Transformerless 5 Volt DC Power Supply
An increasing number of appliances draw a very small current from the power supply. If you need to design a mains-powered device, you could generally choose between a linear and a switch-mode power supply. However, what if the appliance’s total power consumption is very small? Transformer-based power supplies are bulky, while the switchers are generally made to provide greater current output, with a significant increase in complexity, problems involving PCB layout and, inherently, reduced reliability.
Is it possible to create a simple, minimum part-count mains (230 VAC primary) power supply, without transformers or coils, capable of delivering about 100mA at, say, 5 V? A general approach could be to employ a highly inefficient stabilizer that would rectify AC and, utilizing a zener diode to provide a 5.1 V output, dissipate all the excess from 5.1 V to (230×√2) volts in a resistor. Even if the load would require only about 10mA, the loss would be approximately 3 watts, so a significant heat dissipation would occur even for such a small power consumption.
Transformerless 5 Volt DC Power Supply Circuit Diagram:
At 100mA, the useless dissipation would go over 30 W, making this scheme completely unacceptable. Power conversion efficiency is not a major consideration here; instead, the basic problem is how to reduce heavy dissipation and protect the components from burning out. The circuit shown here is one of the simplest ways to achieve the above goals in practice. A JVR varistor is used for over-voltage/surge protection. Voltage divider R1-R2 follows the rectified 230 V and, when it is high enough, T1 turns on and T3 cannot conduct.
When the rectified voltage drops, T1 turns off and T3 starts to conduct current into the reservoir capacitor C1. The interception point (the moment when T1 turns off) is set by P1 (usually set to about 3k3), which controls the total output current capacity of the power supply: reducing P1 makes T1 react later, stopping T3 later, so more current is supplied, but with increased heat dissipation. Components T2, R3 and C2 form a typical ‘soft start’ circuit to reduce current spikes — this is necessary in order to limit C1’s charging current when the power supply is initially turned on. At a given setting of P1, the output current through R5 is constant.
Thus, load R4 takes as much current as it requires, while the rest goes through a zener diode, D5. Knowing the maximum current drawn by the load allows adjusting P1 to such a value as to provide a total current through R5 just 5 to 6mA over the maximum required by the load. In this way, unnecessary dissipation is much reduced, with zener stabilization function preserved. Zener diode D5 also protects C1 from over voltages, thus enabling te use of low-cost 16 V electrolytics. The current flow through R5 and D5, even when the load is disconnected, prevents T3’s gate-source voltage from rising too much and causing damage to device. In addition, T1 need not be a high-voltage transistor, but its current gain should exceed 120 (e.g. BC546B, or even BC547C can be used).
Is it possible to create a simple, minimum part-count mains (230 VAC primary) power supply, without transformers or coils, capable of delivering about 100mA at, say, 5 V? A general approach could be to employ a highly inefficient stabilizer that would rectify AC and, utilizing a zener diode to provide a 5.1 V output, dissipate all the excess from 5.1 V to (230×√2) volts in a resistor. Even if the load would require only about 10mA, the loss would be approximately 3 watts, so a significant heat dissipation would occur even for such a small power consumption.
Transformerless 5 Volt DC Power Supply Circuit Diagram:
At 100mA, the useless dissipation would go over 30 W, making this scheme completely unacceptable. Power conversion efficiency is not a major consideration here; instead, the basic problem is how to reduce heavy dissipation and protect the components from burning out. The circuit shown here is one of the simplest ways to achieve the above goals in practice. A JVR varistor is used for over-voltage/surge protection. Voltage divider R1-R2 follows the rectified 230 V and, when it is high enough, T1 turns on and T3 cannot conduct.
When the rectified voltage drops, T1 turns off and T3 starts to conduct current into the reservoir capacitor C1. The interception point (the moment when T1 turns off) is set by P1 (usually set to about 3k3), which controls the total output current capacity of the power supply: reducing P1 makes T1 react later, stopping T3 later, so more current is supplied, but with increased heat dissipation. Components T2, R3 and C2 form a typical ‘soft start’ circuit to reduce current spikes — this is necessary in order to limit C1’s charging current when the power supply is initially turned on. At a given setting of P1, the output current through R5 is constant.
Thus, load R4 takes as much current as it requires, while the rest goes through a zener diode, D5. Knowing the maximum current drawn by the load allows adjusting P1 to such a value as to provide a total current through R5 just 5 to 6mA over the maximum required by the load. In this way, unnecessary dissipation is much reduced, with zener stabilization function preserved. Zener diode D5 also protects C1 from over voltages, thus enabling te use of low-cost 16 V electrolytics. The current flow through R5 and D5, even when the load is disconnected, prevents T3’s gate-source voltage from rising too much and causing damage to device. In addition, T1 need not be a high-voltage transistor, but its current gain should exceed 120 (e.g. BC546B, or even BC547C can be used).
Transformerless 5 Volt DC Power Supply
Reviewed by haru
on
November 05, 2017
Rating:
Reviewed by haru
on
November 05, 2017
Rating:
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