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I’m self-taught, and this is a little bit of a thought experiment for me to understand Ohm’s Law better.

I have a very simple voltage divider. Given a 15V DC input, each of three 4.7KΩ resistors cuts the voltage by 33%. I started doing some experimentation, and discovered that no matter what voltage I applied to the circuit, the resistors always cut the voltage and amperage by 33% each. enter image description here

But let’s say I wanted to create the same circuit and didn’t know the necessary resistance?

Given a 15V input and desired outputs of 10V, 5V and 0V, how would I calculate the necessary resistance to use? Is it possible to create a voltage divder that does not have proportional drops (e.g., let's say that from this same circuit, I want 14V, 12V, 5V and 0V)? And how does that math work? I think where I’m getting stuck is whether to use input voltage, output voltage, or change in voltage as the V value.

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Here is one way of understanding the problem and thus arriving at the solutions you seek:

  1. You have a voltage V applied across a "black box", consisting of a series of resistors R1, R2 and R3 in this case. The resistances are in series so they add up, thus the Black Box has a cumulative resistance of R = R1 + R2 + R3.
  2. A voltage applied across a resistance causes a current I to flow, thus: I = V / R.
  3. Since the constituent resistors are in series, the SAME amount of current must flow through each of them. There is no alternative path for current to flow from V+ to ground.
  4. A current across a resistance implies a voltage across said resistance, by the same formula as above, thus: V(r1) = I * R1. That is the potential difference between the two ends of resistor R1.
  5. Similarly, V(r2) = I * R2, and so on.
  6. Evidently, one of these resistors, R3, has one end at ground potential, i.e. 0 volts. Thus, the voltage from there to the other end of that resistor is V(r3). The voltage at the next higher measurement point is V(r3) + V(r2), since the voltages add up, and as stated above, reference to ground.

By following this process, the voltages at each of the points of any series resistance network can be computed if either the applied voltage V (15 volts in this case) or the flowing current due to it, is known.

Now, how does one decide what resistances to use? Well, make the total resistance too small, and the current will be high, potentially burning out the resistors or the power supply, or causing the supplied voltage to droop, depending on how ideal we are assuming things to be. Similarly, use too high a resistance, and too little current will flow, thus the readings will be swamped by other noise effects that exist in practical electronics from various causes.

So pick a number that you like, and divide it in the ratio you want the test-point voltages to be. The resistances need not be equal, just as the voltages need not be at 33% each - calculate for any ratio you want.

I hope this helped.

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  • \$\begingroup\$ Better than any other resource I could find. A lot just clicked for me. Thanks! Too bad I can't +5 you. :) \$\endgroup\$ – dwwilson66 Nov 2 '12 at 16:04
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    \$\begingroup\$ @dwwilson66: You can always award a bounty to the answerer :) \$\endgroup\$ – Thomas E Nov 2 '12 at 23:22
  • \$\begingroup\$ If anyone is interested, there is a very useful voltage divider calculator here that searches standard resistor values. \$\endgroup\$ – TimH - GoFundMonica Mar 26 '15 at 22:11
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"Given a 15V input and desired outputs of 10V, 5V and 0V, how would I calculate the necessary resistance to use?"

I think a good way to go about this is to look at one set at a time. The standard voltage divider equation is simple enough, $$\text{Voltage across resistor of interest} = \frac{(\text{Resistor of Interest})}{(\text{Resistor of Interest + Resistor Not of Interest})} * V_{input}$$

When there are multiple nodes, like in the example you've given, just simplify it to the basic resistor divider and find the first voltage. Alternatively, if we're given voltages, we can rearrange this equation to solve for the resistor of interest in terms of the resistor not of interest.

$$\text{Resistor of Interest} = \frac{1}{({V_{input}}\div{\text{Voltage across resistor of interest}})-1}*\text{Resistor Not of Interest}$$

To simplify, in your example for the 10V node, the resistor of interest is the combination of R2 and R3, leaving the resistor not of interest as R1. Once you've found your ratio between (R2+R3) and R1 you can move on to find the ratio for R2 and R3. In this case you can just look at those two as another divider and the input voltage is that first node voltage you've just used as your output voltage. Following this method you'll find that R1 is one third (R2+R3) and that R2 is the same as R3. It makes sense that given equal current flow, an identical drop across each resistor means and identical resistance, following Ohm's law V=IR.

"Is it possible to create a voltage divder that does not have proportional drops (e.g., let's say that from this same circuit, I want 14V, 12V, 5V and 0V)?"

This will be the same process as before, but just plug in different voltages. For the first node:

$$\text{(R2+R3)} = (\frac{1}{(14V\div12V)-1})*\text{R1}=6*R1$$

So the combination of R2 and R3 is six times larger than R1 alone. For the second node:

$$\text{(R2)} = (\frac{1}{(12V\div5V)-1})*\text{R3}=0.71*R3$$

Finally, and this is the most difficult part for most students, just pick a resistor value. This is the engineering part of electrical engineering, you have to make a decision. This one is not too difficult, for the most part larger resistances are better. Larger resistances will reduce current flow while still providing the voltages you need.

There are several other considerations when using a voltage divider in practice. These are great for basic reference voltages or proportionally knocking down a signal voltage in a single direction. For instance a 5V signal being taken down to 3.3V for a microcontroller works well because a voltage divider acts like an attenuation coefficient to the signal, everything gets reduced by the same amount.

If you're proving voltage to a device of some sort, you can sometimes model that current draw as a resistance, assuming it's always constant (R=V/I). This device resistor, or load, is usually the resistor of interest or parallel to the resistor of interest. I would not recommend this at any time however as the node voltage will change depending on the current draw of the load.

"And how does that math work?"

See equations above.

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The math is one of simple linear proportions. The key is that the same current (I) flows through all the resistors, and I = V/R. So one way to look at current is that it is "volts per ohm". Each ohm of resistance in the divider gets the same number of volts as every other ohm. The voltage drops therefore follow the ratios of the resistors. The voltage on each resistor is the "volts per ohm" (current, same everywhere) multiplied by its ohms. If the ratio of the resistances is 4:3:1, then the ratio of voltages is 4:3:1. Simple.

Voltage dividers are disturbed by loads. As soon as you start drawing current from the various voltage taps along the divider, the voltages will change. This is because the current is then longer the same everywhere in the divider.

Voltage dividers with lower resistors are less easily disturbed ("stiffer") than voltage dividers with higher resistors, but draw more current.

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Kaz has it right. If you have 15 and want 14V, 12V, 5V and 0V then each resistor drop is 1,2,7,5 [V} so the resistor ratios are the same. then add up all the values and take a ratio of all of them to choose the current as it is the same for each. (assuming no external load)

Thus for each R = 1+2+7+5 [Kohm] = 15 KOhm since 1mA is shared.. To choose any other current simply scale the resistors equally. e.g. choose 30uA so R = 15V/30uA = 0.5MΩ and each value is {1/15, 2/15, 7/15, 5/15 } * 0.5MΩ i.e. the result is V+ to 33KΩ, then 67KΩ, 233KΩ, 167KΩ to ground ( which add up to ~*0.5*MΩ)

So choose total current, then the voltage drop is proportional to R and of course equal drop is equal resistors.

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While you should work through Ohm's law and do the math for a complete understanding, you can also do this by inspection, which is how it's done after you've gotten the underlying theory. In your original circuit, +5V is 1/3 of the input voltage, so R3 should be 1/3 of the total resistance (i.e., R1 + R2 + R3). Similarly, 10V is 2/3 of the input voltage, so R2 + R3 should be 2/3 of the total resistance. All you need to do now is decide how large the total resistance should be, and the three values just fall out. If the total resistance is 4700 ohms, then R3 is 4700/3, or 1533; R2 + R3 is 4700 * 2 / 3, or 3066, so $2 is 1533; and R1 is the rest, 4700 - 1533 - 1533, or 1534 (yes, off by one because of rounding).

Or if you need a particular resistance for, say, R3, you can start there: the total resistance is 3 * R3, and from that you can figure the values of R2 and R1 just as above.

When you need other voltages, just apply the corresponding fractions. Let's do your example of 14V, 12V, and 5V (I'm ignoring 0V because it's trivial). Since you want three voltages instead of the two in the original example, you need four resistors instead of the three in the original. 5V is 1/3 of the input voltage, so R4 would be 1/3 of the total resistance. 12V is 4/5 of the input voltage, so R3 + R4 would be 4/5 of the total resistance. And 14V is 14/15 of the input voltage, so R2 + R3 + R4 would be 14/15 of the input voltage. Again, pick the total resistance, and the individual values fall out.

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