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I tested a 470 ohms resistor with 35mA DC current flowing and my resistor is heating up.

On my final design I have created a voltage divider circuit using

R1=10k ohms 
R2=220 ohms  
maximum input voltage is 220 Vdc  
maximum DC current would be 21.5mA DC

I've only tested a maximum of 20Vdc as an input because I don't have available 220Vdc. I wonder if my R1=10k ohms and R2=220 ohms resistor will heat up on maximum DC current of 21.5mA

Is there a way to calculate whether it will heat up or not?

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  • \$\begingroup\$ calculate the wattage/power instead of the amperage/current \$\endgroup\$ – Skaperen Aug 13 '15 at 10:47
  • \$\begingroup\$ how much power or current is needed by the device you are feeding the divided voltage to? \$\endgroup\$ – Skaperen Aug 13 '15 at 10:52
  • \$\begingroup\$ I feed the output voltage to a micro-controller like arduino to measure the input voltage. I don't know how much power it is but the output voltage must not exceed 5V and 40mA to protect the micro-controller. \$\endgroup\$ – Jam Ville Aug 13 '15 at 12:25
  • \$\begingroup\$ It will heat up if the current is greater than 0. I expect you want to know how much it will heat up. \$\endgroup\$ – user253751 Aug 13 '15 at 12:58
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The temperature rise of the resistor will depend on many things - its thermal impedance to air, what it is soldered to, what type of resistor it is and how much airflow there is across it. However, we can get a good idea of whether we are designing within the recommended specification of the resistor in question by calculating the power consumption.

Most cheap through-hole resistors (which I am going to assume that you are using) are rated at 0.25W - they tend to look like this:

0.25W metal film resistors

The power dissipated in a resistor can be calculated in many ways, but the most useful for us here is

\${P=I^2R}\$

Calculating that for the 220 Ohm resistor gives \${P=0.0215^2\times220=0.102W}\$. You can see that this is close to the maximum rated 0.25W, but well enough within for the temperature rise not to damage the resistor.

Now let's examine the same for the 10kOhm resistor: \${P=0.0215^2\times10000=4.62W}\$! This is way above the power rating for this resistor, and it will very quickly heat up, glow, and then burn off the resistive element, releasing smoke and causing your circuit to fail.

So how do we fix this? There are two options.

  1. We could choose a 5 Watt resistor so that the power consumption was within limits. This will still work, but the 5 Watt resistor will get too hot to touch and will need to be mounted so that it is exposed to airflow and away from wiring and other components so that it cannot damage anything else.

  2. We could increase the values of the resistors used. If you chose a 470k resistor to replace the 10k resistor you are currently using and a 10k resistor to replace the 220r resistor, your divider ratio would be almost the same (you can calculate this yourself) but the power dissipation in the 470k resistor would be only 0.098W - well within specification again.

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  • \$\begingroup\$ It is important to note @Andyaka's answer below too - he explains how you can use multiple resistors to achieve a higher power rating and explains that there is a design risk is arbitrarily choosing higher resistor values, depending on what the divider connects to. \$\endgroup\$ – stefandz Aug 13 '15 at 8:23
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The important resistor for heating up is the 10k resistor. Current is 21.5 mA and this will cause a power of 4.63 watts in the 10k resistor. I would recommend a 10 watts resistor for 10k OR, if you can use 100k and 2k2 then consider doing that.

Using a 100k resistor means the power will be slightly below 0.5 watts so you might get away with a 0.6W resistor. You can always put two 50k resistors in series - they will share the power. You could also put two 200k resistors in parallel and achieve the same result.

Choosing a resistor that is as high as possible without compromising performance is the name of the game. As we do not know what the potential divider feeds this upper limit cannot be calculated/estimated.

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You are talking about burning almost 5W in order to measure a voltage. You need some way bigger resistances, I'd say at least 1M and 22k. If your micro-controller has a reliable, exact impedance on the measurement gate, you could simply use that instead of R2, along with an R1 suited to the value.

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