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# Classic Linear 5v Supply Using 6.3vac Transformer

2017-03-03 22:05
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A classic method for producing a regulated 5v DC supply is shown below.  This circuit consists of an iron core transformer, a bridge rectifier, a filter capacitor and a voltage regulator.  Many people are tempted to use a very popular 6.3v transformer for this 5v supply but they will often discover that there just isn’t enough voltage from the transformer to make the circuit work properly under all but very light load conditions.  Higher transformer voltages will work but at the expense of much more power being dissipated in the voltage regulator.

Most transformers are specified for 120vac inputs.  At 110vac, the output of a 6.3vac transformer may only be 5.8vac.  To insure operation under nearly all reasonable conditions, I will often design a circuit so it would operate properly even with a low 105vac line voltage.  Under that condition, the 6.3vac from the transformer may only be 5.5vac.

If a typical bridge rectifier and filter capacitor were used, there would not be enough peak voltage to insure good regulation.  But, all is not lost.  This classic transformer design could still be  used but only if you pick the right parts.

For starters, the conventional bridge rectifier can be replaced with four power schottky diodes. A typical bridge rectifier using standard silicon diodes would have a voltage drop of about 2 volts, while a schottky diode bridge would drop only one volt. Although  this isn’t a lot, it sure helps.  Next, the main filter capacitor size could be increased, to decrease the ripple voltage across it.  One way to calculate the ripple voltage is with the equation: dv/dt = I/C.  dv/dt is the voltage change across the capacitor.  I is the DC load current and C is the capacitance.  For 60Hz power lines, the dt value would be 0.008 seconds. For 50Hz power lines, use the value of 0.010 seconds.

So, if we picked a big 20,000uF capacitor, with a current of 1 amp, then with a 60Hz frequency, the ripple voltage (dv) would be about 0.40v and 0.50v for a 50Hz system.  At 105vac the transformer secondary would be 5.5vac.  The peak voltage would be 1.41 X 5.5v or 7.7v.  Subtract 1 volt for the schottky bridge and 0.40 for the ripple voltage and you are left with 6.3v DC for the input to voltage regulator.  This is only 1.3v above the desired 5 volts but it should be enough if you use a regulator with a low input to output voltage drop.  I suggest using a LP3872ES-5.0 voltage regulator from National Semiconductor.  This device only needs about 5.3v at the input side to maintain 5 volts at the output, with a 1 amp load.  In my design, I suggest using a 6.3vac transformer rated at 2 amps and two big 10,000uF filter capacitors at the bridge output. Although the voltage drop across the regulator will be small, I would suggest mounting the regulator on a heat sink, rated for about 5 watts of dissipation.