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This is a two-part follow-up question (which wouldn't fit within the 600-character limit for comments) to Using LEDs in a bridge rectifier circuit. I appreciated Anindo Ghosh's response to that question with its explanations and recommendations for getting the best result if my main purpose was to build a good rectifier.

Anindo was correct, however, when he inferred that I want to use the light from the LEDs in the bridge rather than its d.c. output. So....I have a couple of follow-up questions:

  1. for the current-limiting resistance needed to protect the LEDs, should I put a single resistor between one leg of the generator output and its bridge connection (with the resistor value calculated from the voltage drop across the sum of the forward voltages divided by the sum of the forward currents of all 4 LEDs), or should I put a resistor between each of the 4 LEDs, all with the same value (calculated from the voltage drop across a single LED divided by the forward current of a single LED)?

  2. would the LEDs in the bridge produce light without a d.c. load (e.g. another LED or a resistor) on the bridge?

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A bridge design is odd for this use, however it'll work if you use a single resistor as the load of the bridge. That resistor will determine the current through the LEDs, and thus the brightness.

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  • \$\begingroup\$ Actually, that's not the best place to put the resistor, as it puts excessive reverse voltage across half the LEDs on each half cycle. It's better to put the resistor in series with the generator, as the OP suggested. \$\endgroup\$ – Dave Tweed Nov 9 '12 at 0:44
  • \$\begingroup\$ Adam- Thanks for your interest in this, and in my original question. I do want all 4 LEDs lighted simultaneously, so I'll use the bridge circuit. Should I use a single current-limiting resistor (with value calculated as I described) or 4 such resistors (with values calculated as I described)? You say I could vary the brightness of the LEDs by adding a resistive load at the output and varying its value. Without any load would the LEDs be dark, or at their maximum brightness? \$\endgroup\$ – Chuck Nov 9 '12 at 0:53
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The resistor value should be based on the voltage drop from the generator to two of the LEDs, since that's how many LEDs that will be conducting on each half cycle. Also, since the LEDs are in series, you do not add the currents; the same current is flowing through both LEDs.

So, the final resistor value is (VGENERATOR – 2 × VLED) / ILED.

This resistor should go in series with the generator, as you suggested. The "load" on the bridge should be a dead short connecting the corner with the two anodes to the corner with the two cathodes.

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