Simple electronics question - voltage divider?

How do I get 1.5v off a 5v rail?

I have a circuit that switches on a 5v rail that I want to use to power a 1.5v device. Do I use a voltage divider between 0v and 5v? How would I do this?

Embarrassingly simple - but I had not luck with the reading books or internet.

It’s all explained here:
http://www.doctronics.co.uk/voltage.htm

A far better option would be a regulator, particularly if you’re unsure of the supplied current. Here’s a sample circuit:
http://www.zen22142.zen.co.uk/Circuits/Power/lm317.htm

I don’t know the current, but a voltage regulator seems like overkill. Won’t the current depend on what is hooked up to the 5v?

Using the formula in the link R[sub]T[/sub] = 0.7/0.3 R[sub]B[/sub].
So I know the ratio, what about the values? Something like…

R[sub]T[/sub] = 2.3k ohm
R[sub]B[/sub] = 1k ohm ???

or a different order of magnitude?

Your R values work, but the amount of usable current you can pull is limited. Effectively, you have a 2.3 kOhm limiting resistor in series with your load AND a 1 kOhm load in series with it, otherwise the voltage drop across the load will be too large and you’ll no longer have 1.5 V available. If your load current is VERY small–say a few milliamps max., you might get away with it. Driving a single LED, for example. I won’t go into minimum-loss resistor voltage divider design considerations, because you’re probably better off with a voltage regulator anyway. It’s simple to work with, and the 78xx series of regulators are cheap and can source up to 1 amp without external pass transistors.

Q.E.D. is correct.

Voltage dividers aren’t as simple as they look. Sure, they “divide” a voltage. But they’re not regulated. In order to make a voltage divider work properly, you need to know the following:

  1. Source impedance.
  2. Source voltage.
  3. Range of load impedance.
  4. Acceptable load voltage range.

This is exactly why resistor-network voltage dividers are used almost exclusively in bridge circuits, where all these parameters can either be carefully controlled or included as design considerations. Bridge circuits are used whenever accurate measurements are necessary. The Wheatsone bridge, which is used to measure resistance very precisely, is a good example of this type of circuit.

I hope that 1.5 volt device isn’t a diode. If it is, you want to current limit it, not try and drive it with a constant voltage. Diodes go from dark to poof within a very narrow voltage range. You don’t want your LED to become a DED.

If it’s a 1.5 volt lamp (like a flashlight bulb) then you can just use the lamp as one resistor in the voltage divider, and all you need is another resistor.

For most other things, you want a voltage regulator, as previous posters have said.

The problem with a voltage divider is that the load is in parallel with one of the resistances of the divider, which means that variations in the load cause the division of the voltage divider to vary. You need to have the resistances of the voltage divider be much smaller than the effective impedence of whatever device you are driving for a voltage divider to be effective, and then you often end up wasting a lot of energy as heat because of this.

The 1.5v device is one of those little ‘tunes in a chip’ with a little speaker. The idea is that a RF reciever switches a relay which turns on the melody for a few seconds. I have the reciever with the relay already running some LEDs. There is another free relay I want to use for the melody.

Looks like I should buy a regulator.

I think a regulator is the way to go. If you insist on a voltage divider you have to find out what the current of your chip is while it is playing the music. However you probably ought to have some idea of how much current the chip uses in order to get the right regulator.

Then you figure an equivalent resistance of the chip by dividing that current into 1.5 volts. Make the total resistance of your divider equal to 1/2 that. The bottom resistor of the divider should be 1/2 the top resistance. In other words.

Rtop + Rbottom = Requivalent/2

Rtop = 2Rbottom

If you proportion the divider this way the voltage when the chip is playing is 1.5. If the chip draws no current when not playing the voltage across the chip will be 1.667

I don’t know what resistors cost these days, but the regulator might not be a lot more than two of them.

And by the way, you will probably have to use at least 1 W resistors. If the chip takes 25 ma when playing the dissipation in the top resistor is about 1/2 W. Another good reason to use a regulator.

Been thinking about it. Here are some options:

1. Use an adjustable voltage regulator.

Advantages: A fairly inexpensive and common approach. I think even Radio Shack carries the LM317.

Disadvantages: You’ll need to add a couple of “programming resistors,” plus a protection diode. And all regulators specify a minimum voltage across the regulator (V[sub]in[/sub] – V[sub]out[/sub]). The LM317, for example, is usually setup to ensure V[sub]in[/sub] – V[sub]out[/sub] ≥ 5 V, which means the input voltage should be ≥ 6.5 V if the output voltage is 1.5 V. But I think the LM317 can be operated with as little as 3 V across it (i.e. V[sub]in[/sub] – V[sub]out[/sub] ≥ 3 V), which means you can probably use it to knock 5 V down to 1.5 V without a problem.
2. Use a 1.5 V regulator. One example is the National Semiconductor LP3992.

Advantages: Most precise; the regulator is trimmed to provide a precise 1.5 V. It also does not require “programming resistors.”

Disadvantages: Availability will be a problem; you definitely can’t buy this at Radio Shack. Input requirements are somewhat finicky. (The LP3992, for example, doesn’t want to see an input voltage above 5.2 V.) And the package might be surface mount, which means it will be difficult to solder.
3. Use 5 silicon diodes in series.

Advantages: Very simple. Very cheap.

Disadvantages: Unlike a voltage regulator, there’s no closed-loop regulation. But the regulation is better than a voltage divider.

A “tune in a chip with a little speaker” will use hardly any current if it’s designed to run off a little button cell, so a low-power shunt regulator (like a zener) will suffice. Trouble is, zener diodes are generally only available down to about 2.7V.

I’d be tempted to use two forward-biased signal diodes connected in series as a shunt. Biased at 5mA two series 1N4148s (say) will drop 1.2V to 1.4V, which should be plenty to drive a 1.5V low-current circuit like a tuneful chip.

To do this, connect a 750 ohm resistor (any practical wattage, the dissipation is negligible) in series with the two series diodes, and tap your 1.5V supply from the resistor/top diode junction. Ideally you’d decouple the supply with a small ceramic capacitor (usually 10nF to 100nF) in parallel with the 1.5V output, but in this application you should get away with omitting this.

The LP3992 is in stock at Digi-Key, a giant on-line electronic parts supplier. They also have a very good search engine - they’re the google of electronic parts (for the parts they stock, at least). If you type “voltage regulator 1.5v” in their search box, you’ll pull up all 465 suitable devices! Most of them are surface-mount devices, though, so be prepared for some small parts.

If your chip can operate at 1.25V, you have a lot more options - that’s a common voltage reference value. A good choice would be the ZR431LC02L, a 1.25V reference in a nice TO-92 package (looks like a small transistor). It’s $1.13 each at Digi-Key.

If your chip can take 1.8V, you could just use a 1.8V zener diode - just feed it with a resistor to the 5V rail (resistor sized to handle the current drawn by your chip). This isn’t the most efficient, but is nice and cheap if the current isn’t high (in which case the efficiency may not matter). This is also available at Digi-Key.

A couple of other dirty tricks:

Use a separate 1.5V battery for the chip -since you’re using a relay, you could have it switch in a separate battery just for this. If the battery pack feeding the 5V supply is composed of multiple cells, you could pick off one for this.

If an LED is also lit during the playback time, you might be able to parallel the chip across that, assuming the LED drive current is high enough (maybe not if your chip is driving a speaker). This would probably require some tweaking if the LED voltage is too high (maybe a diode in series with it to drop some voltage to the chip), but could be an option if you didn’t have a relay output.

If it were me, designing a one-off project, I’d either pick off a single cell in the battery pack (assuming one is available), a series bunch of diodes like Crafter_Man suggested, or use a 1.25V regulator (since I already have a bunch handy!)

Arjuna34

Using a Zener diode & resistor works well when you need to generate a reference voltage (e.g. an offset voltage for an op-amp circuit). But IMO a Zener diode and resistor is not a good technique when you’re tying to power something. You either have poor regulation, and/or (because it is a shunt device) the efficiency is poor.

My advice is to go with a LM317, which (I believe) you can get at Radio Shack. Or you can go with the “quick and dirty” approach and use five silicon diodes in series.

I agree. The LM317 is indeed available at Radio Shack and really is the proper solution to the problem.

http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm

For V[sub]out[/sub] = 1.5 VDC,

R[sub]1[/sub] = 100 Ω
R[sub]2[/sub] = 20 Ω

R[sub]1[/sub] and R[sub]2[/sub] were chosen to ensure the regulator will output at least 10 mA, even with no load. But this comes at a price… when you’re driving a load, R[sub]1[/sub] and R[sub]2[/sub] will always be wasting power (18.8 mW). If you will always have a load connected to the output of the regulator, and if you’re confident the load will always draw at least 10 mA, then I would increase the values of R[sub]1[/sub] and R[sub]2[/sub] to increase efficiency. Consider increasing each by an order of magnitude, i.e.

R[sub]1[/sub] = 1000 Ω
R[sub]2[/sub] = 200 Ω

You may even be able to go higher.

I love over-engineering as much as the next engineer, but remember, this is a little circuit to power something like a tuneful birthday card, which probably uses a few hundred microamps at most. Zener-type shunt* regulators are inefficient under light load conditions, but so what? There’s plenty of power to spare, and this particular toy tune circuit doesn’t require rock-solid regulation. One resistor and two signal diodes will work as well as anything else, and it’ll cost about 10 cents and takes 30 seconds to make. You could use a bandgap diode and an instrumentation amp with foldback current limiting. It’d be nice, but expensive and unnecessary.

If you want an efficient (90%), cheap and fiendishly clever 2-transistor regulator, you could do worse than the hippyware circuit from Roman Black.

You don’t have to use zener* diodes; at a pinch you could use forward biased diodes for small voltage drops, or even an LED if you want your circuit to glow.

**What are called ‘zener’ diodes are generally avalanche diodes above about 5V, and only true zener diodes below about 5V. True zeners have a negative temperature coefficient, while avalanche diodes have a positive temperature coefficient. A bit of both is used for 5V (approx) ‘zeners’, and the temperature coefficients for this unique diode cancel each other out.

Fridgemagnet: Yea, a Zener will work. But it is much less efficient than an LM317. The adjustment pin on an LM317 only consumes 50 uA, while at least 5 mA are typically needed to bias a Zener. If this is a battery-operated device, then every milliamp is important.

Here’s an updated list:

1. Use a series resistor.

Advantages: The simplest and cheapest method. Efficiency is good (though certainly not as good as a switching converter).

Disadvantages: Very poor regulation. Will only work well when the load resistance doesn’t change and the source voltage doesn’t change.

2. Voltage divider.

Advantages: Simple – only 2 resistors needed. Better regulation than using a series resistor (#1 above).

Disadvantages: Input and output regulation can be good if you make a “stiff” voltage divider, i.e. if you choose low resistance values for the two resistors. But this has the drawback of significantly decreasing the efficiency of the circuit.

3. Shunt Zener regulator.

Advantages: Simple – only 2 components needed. Good input and output regulation. Better regulation than using a resistor-resistor voltage divider (#2 above).

Disadvantages: The Zener requires current to operate. So this solution is less efficient than using a series regulator.

4. Use 5 silicon diodes in series.

Advantages: Very simple. Very cheap. Good output regulation.

Disadvantages: No input regulation. Consumes quite a bit of board space.

5. Use a Zener diode in series.

Advantages: Very simple. Very cheap. Low parts count.

Disadvantages: No input regulation.

6. Use an adjustable voltage regulator.

Advantages: A fairly inexpensive and conventional approach. Even Radio Shack carries the LM317. Excellent input and output regulation. Temperature and over-current protection. Almost the same efficiency as using a series-regulator (#1, #4, and #5 above).

Disadvantages: You’ll need to add a couple “programming resistors,” plus a protection diode.

7. Use a 1.5 V regulator. One example is the National Semiconductor LP3992.

Advantages: Most precise; the regulator is trimmed to provide a precise 1.5 V. It does not require “programming resistors.”

Disadvantages: Availability may be a problem; you can’t buy this at Radio Shack. Input requirements are somewhat finicky. (The LP3992, for example, doesn’t want to see an input voltage above 5.2 V.) And the package might be surface mount, which means it will be difficult to solder.

8. Switched-mode boost converter.

Advantages: The most efficient approach.

Disadvantages: Complicated. High parts count. Noisy.

Since we’re in GQ:
[nitpick]
Going from 5V to 1.5V involves stepping-down, so it would be a buck converter, not a boost converter.
[/nitpick]

I vote for the LM317. The LM317T has a typical dropout voltage of less than 2 Volts at load currents less than 1 Amp, so going from +5V to +1.5V is fine. The low average current draw of a “tune-on-a-chip” device means that efficiency shouldn’t be an issue.

Of course. :smack: Was typing too fast.

I agree. The LM317 has the best combination of simplicity, efficiency, and regulation. The only way to get more efficient is to implement a switching buck converter. The ADI ADP3050AR would do the trick. Though for this application it may not be warranted; if the load only consumes a few milliamps, I would go with a linear regulator.