# Voltage divider resistance/wattage sizing

I am working on a small project currently -- I recondition hybrid car battery packs with a grid charging system, and I am working on a voltage sensing circuit that's monitored and available via a network. The voltage sensor I currently have is only capable of measuring 250VDC, and some battery packs exceed this.

I was going to make a simple voltage divider with like-sized resistors to just cut the voltage in half, which would bring me well within the 250V range. The voltage sensor is obviously an extremely low load, so what value resistors would I need for this? And what wattage rating should the resistors have?

Thank you.

Edit: here is the link to the voltage sensor I plan on using. http://www.yoctopuce.com/EN/products/usb-electrical-sensors/yocto-volt

Also, maybe I'm thinking incorrectly, but if I connect two resistors in series(and use the center tap as the half voltage) to a 250vdc battery, aren't the resistors just going to blow? I really feel like I should know this :/

• You also need to take into account the voltage rating of the resistors (more so than wattage since they can be rather high resistance resistors). Also please be careful, 250VDC will possibly kill you if you do something wrong. – Wesley Lee Mar 18 '19 at 8:12
• The voltage sensor is obviously an extremely low load That is only true if you design it to be a low load. A resistor divider consisting of two low value resistors (like 2x 125 ohm in series) would not make a low load but would divide the voltage just fine, while dissipating 250 W. If you use very high value resistors, for example 2x 1 Mohm in series, dissipation is only 0.032 Watt but the input impedance of your sensor (it might be 10 Mohm) will then influence the voltage division ratio making the measurement less accurate. – Bimpelrekkie Mar 18 '19 at 8:21
• Consider the accuracy of the resistors, and consider including a potentiometer in series to tune it. Strings of lower voltage resistors can be used in place of higher voltage ones. – K H Mar 18 '19 at 8:21
• Can you tell us the maximum input current of your voltage sensor? It's a good place to start, since the current on the voltage divider should be 100 or 1000 times that for accuracy. Hopefully it may be as low as uA or nA range. If it is at least as low as 1 uA, you can use 1 mA for the divider, dissipating up to .5 of a watt at 1000:1 and 500V. You can also connect the divider for only long enough to take a steady measurement, If you only measure for 1mS out of every second, you can still use lower wattage resistors if you get stuck with a high wattage divider. – K H Mar 18 '19 at 8:33
• Sorry for a double comment, but also here is a link to the voltage sensor. yoctopuce.com/EN/products/usb-electrical-sensors/yocto-volt – Matthew Toye Mar 18 '19 at 15:42

You can use a voltage divider made of two resistors so that your voltage sensor can safely read the voltage.

Here are its design equations:

$$\begin{split} &\frac{V_\mathrm{in}} {Z1 + Z2} \cdot Z_2 = V_\mathrm{out}\\ \\ &\frac{(V_\mathrm{in} - V_\mathrm{out})^2}{Z_1} =\text{ Power dissipated by }Z_1\\ \\ &\frac{{V_\mathrm{out}}^2}{Z_2} =\text{ Power dissipated by }Z_2 \end{split}$$

If $$\V_\mathrm{out} < \frac{V_\mathrm{in}}{2}\$$ I suggest choosing $$\Z_1\$$ first since it will need to dissipate more power than $$\Z_2\$$.

Else choose $$\Z_2\$$ first it will need to dissipate more power than $$\Z_1\$$.

Edit:

Since the input impedance of your voltage sensor isn't $$\\infty\$$, there will be a voltage drop at $$\V_\mathrm{out}\$$. To minimize this voltage drop, you need to use the lowest $$\Z_1\$$ and $$\Z_2\$$ value possible. The lower these values get, the more power $$\Z_1\$$ and $$\Z_2\$$ need to dissipate. You have to make a compromise between power consumption / dissipation and precision. One thing is for sure, $$\Z_1 = Z_2\$$ because you need $$\V_\mathrm{out}\$$ to be $$\\frac{V_\mathrm{in}}{2}\$$. Here are the minimum $$\Z_1\$$ and $$\Z_2\$$ value if you are using $$\\frac{1}{4}W\$$, $$\\frac{1}{2}W\$$ or $$\1W\$$ resistor:

$$\\frac{1}{4}W\$$ resistor:

$$\Z_1 = Z_2 = \frac{{V_\mathrm{out}}^2}{Power} = \frac{{250V}^2}{0.25W} = 250k\Omega\$$

$$\\frac{1}{2}W\$$ resistor:

$$\Z_1 = Z_2 = \frac{{V_\mathrm{out}}^2}{Power} = \frac{{250V}^2}{0.5W} = 125k\Omega\$$

$$\1W\$$ resistor:

$$\Z_1 = Z_2 = \frac{{V_\mathrm{out}}^2}{Power} = \frac{{250V}^2}{1W} = 62.5k\Omega\$$

• Thank you for that. So if I wanted to account for a voltage range of 350(or close to it) I can use 120k - 1 watt resistors and that would give me roughly 346 volts before blowing the resistors? – Matthew Toye Mar 18 '19 at 22:00
• @Matthew Toye No if Vin = 350V then Vout would be 350V / 2 = 175V. So for a 1W resistor Z1 = Z2 = 175V^2 / 1W = 30.625 kOhm minimum. – A.Lacasse Mar 18 '19 at 22:14
• Thank you, and my apologies for all the questions.. so a 5w 6.2k resistor would work? Would this be more efficient than a 1w 30k resistor? Is heat dissipation an issue with a voltage divider like this? This circuit will be hooked up for 5 days straight. Promise this is my last question, lol. – Matthew Toye Mar 18 '19 at 22:51
• @Matthew Toye I don't think a 5W 6.2 kOhm is needed for your application. In fact the precision difference between a 1W 30 kOhm and 5W 6.2 kOhm isn't required for what you're doing. You also have to keep in mind that the lower the resistor value are, the more current are flowing through them. This current is supplied by your battery or charger and will discharge your battery or slowdown the charging time. The heat dissipation isn't an issue if you respect the power rating of your resistor. – A.Lacasse Mar 18 '19 at 23:10
• Awesome! Thank you so much!!!! – Matthew Toye Mar 18 '19 at 23:15

The important spec for your sensor is the line:

Input impedance: >660KΩ

So as long as your resistors are well below 660 kΩ, they will not distort the measurement.

You didn't say what you expect your max input voltage to be, but assuming it's 400 V, you could use two 80 kΩ resistors. This would give

$$\\text{Power} = \frac{V^2}{R} = \frac{(400\text{ V})^2}{2\times80\text{ k}\Omega}=1.0\text{ W}\$$

This would be split between the two resistors, each dissipating 0.5 W.

The resulting voltage division assuming worst case of 660 kΩ input resistance is:

$$\\text{Error} = \frac{660\text{ k}\Omega\parallel 80\text{ k}\Omega}{(660\text{ k}\Omega\parallel 80\text{ k}\Omega) + 80\text{ k}\Omega} = 0.4714 \$$

This is a 2.9% error from ideal division by 2.

• Hum I'm afraid you forgot squaring the voltage, 20 milli(amperes) is just the current. Power is going to be 4W each resistor. – carloc Mar 18 '19 at 20:32
• @carloc - Thanks for catching that! I've updated my answer, also changing from 10k to 80k resistors. – Justin Mar 19 '19 at 12:19