Simple Voltage Divider

When building a circuit, there may come a time when you only need a fraction of the voltage that is in the input.  Dividing this voltage is simple and requires a few basic components.

You are going to need two resistors of the same value, and a battery.  The reason we are using two resistors of the same value is so we will cut the voltage in half.  Later I will describe the equation used to calcuate the resistor values needed to achieve a different output such as 3/4 of the input, or 1/8. 

I was inspired to write this instructable because I needed 4.5 volts to charge batteries in a circuit, and I had a rechargable 9v battery.  I used two 1k ohm resistors to dial this down to the 4.5v needed for my project.  It worked perfectly and I still have the circuit now.

As you can see in the picture, I used two 2 ohm resistors, but you can use any value resistors to cut the voltage in half as long as they are equal.

Here is the equation to calculate the Output voltage and the resistor values.

Output Voltage=  r2    
                            r2+r1   x Input Voltage

I hope you enjoyed building a voltage dividing circuit, and questions and comments are appreciated so please post them below!

-Doctordv

Comments

[splud]

This article showed up in a feed for me recently, and I was taken aback by the anchor image with 2 Ohm resistors for a resistive divider: that is for most intents and purposes going to behave as a dead short.

It is important to note that generally, resistive voltage divider circuits are useful for reference voltages, not anything of significant current (which I'd personally qualify charging a battery as, rechargeable coincells perhaps being an exception). That fun anchor image with two 2 Ohm resistors: I = V/R = 9v / (2 + 2) = 2.25 AMPS would be flowing from Vcc to GND. The power dissipation wattage (not even factoring a desire to pull current to something hanging off the 4.5V tap) would be P = IIR = 2.25 * 2.25 * 2 = 10.125W PER RESISTOR (here I used the resistance of just one resistor, because each of the resistors would individually need to handle that much dissipation - if you had a config where the resistors were unequal, you might find one resistor needs to dissipate more than the the other, and perhaps you need 5W for one and only 3W for the other - math it out). That's quite a bit of heat, and given that the standard hobby resistor is just 1/4W (0.250 Watt), that is over 40x the max rated dissipation of the type of resistor most hobbyists are likely to have in their kit. Even if the original supply had been a 9V wall wart, chances are its voltage would drop significantly under that load if the PSU didn't trip on over-current altogether. If you instead used 1K resistors: 9v / (1000 + 1000) = 4.5mA constant draw, and each resistor would dissipate just 20mW - but you'd also have virtually no current available to the load. To be fair, the written article mentions 1K, but that anchor image is quite something else, and in another paragraph, the article indicates the author used 2 ohm.

The equation for output voltage from a given divider bears reformatting, since as presented in the article, there is no diagram associating R1 and R2 to specific resistors (and when they are not equal, it is important which is which), and the multiplication by Vin is applied to the result of the resistor math, not just to the divisor as the article formatting might be interpreted to be, thus, let's use grouping to make it clearer:

Vout = ( Rhi / ( Rhi + Rlo ) ) x Vin

Where Rhi is the upper resistor (generally connected to Vcc, positive supply voltage), and Rlo is the lower resistor (generally connected to GND, or battery negative voltage). A rule of thumb here: if you are seeking an output voltage that is above half of the input voltage, Rlo should be the larger value resistor, and if below half, Rhi should have the larger value of the two.

A complication to the math is that when you actually draw current at the tap point (versus it being used by a high impedance input as a reference voltage say into an OpAmp), you are introducing another path to ground, with an additional load which is often modeled as a resistive load, which is generally parallel to Rlo, which will effectively reduce the low side of the divider, which will lower the output voltage. Yet another reason why resistive dividers are best limited to producing reference voltages.

As a rule, resistive voltage dividers are best used with fairly high value resistors to limit the current (and consequently power dissipation), and are most often used with say an ADC on a microcontroller (where the voltage is serving as a feedback, and the divider is needed say because the voltage you're measuring exceeds the input voltage of the µC - for example, the µC is itself controlling whether a battery is actively on charge or not, but the µC voltage itself is 3v3 - it cannot directly measure > 3.3 volts), or as a voltage input (again, not power supply) for an OpAmp, comparator, or an adjust pin on a linear regulator, such as an LM317.

An OpAmp + voltage divider, a voltage reference (a zener diode, a TL431, or other adjustable precision reference) and transistor could provide you an adjustable voltage for a project, where the transistor could be affixed to a heatsink if necessary. It would also be easier to find a transistor rated for the power dissipation, and since the resistive divider would not need to output much current at all (being used for a reference voltage only), they could be common 1/4W (or substantially smaller really) rather than some higher wattage one that wouldn't be in a typical parts kit (imagine needing to stock a bunch of 3, 5, 10W resistors in multiple values) . Undoubtedly, more parts, but actually capable of delivering current and regulating voltage, and the parts you would need to vary with the individual application wouldn't need to be special. As VooDoo magic as an OpAmp sounds, they are versatile ICs.

Another downside to trying to use a voltage divider for a supply: if the input voltage varies, so does your output (which is desired when using a divider for sampling/feedback, but not as a supply). This would certainly be crucial if you had a circuit and you wanted to be able to power it from 9V or 12.6V (two common voltages), or even a small solar module which could vary depending on how much sunlight it is receiving. The divider ratio (50% in this example) would remain what it is, and consequently you would get 4.5V and 6.3V respectively. Equally important - if you're running from battery then the battery voltage is dropping over time as you deplete its charge, after some use (especially if there's high dissipation), you may have
say just 8.6V on the cell, at which time you would have only 4.3V on the output rather than
the originally desired 4.5V. The converse is true as well - very fresh batteries can slightly exceed their rated voltage (or a 12V automobile circuit could be almost 15V when the engine is running and the alternator is charging the battery). Taking the 9V example, a fresh from the package cell could be 9.1V, resulting in a 50:50 divider presenting 4.55V, which might exceed the voltage your circuit can safely manage. In either case there may be adverse impacts on your circuit. 9V through a proper low dropout linear regulator would at least maintain the desired output voltage (4.5V in this example) until the battery voltage dropped to say about 6V or perhaps lower (at which point, the 9V battery is considered thoroughly depleted anyway), and also burns off power with a linear relationship with how much current is being drawn through Vout.

More bad news with this approach, specifically relating to the suggestion that it was being used to charge another battery (but admittedly, not detailed in this article): What happens when the supply battery, subjected to the divider results in a lower voltage than the target battery? How about the impact on the target battery which is presumed to be connected to the divider output and circuit ground? Rlo is always in circuit, so the to-be-charged battery would be DISCHARGING over that. It would be necessary to use a diode with anode at the divider output and cathode at the charge battery + (which results in a forward voltage drop across the diode which needs to be accounted for in the divider math), otherwise the target battery will discharge to ground, which is probably not desirable. This needs to be considered - a normal load - a little motor, a µC, LED, etc wouldn't be a voltage source and wouldn't risk discharge - but a battery (or even a capacitor) would.

Bottom line: resistive divider circuits can be handy, but it is important to understand their current limitations and where they are appropriate to use - generally not for powering a circuit. Work the math.

[Donald Stables]

how much heat does resistors create

[Jack A Lopez]

P=V*I, and for a resistor, V=I*R, because of Ohm's Law, so the power dissipated (as heat) by a resistor is:

P = V*I = I^2*R = V^2/R

That result is general. So if you have some network with a bunch of resistors in it, you can calculate the power dissipated by each resistor, provided you know the voltage V across (or current I=V/R through) each resistor.

[youn_link]

this circuit is comonly used to generade a simetric supply; the ground is bad, this have to be were the 4.5v is, and the ground one will be the negative of the batery also is common use a couple of capacitors to keep the volts

[lemonie]

but you can use any value resistors to cut the voltage in half as long as they are equal.
9V, 4 Ohm; if the battery will supply 2 amps you'll be generating 18 watts of heat in this.

How much current do you get from your charger-divider?

L