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  • Where is the best to place resistor in full wave rectifier circuit?
  • What is the difference in usage of the resistors in these two circuits?

(values are not precise)

Three side questions:

  • what about removing one resistor and replacing the left one with bigger value in the first circuit?
  • is there any disadvantage of relatively big capacitance of the cap (instead of size) or the wave will be just smoother?
  • when there is the DC fan, do you recommend any snubber or flyback diode in this circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

Thank you!

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    \$\begingroup\$ Why do you want to put a resistor somewhere exactly? the only reason I can think of is to reduce inrush current for initially charging the capacitor, in which case puting it behind the capacitor makes no sense. More capacitance will give you, in general, a smoother DC, but you need to realize that it's also bigger and more expensive, and your application might not require it. \$\endgroup\$
    – Joren Vaes
    Jun 20 '17 at 9:18
  • \$\begingroup\$ I'm maybe totally wrong (hobbyist), but to limit the voltage. I have AC source (that says 12V AC) where my multimeter (connected after rectifier instead of the fan or parallel to fan) reads more than 12V DC (around 14-17V DC) so I would like to make it suitable for my fan. \$\endgroup\$
    – pejey
    Jun 20 '17 at 9:36
  • \$\begingroup\$ If the AC source is just a transformer, then you should expect to see more than 12VAC when no load is applied. As soon as you apply load the voltage goes down, how much depends on the transformer. Also keep in mind that the oltage drop over your diodes depends on the voltage. At load, you are looking at dropping 1.5V. The transformer is also rated in RMS voltage. This means the peak voltage will be sqrt(2) times higher. For 12V this gives us 17V. \$\endgroup\$
    – Joren Vaes
    Jun 20 '17 at 9:44
  • \$\begingroup\$ yes, transformer. Thanks for the calculation/explanation. With the fan connected, I am getting about 14V and I would not like to burn it. To get the lower voltage, I need higher load so the second circuit with the resistor as a load could work? \$\endgroup\$
    – pejey
    Jun 20 '17 at 9:52
  • \$\begingroup\$ Use two 6.1V zeners in series to make a stiff 12.2V power source for your fan, of course if you still get more than 12V with the load connected. \$\endgroup\$
    – dirac16
    Jun 20 '17 at 9:53
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The answer depends on what the purpose of the resistor is. Most of the time, you use none at all.

Note that in both cases, the resistor is in series with the load, so will dissipate power proportional to the square of the load current. Gratuitously dissipating power is a bad idea. If you're going to spend the power, you need to have a good reason. You have not provided any.

In the top case, the resistors will limit the inrush current a bit when the cap is first charged up. Sometimes it is necessary to avoid large inrush currents.

However, R1 and R2 are in series. Electrically, you'd have the same thing by replacing one of them with a 40 Ω resistor and getting rid of the other. Using two resistors in series can be a legitimate way to get a higher power resistor. If you're already stocking the two smaller resistors, it can be all around cheaper to use them instead of stocking a whole new power resistor for a low volume product. In that case putting one in each leg may be useful too, but only to put them physically distant enough from each other so that they don't mutually heat. Electrically there is no advantage.

If it is really necessary to limit inrush cheaply, the usual solution is a negative temperature coefficient thermistor. You size it to limit inrush at room temperature. It will heat up a bit and have lower resistance during normal operation. That still eats power during operation, but not as much as a fixed resistor.

More sophisticated circuits turn on a FET slowly across the inrush limiter. After a few line cycles, the FET is fully on and there is very little resistance in series with the power feed.

The bottom circuit is just plain silly. The resistor doesn't limit inrush at all, just gets in the way of providing power. You haven't told us of any problem it is intended to solve.

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  • \$\begingroup\$ I was up to limit the voltage to 12V DC (7-12V would be fine) going to the load (12V DC fan) because 12V AC transformer gives me around 14V when the load is connected (~17V without the load) and it could harm the load after some time. I was looking for resistor solution because I have plenty resistors around. \$\endgroup\$
    – pejey
    Jun 20 '17 at 11:44
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Option number one is best if you want to limit inrush current into the smoothing cap.

Check the data sheet of the fan to see what actual voltages it can handle. Chances are it can deal with the 14V just fine.

If you want to regulate voltage get a voltage regulator. At 0.15 A a resistor in series dropping 3 V is going to be dissipating 0.5 W of heat.

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  • \$\begingroup\$ Oh yes, I think I could smell the resistor when it was in the second circuit. But the first circuit with two resistors was actually dropping the voltage down without noticeable heating. \$\endgroup\$
    – pejey
    Jun 20 '17 at 10:55

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