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I have a transformer that says 6V 600mA 3.6VA, AC current (but the voltage measured is actually 8.65Vac).
I will add a 4 diode rectifier WITHOUT capacitor (I say without in order to simplify the calculus). I want to connect LEDs (200mA per LED). So, I should be able to connect 3 LEDs.

I want to know how to calculate the series resistor for each LED. But I don't know the REAL output voltage of the transformer. Will the voltage drop after I add each LED (this is what I think it will happen) or only after I reach the maximum power consumption?

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  • \$\begingroup\$ More than you expect. 9V * 600mA = 5.4W so a 3.6VA transformer can't supply that much. \$\endgroup\$ – Brian Drummond Dec 24 '16 at 14:56
  • \$\begingroup\$ This is just an exercise. This is where I want to use this information: electronics.stackexchange.com/questions/276665/… \$\endgroup\$ – WeGoToMars Dec 24 '16 at 15:33
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    \$\begingroup\$ The voltage rating of the transformer is under full (resistive) load. \$\endgroup\$ – Spehro Pefhany Dec 24 '16 at 17:08
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With a transformer that small, the load regulation will be very poor, maybe 20% (that is, loaded voltage = 80% of unloaded voltage), or could even be worse. The voltage will drop steadily as you add more load.

Your best bet is to measure it under load. You can do this safely by measuring the voltage off load, then compute the dropper resistors for that voltage. Attach the LEDs. The transformer voltage will drop under load, so having the wrong values will be safe. Take new measurements under load, rinse and repeat.

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  • \$\begingroup\$ So, the answer to my question is "the voltage will drop after each LED"? \$\endgroup\$ – WeGoToMars Dec 24 '16 at 15:00
  • \$\begingroup\$ Is the resistance of the secondary coil the internal resistance of the transformer? In this case I could do some calculus about how much it will drop. \$\endgroup\$ – WeGoToMars Dec 24 '16 at 15:01
  • \$\begingroup\$ It's the resistance of the primary, the secondary, the coupling coefficient between them, and losses in the core, well, these are the principal loss factors, there are others. So measurement will be more practical. Yes, the voltage will drop as you add each load. How will you connect even the first LED if you don't know how to calculate its dropper resistor? \$\endgroup\$ – Neil_UK Dec 24 '16 at 15:04
  • \$\begingroup\$ The resistor is 25 ohm / 1W power (for a 3.1V LED @0.2A) if the Vcc output is 8V (I took in consideration the drop on 1 rectifier diode). \$\endgroup\$ – WeGoToMars Dec 24 '16 at 15:29

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