I have a DC circuit powered by a 9v battery and I need to power a 1.5v timer from the same battery. Is there a simple way to reduce the 9v to
1.5v? I have seen the voltage divider circuit in the Forrest Mims book from Radio Shack: Two resistors in series are connected to the two terminals of a battery. V+ out is taken from between the two resistors, V- from the neg. battery contact. V out is determined by the values of the resistors. I have no problem doing the math, but wouldn't that circuit short out the battery?
Well, yes, that circuit will draw current from the battery all the time. Depending on the values of the resistors, this may degrade the battery quickly.
Another way is to set up circuit with the battery, a resistor and the timer in series. The appropriate value for the resistor depends upon the resistance of the timer. The idea is to pick a resistor that absorbs 7.5 v and leaves 1.5 v for the timer. This, too, is not the most efficient possible circuit since most of the energy from the battery will be consumed by the resistor.
Hi, Greg. Congratulations on getting the Forrest Mims book -- it's probably the best total newbie start to learning electronics. You can easily do what you're suggesting, and it won't short out the battery (view in fixed font or cut&paste to M$ Notepad): | | .--------. . | | | | | .-. | | | 750| | |I | | | | | | | '-' V |9V | | | +| o-----------. | --- | | | - .-. | | | 150| | | _ | | | | | |I1 / \\ |I2 | | '-' | ( T ) | | | | V \\_/ V | | | | | | | | | '--------o-----------' | (created by AACircuit v1.28.5 beta 02/06/05
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This will provide you with a steady 1.5V at the timer -- as long as it doesn't use very much current.
Look at the circuit above. I = I1 + I2. If I2 is near zero, then there's about 900 ohms load across your battery, and I = 10mA (Ohms Law
-- 9V / 900 ohms = 0.01 amps = 10mA). If you've got one of those small LCD watch battery clocks, they generally use about 1mA of current. That won't affect the voltage divider too much -- a 1mA timer load will bring the divider voltage down to 1.35V or so. You should still be OK here.
little timer isn't quite steady -- it probably requires more current in very short pulses for the timer crystal and the display IC. This is probably going to interfere with the timer operation. If you want to use this circuit, you really should go to Radio Shack and purchase a
10uF (microfarad) cap, and put it in parallel with the 150 ohm resistor like this:
Be sure to attach the lead of the cap marked "+" on the side of the circuit where the 750 ohm and 150 ohm resistors meet. This will work for a fleapower load. But there are two problems here.
First, if your little clock has a piezo alarm, current draw on the 1.5V is going to go way up when the alarm goes off. That will bring the voltage way down, and probably stop your clock. Even worse, if you've got one of those quartz clocks, their average current draw is more like
10mA, which is completely unacceptable, alarm or not. Not only that, but the quartz clock draws current in spikes at the moment the hands move, which will ruin things for sure.
The second problem is also a biggie. For the above circuit, you're burning 10mA to get a useful 1mA at 1.5V with 10% regulation. In other words, you're using 90 milliwatts of power (9V * 10mA) of power to get a useful 1.5 milliwatt (1.5V * 1mA). Your batteries aren't going to last nearly as long as they could.
If this is just an experiment, you should be OK here. If you're planning on running this permanently and don't own stock in Ray O Vac, you can get an IC from Radio Shack that will make your voltage regulator more efficient. This IC is required if you've got a quartz clock -- you might be able to make the ohms of the resistors a lot lower to get better regulation for a larger load, but the 9V battery can't crank that much current.
If you'd like, please feel free to post again for more advice. Be sure to describe what kind of timer you're using, how much current it needs if you know, and also if it has an alarm output and what type.
You might use a small 3 terminal voltage regulator, like the LM317LZ, with the output voltage set with a couple resistors.
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If you can live with 1.25 volts out, instead of 1.5, you need only a minimum load resistor. The data sheet shows a 240 ohm resistor from output to the ADJ pin, to handle the absolute worst case for minimum load current, but it will probably work fine with a 560 to 1000 ohm resistor there (for lower battery load). The data sheet says that the typical minimum load current is about 1.5 mA with a 3 to 15 volt input. That would be 1.25/.0015 = 833 ohms, especially true if this thing will never see its full rated temperature.
Hi, John. If the OP wants to go to Radio shack, he'll have to settle for an LM317T:
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Radio Shack page:
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Catalog #: 276-1778 $2.29 USD
The LM317 IC is a 3-pin linear voltage regulator available at Radio Shack, that will maintain a steady output voltage for a varying load. Not only that, but it will only use 3.5mA extra, instead of the 10mA we were talking about for your voltage divider. Not quite as good as the LM317L, but still OK for battery use.
If the OP gets one of these, he can follow the pinout on the package after reading the data sheets, and put this fairly simple circuit together:
He'll need a 360 ohm resistor, a 68 ohm resistor, and two 10uF caps (all of which he should be able to get at Radio Shack). And as John says, if your timer can live with 1.23V, you can forget the 68 ohm resistor, and just connect the ADJ pin and the other side of the 360 ohm resistor to GND. But whether the load is an LCD clock or a quartz clock, you really should include the 360 ohm resistor to GND -- either load is pulsed current, and you should always make sure there's at least 3.5mA going through the LM317.
The great thing about this circuit is, he will be able to get a steady
1.5V output over the entire current range of his transistor battery (from zero to 100mA). Also, he only "wastes" 3.5mA. This would mean that, for a 1mA load, he'll only need 4.5mA from his battery rather than 90mA.
The OP will always need the load resistor of 360 ohms for low output current. He needs a minimum current from the LM317 OUT pin of 3.5mA, or the output voltage will rise above its resistor-programmed point. If his load never goes below 3.5mA, he can use higher value resistors to set the programmed output voltage. And of course, if he only needs
1.2v, and his load current is greater than 3.5mA, he can forget both resistors entirely. But connecting it directly to GND is only if he only needs 1.2V.
Assuming the OP is using an LCD timer clock without alarm, he may only need 1mA. Using an LM317 may only then double his battery life, which may or may not be a good investment. But if he's using a quartz clock, or if he's got an alarm function, he probably won't have a choice, if he wants his battery to last a while. Getting good regulation for even a 10mA load with a voltage divider will mean cranking 100mA from the battery, which should kill it quickly.
Possibly the OP could describe his timer a little better.
Since your source is a battery and doesn't require filtering and if your current requirements are a only a few milliamps I would simply use a 7.5 volt Zener diode in series with the timer.
Unless your current requirements are low all the solutions presented here, including mine, are very wasteful of battery power. An efficient switcher design however would be a complicated solution to your problem.
Thanks folks for a very informative set of solutions. It looks like I'm making this more complicated than need be. It would be much simpler to add an AA battery in a holder for the timer.
The timer is not a quartz device. It's an inexpensive pocket timer with LCD display and a piezo buzzer alarm. I am going to disconnect the buzzer and the leads will go to turn on another circuit (powered by the
9v batt) via an SCR.
I don't know the current draw of the timer. I'm still trying to get my hands on some of them from an outfit called OzoneLabs in Canada. I can't get them to respond to my emails or phone messages. Little timers like these are hard to find (countdown from 99 hours).
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