Born2bwire
25+ Member ;-)
- Joined
- Nov 11, 2001
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Quote:
Whoops, made a mistake in the math. The error should be:
abs(1+2*(E1*E2-1)/(2+E1+E2)).
So it would appear that the final error is going to be equal to or less than the larger error of the two resistors. In your second circuit though there still is an error of 1%.
Another reason to go parallel that I thought of is that you split the current running through each resistor by half while increasing the resistance by 2. So the power dissipated will reduce by a factor of two. So theoretically, you could use cheaper parts. For example, instead of a 1K 1/2 W 5% resistor, you could use a 2K 1/4 W 1% and a 2K 1/4 W 10% and achieve the same power specs with a maximum error of 5.7% Though in real life I doubt there is any such combination of two resistors that would actually be cheaper than the single resistor that you are replacing.
Originally Posted by Denim You have the right idea here, but the explanation could use some additional info. It's possible that 2 resistors in parallel will achieve a final resistance closer to the desired value, but the errors need to cancel out each other. That is, one resistor having a higher reading, and the other having a lower reading. If they are both higher, or lower, then the total resistance can still be off as much as a single resistor. Here's the math -- Desired resistance - 50 ohms Circuit 1: 100 ohm (measures 110 ohms) & 100 ohm (measures 110 ohms) in parallel = 55 ohms. Resistors off by 10%, circuit off by 10% also. Circuit 2: 100 ohm (measures 110 ohms) & 100 ohm (measures 90 ohms) in parallel = 50 ohms. Resistors off by 10%, circuit achieves desired resistance because the offsetting differences cancel each other out. As for resistors being polarity sensitive, I haven't seen one yet. |
Whoops, made a mistake in the math. The error should be:
abs(1+2*(E1*E2-1)/(2+E1+E2)).
So it would appear that the final error is going to be equal to or less than the larger error of the two resistors. In your second circuit though there still is an error of 1%.
Another reason to go parallel that I thought of is that you split the current running through each resistor by half while increasing the resistance by 2. So the power dissipated will reduce by a factor of two. So theoretically, you could use cheaper parts. For example, instead of a 1K 1/2 W 5% resistor, you could use a 2K 1/4 W 1% and a 2K 1/4 W 10% and achieve the same power specs with a maximum error of 5.7% Though in real life I doubt there is any such combination of two resistors that would actually be cheaper than the single resistor that you are replacing.