the point that the relay drops out ( Idrop or Edrop)

Diodes can then be placed in series to set the dropout point in .6V increments or Zeners can be used to lower the parts count.

As you have discovered most "15" volt relays will pick (energize) at less then their rated value and drop (deenergize) at a somewhat lower value. automotive relays generally will pick at less then the 12.6 volts of a car battery.

You might consider using a common variable voltage regulator with the output connected to the low side of the relay coil. with the test voltage set at the desired amount raise the supply voltage until the relay drops out.

commercial battery discharger / cyclers are often designed with a solid state "constant current" load rather then a simple resistance. this maintains a constant rate of discharge rather then a tapering one as battery voltage sags. this makes it easer to calculate mAH (mili ampere hours) by simply timing the discharge cycle.

• posted

Give your kid a chunk of 2x4 and some drill bits and let him or her make wood shavings until the drill doesn't run anymore? :) (Nope, I Googled you up... hmm... give the new trainee a fresh blade and the Chief's car? Another option would be to see if DeWalt makes a flashlight that fits your battery; rotate the batteries into the flashlight and make sure the flashlight gets used regularly.)

[discharging a battery and getting a relay to shut off at a certain voltage]

Yes, but not always by pencil and paper. The voltage that the relay needs to operate will vary with the design of the relay and even from unit to unit of the same model relay. If this is the typical 1" (2.5 cm) cube automotive relay with four or five prongs sticking out of the bottom, rated at 12 V nominal, I have found that these _usually_ pick up (close) at around 8-9 V and _usually_ drop out (open) around 6-7 V. So, you need a resistor that will limit the voltage at the relay to around 6 V or less when the battery is at 15.0 V.

If your reading below of 68.4 ohms is correct, and if you assume you have to drop the voltage at the relay to 6.0 V to get it to open up, you can calculate:

At 6.0 V, the relay will draw (6.0 V / 68.4 ohms) or 0.0877 A of current. This same current will pass through the series resistor. You want to drop (15.0 V - 6.0 V) or 9.0 V across the resistor, so the resistor should be valued at (9.0 V / 0.0877 A) or about 102 ohms. This resistor will be dissipating (0.0877 A * 0.0877 A * 102 ohms) or 0.789 watts of power, so it will need to be rated 1 watt or more. A Radio Shack

271-135, 100 ohm 10 watt resistor, would be plenty.

(As an aside, the 330 ohm resistor should have been more than plenty - that is, the relay should have opened up well before the battery got down to 15.0 V. Did you have it hooked up correctly - something like this: battery positive---330 ohm resistor---relay 86 battery negative---relay 85 or this: battery positive---relay 86 battery negative---330 ohm resistor---relay 85 ? If not, hook it up like this and try again.)

If you want to get it closer, probably the simplest thing to do is to buy a variable resistor (aka potentiometer, pot, or rheostat) and use it in addition to the fixed series resistor. This lets you adjust the relay operating point. Assuming your 68.4 ohm reading is correct, and assuming that the relay will open up at 6 V, you'll need something in the ballpark of 75 to 125 ohms at about 0.8 watt. At Radio Shack, buy a 271-265 25 ohm 3 watt rheostat, a 271-1109 150 ohm 0.5 watt resistor (pack of 5), and a 271-1105 47 ohm 0.5 watt resistor (pack of 5). Wire it up like this - "-" and "|" are wires, "*" is a connection: (use fixed width font)

one end of pot---*---150 ohm resistor---*---battery positive | | *---150 ohm resistor---*

middle of pot---relay 86

other end of pot---nothing

relay 85---battery negative

Put the pot in the middle of its rotation. Hook it all up to a battery that you have previously discharged to be at 15.0 V. Turn the pot back and forth... you should hear the relay click as it operates. Turn the pot to where the relay just drops out (opens) at 15.0 V, and then turn it a few degrees more for insurance. That's a good setting for the pot.

If you can't get the relay to drop out (open), you need to add more resistance to the circuit with the 47 ohm resistors you bought. The only thing that you'll change is the top line of the above drawing:

one end of pot---*---150 ohm---*---*---47 ohm---*---battery positive | | | | *---150 ohm---* *---47 ohm---*

Do the same thing... hook it up to a battery that is at 15.0 V and turn the pot back and forth until you find the spot where the relay just drops out (opens), then a little more.

Another way to reduce the voltage to the relay (as has been mentioned) is by using diodes. An advantage here is that the voltage drop doesn't depend on the current, like it does with a resistor. Each diode has a fixed voltage drop of about 0.6 V. At Radio Shack, buy two packages of

276-1122, 1N914/1N4148 diodes. At a guess, you'll probably need to drop about 9 V, which is about (9 V / 0.6 V) or 15 diodes in series.

Notice that the diodes have a stripe on one end. In the schematic, ">|" is a diode; the "|" end of the diode is the end with the stripe, and the ">" end of the diode is the end without the stripe.

Wire it up like this:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 battery positive--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|---relay 86 battery negative---relay 85

In this case, start with a fully charged (18 V) battery. Hook up the above circuit and start discharging the battery - keep the voltmeter connected across the battery so you can watch the voltage.

If the relay opens up before the battery gets to 15.0 V, you need fewer diodes in the string. For example, if the relay opens up when the battery voltage is 16.0 V, you need about 1.0V less of voltage drop - try taking two diodes (1.2 V total) out of the string. Hook the relay back up to the battery and see if it picks up (closes) - if it does, keep discharging the battery, with the smaller string of diodes hooked up with the relay, to make sure the relay drops out at around 15.0 V. If it doesn't close, charge the battery back up to 18.0 V and try again.

If the battery has discharged to 15.0 V and the relay still hasn't opened up, you need more diodes in the string. Disconnect the relay, add one diode to the string, charge the battery back up to 18.0 V, and try again.

That sounds about right. The ones I have measured are around 90 ohms, give or take. The horseshoe is the Greek letter omega, the symbol for resistance in ohms.

That sounds wrong for a 330 ohm resistor. Were you measuring across the two ends of the resistor only, without the battery hooked up? When you are measuring resistance, you can't have any other source of power in the circuit (your meter has an internal battery that it uses to make the measurement) or your readings will be wrong.

You can buy chargers that do all this stuff for you (

, but it sounds like you'd rather tinker around with it, which is fine. You might also post your question to sci.electronics.basics for more ideas.

Matt Roberds

• posted

That may be an understatement! His resistor is 330 ohms, yet measures 74.8? But I don't think there is any value of R in series with his relay that will work. Check your numbers. Using them (minimum pick up 8V), I got a minimum pick up current of a bit over 117 ma (8/68 = .11764). WIth your 100 ohm R in series and even if the battery is at 18 V, the relay won't get enough current to energize: 18/168 = .10714.

If the OP's R was really 330, the relay would never have energized. There would be a bit less than 3 volts across it, even with the battery fully charged. 'Course all of this is based on a 68 ohm relay coil - and that value is questionable.

Ed

13 14 15

positive--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|--->|---relay

86
• posted

I was modeling the relay coil as a resistor and shooting for a certain voltage, letting the current do what it wanted. I have played with one of these relays across a variable voltage supply, but I haven't tried regulating the current through one. Obviously limiting it to (say) 10 mA probably isn't enough to operate it, but I don't have a good feel for what they really take.

Another possibility: many of these automotive relays have a resistor or a diode across the coil to help limit spikes when the relay switches off. Having a resistor in parallel might explain the coil resistance value. If it has a diode in series, and the diode is beefy enough, and the OP was unlucky enough to get the connections wrong, he might have been discharging the relay through the diode...

Matt Roberds

• posted

The problem is that they don't operate like a resistor. They require higher current to pull in than they do to stay pulled in. And being essentially electro-magnets, they don't care what voltage they "see" - they operate on current. As long as the current is within spec, the relay coil is happy. (Within reasonable limits - we're not talking about putting a 12 volt relay in a 10000 volt circuit where the relay coil insulation would breakdown.)

After my earlier posting, I experimented with a number of relays in my junk box, including one cube type automotive relay, after posting. None of them would work with a series resistor to do what he wants. I may have the same relay he has. My automotive cube relay measures

64 ohms. It pulls in at 56 mA and drops out at 42 mA. The smallest resistor you can use and still have it pull in at 18 V is ~257.42 ohms. The problem then is that it will draw 46.667 mA when the battery drops to 15, and won't drop out. In fact, it won't drop out until the battery reaches 13.5 volts. If you compute the other way, first figuring out what resistor to use to get it to drop out at 15, you need a 293.14 ohm series resistor. The problem there is that it will draw only 50.4 mA, which is not enough to energize the relay.

He needs a different circuit than just his relay and a resistor. I believe he needs a different circuit than just about *any* relay and a resistor. He got sucked into the simple concept of a resistor + a relay would work for him. It sounds elegant, but it's bullshit. If you analyze the requirement, it becomes apparent: He wants to discharge an 18 volt (nominal) pack to 15 volts. He specified that 14.98 was too low. The pack that he wants to discharge could be at anything between 15 and 18 volts, so his discharger needs to energize at anything above 15 volts and de-energize (ie draw NO current) at anything below 15 volts. A relay and series resistor won't fill the bill - and will merryly discharge his battery pack to 0 volts, whether the relay is energized or not. If he is going to use the relay/resistor combo, at a minimum he needs to be told to connect it through its own normally open point so that it switches itself out of the circuit when the relay drops out. He then needs to add a switch, or change to an open frame relay which he can manually operate. He'll also have to settle for a discharger that works at the limits of the circuit design, instead of at the optimum points. It's a BAD solution.

I sent him E with a circuit diagram for a proper circuit, which I also posted on a.b.s.e. He hasn't responded. The circuit is simple, but may be beyond his experience.

.
• posted

You're right of course about the difference between pickup and dropout current. But one technique that we use on large DC contactor's for motors is to have a spare NC contact on the relay wired across the resistor. When first energized, the contact shorts the resistor so you get good pick up current. And once the thing picks up, the contact opens inserting the resistor into the circuit. So now the drop out will work where you want it. Size the resistor for the drop out current.

Of course, this uses up a relay contact, so now you need a different relay. Or two relays working together, or????

daestrom

• posted

Right! Somewhere the op got the idea that a resistor in series with a relay is all he needs to do what he wants. Not so. If a relay and series resistor are to be used, he needs: 1) a means to completely open the circuit when the relay drops out, so that it does not continue to draw current; 2) a means to include "dropout" resistor; and 3) a means to energize the relay.

Here's a way that does all of that: Vcc--resistor--normallyopenrelaycontact--coil--gnd Then a second path to the coil to start the discharge: Vcc---normallyopenpushbutton--coil--gnd

This is still BAD, however. The discharge relay is energized on the hairy edge as Vbatt drops. A slight mechanical vibration can cause it to drop when it is close to the drop out point. The relay point cannot be used as a tap point for a discharge load, as connecting a load there will increase the drop through thru resistor, thus making the thing drop out above 15 volts. And it doesn't draw enough current in the absence of an external load to discharge the battery in a reasonable time, nor does it have any means of indicating that the discharge is completed.

The function screams for a better approach. One way is a voltage reference/comparator to drive the relay, which is in the circuit sent to the op and posted on a.b.s.e.

Ed

• posted

you did pretty good if your voltage dropped off at minus.02Volts, but you want perfection as we all do };-)

I used to hammer these type of problems with my buddies at the basketball court after hours, you see all you need to do is visualize the devices as teensy weensy people (lets call them nanofolk) we all know that they will let through current =3D(electron folk) or they won't.

{basically, semiconductor theory., may apply to resistive matter & solid state devices.}

All you need to do is give a nano sized voltmeter to the most technologically inclined nanofolk in the batch, all you'll need is one of them with a tester or panel meter, hence, when the required 15.00vdc has been attained by your battery community the nanofolk inside your charger community will just stop letting the current through by blocking the other electron folk from passing. thus solving your resistive problem.just make sure you design a nice community environment with all the amenities for your nano folk tenants };-) otherwise they will slack off, or totally abandon your nano community leaving your inanimate battery charger on it's own imperfections again.

=AEoy

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