How do you choose the resistors in a potential divider?
If I have a potential divider where R1 == R2 how do I choose the resistor value. Obviously any value will work but if I choose 1ohm the input will be heavily loaded. But what will happen if I choose 100k or 1M?
How do I work out which value to use?
Posted Oct 31, 2011, 8:29 PM UTC
67
You choose by how much input current the circuit using the divided voltage needs. The amount of current the divider can supply will be limited by the upper resistor. You need to make that resistor small enough so it can supply the amount of current required by whatever's using the output of the divider.
There are multiple ways to do this, but the easiest IMO is to just to decide that you're going to have equal currents moving through the bottom of the resistor, and coming out of the divider. Then, because you know the total current (2* I(bottom resistor)), and you know the voltage drop (difference between supply voltage and divider output voltage), you can just use V = IR to find the top resistor.
Let's do an example. Your supply voltage is 5V. Your voltage out of the divider is 1V. And you want 1mA of current coming out of the divider.
You choose the bottom resistor current to be the same as the divider output current, 1mA. Therefore in total, there will be 2mA going through the top resistor. (1mA out of the divider, 1mA through the bottom resistor.) So now we know I: I = 2mA.
How about V? Well, supply voltage is 5V. And Vout is 1V. So the voltage drop across the top resistor will be 4V. So we know V.
Now we just do V=IR:
This is the value of Rtop. Rtop = 2k.
How about the value of Rbottom? Well, we know the current through Rbottom - we chose that to be 1mA. And we know the voltage drop across Rbottom, it's 1V above ground, so 1V - 0V = 1V. Now we just do V=IR again...
So there you have it. Rtop = 2k, Rbottom = 1k. 2 mA of current flows through the 2k resistor into the middle of the divider. Half of it (1mA) flows out of the divider, and half of it (other 1mA) flows through Rbottom into ground.
Of course, this is the bare minimum. If whatever's on the output of the divider suddenly starts drawing 1.5mA instead of 1mA, the ratio of currents between the bottom resistor and divider output is no longer equal. Consequently, the top resistor is now wrong, and output of the voltage divider will crash down to 0.666 V. Similarly, if the output of the divider starts drawing only 0.5 mA the voltage will shoot up to 1.333 V.
This is a pretty lousy voltage reference. Any little variation in output pretty much kills it. We can improve on it, however. But it comes at a cost: we must increase the amount of wasted current through the bottom resistor.
Let's say that instead of a 1:1 ratio of current out of the divider and through the bottom resistor, we want a 10:1 ratio. There will still be 1mA coming out of the divider, but now there will be 10mA flowing through Rbottom. For a total of 11mA through Rtop. Same output voltage, 1V. So Vdiff across Rtop will still be 5V - 1V = 4V. V=IR again:
Rbottom? Easy. We know there will be 10mA flowing through it and it will have a voltage drop of 1V, so V = IR, 1 = 0.01 * R, R = 100.
So we set up the 363.63 Rtop and 100 Rbottom and our 1ma constant current draw from the divider. And sure enough, the output of the divider is 1V. However just as we predicted, there is 10mA flowing through Rbottom, and only 1mA coming out of the divider. So we're wasting ten times as much current into ground as we get out of the divider.
However, Rtop is smaller now. Consequently, it is less sensitive to variations in current. So if we increase the draw out of the divider to 1.5mA, the voltage only decreases to 961 mV. Only a 4% drop in voltage - which might be within the tolerance of chip that's designed to handle a 5% variation in input voltage. And what if the load current decreases to 0.5 mA? The voltage only moves to 1.04 V. Again, only 4% error.
So there you go. You can play with the math all day long to get whatever kind of error tolerance you want. But remember, it always comes at the cost of dumping lots of wasted current into ground through Rbottom. No free lunch.
...
Which is exactly why man invented zener diodes. Zener diodes act like voltage-variable resistors with a specific set point. When the input voltage goes above their set point, they act like small value resistors - dumping excess current to ground and maintaining the voltage very close to their set point. When the voltage goes below their set point, they act like large value resistors, passing little or no current to ground, and allowing the voltage to fall as far as it likes. In short, they ceiling clamp the voltage near a maximum value.
And if you combine a zener diode with an NPN transistor (which multiplies the zener's current handling capability by the transistor's Beta, at the cost of a 0.65 V voltage error due to the Vbe drop across the transistor's base-emitter junction) then you have a surprisingly workable voltage regulator with just a couple, three components.
That was one of the clearest answers I've ever read.. Well said!
Thanks, very useful.
I'm going to use a potential divider as a virtual ground for a transducer which will feed into an op-amp. Basically to convert the +/- AC output into one that is +ve.
Most welcome.
In general, opamps these days have tiny input currents. Usually something in the ballpark of tens of nanoamps. With an input current that tiny, a divider made out of resistors in the tens of k-ohm range is usually way more than good enough.
But if you want to be sure, look at the data sheet for the opamp you're using and find the "input bias current" rating. Then you can do the math and find the exact answer.
This is absolutely amazing! I've always questioned this, but never spent much time dwelling on it. We even went through several weeks learning voltage dividers in class and the prof never explained it this well.