11
\$\begingroup\$

I'm restoring an old tube radio. When looking into the schematics I noticed that they drew 2 resistors in parallel instead of 1 resistor, between the power transformer and the rectifier tube. Is there any benefit from installing 2 parallel resistors instead one resistor?

\$\endgroup\$
  • \$\begingroup\$ Maybe the combined value was not available in a single resistor? \$\endgroup\$ – Arsenal Oct 29 '15 at 9:50
  • \$\begingroup\$ Also, you statistically get a better chance at being closer to the actual value you are looking for. \$\endgroup\$ – njzk2 Oct 29 '15 at 18:26
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/25883/… \$\endgroup\$ – Fizz Oct 30 '15 at 18:40
17
\$\begingroup\$

Two resistors have an increased ability to dissipate power/heat, up to twice the time one can, when mounted with enough space between them.

Also for a few applications it is easier to find two standard value resistors that together form one resistor value that is not among the standards you already have in your BOM (or are available).

Lastly for the gray beard hand chosen value resistor, it might have a higher yield to use two that are slightly off than to find the one that matches perfectly.

\$\endgroup\$
  • \$\begingroup\$ Is there also a possibility to reduce the effect of a failing resistor from a fatal fault to a fault which allows some sort of recovery or safe fault mechanism? In our designs we often have to put 2 resistors in series (not in parallel) to prevent a single failure to lead to a critical state, but each of the resistors must be able to handle all the power. \$\endgroup\$ – Arsenal Oct 29 '15 at 11:00
  • \$\begingroup\$ @Arsenal: It might be an added benefit, but I don't think it is the prime reason to initially add two resistors. When one fails short, the fault is still fatal. When one fails open, the resistance is doubled, but I can hardly imagine a scenario where this would not render a device as equally unusable as with a total open. If you design for safety, then you likely design for complete open or short, and anything in between is not going to be much better, possibly even worse. \$\endgroup\$ – PlasmaHH Oct 29 '15 at 11:07
  • \$\begingroup\$ I was curious, I had to check the pricing on this one. RS sells individual 1W resistors for 0.03GBP (bulk price is about half that.) they sell 10W resistors in 5-packs for 0.83GBP (bulk price is about half that.) So the saving is considerable, provided you don't take into account a nice large board to mount them on (which wouldn't be an additional requirement in the OP's case). Disclaimer: my prices are based on the cheapest available, with no consideration given to the resistance value. \$\endgroup\$ – Level River St Oct 29 '15 at 12:41
  • \$\begingroup\$ @steveverrill: It would probably more realistic for the OPs case if you compared 500mW to 1W or similar, but yeah, I would subsume that all under "availability". \$\endgroup\$ – PlasmaHH Oct 29 '15 at 12:45
  • 1
    \$\begingroup\$ @PlasmaHH: Many tube circuits are supplied by voltages far in excess of what they're designed to work with. Consider a tube which is intended to drive ~2mA into a 1k load at ~2 volts. Remove the load resistor and it will feed ~2mA at 70+ volts. \$\endgroup\$ – supercat Oct 29 '15 at 19:16
6
\$\begingroup\$

The benefit is the power dissipation.

If you have a voltage V and a current I the power dissipation across a resistor is

$$ P=V \cdot I$$

Now if you put two resistors in parallel you divide the current in each resistor by 2 (if they have the same value) and then the power dissipated by 2.

Sometimes it is cheaper to put 2 resistors in parallel to dissipate more power than only one bigger resistor.

\$\endgroup\$
  • 1
    \$\begingroup\$ If you put two resistors in parallel in order to halve the power dissipation in each one, then you must increase the values of the paralleled resistors to twice that of the single resistor. BTW, it's \$"P=VI" \text{or}"P=EI" \text{, not}"P=UI".\$ \$\endgroup\$ – EM Fields Oct 29 '15 at 10:45
  • 1
    \$\begingroup\$ @EMFields U is a commonly used symbol for voltage in Germany at least, as there is no way to confuse it with V as Volt, so P=UI and U=RI over here. \$\endgroup\$ – Arsenal Oct 29 '15 at 10:53
  • \$\begingroup\$ @Arsenal: What's to confuse? If V is voltage, then P=VI and V=RI. \$\endgroup\$ – EM Fields Oct 29 '15 at 10:59
  • 1
    \$\begingroup\$ @EMFields I guess they tried to prevent some confusion like V = 1 V so 0 = 0 or something like that. I don't know, it's just that on some parts of the world U is voltage so P = UI as well and it's hard to get away from stuff you are trained to use for years. \$\endgroup\$ – Arsenal Oct 29 '15 at 11:04
  • \$\begingroup\$ U is a pretty common symbol for a voltage variable in Brazil too. \$\endgroup\$ – ricardomenzer Oct 29 '15 at 12:06
1
\$\begingroup\$

As others pointed out, the main reason increased is heat dissipation. I had a sound amplifier that had a huge power resistor which every week was burning out literally. One day I replaced it with 2 resistors and since then never failed.

Another experience of mine is regarding PCB design. I wanted to order minimum variety of components. So the pick and place machine won't charge me for extra rails. I had many 5k resistors in the design. Anywhere that I had 10k or 20k resistors I simply used 5k in series. A bit stupid but saved me few cents!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.