Wiring Diagram circuit

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Inductorless 3 5 Volts Converter

By configuring a comparator and a transistor to control the oscillator in a charge pump circuit, you enable the pump to generate a regulated output of in principle any desired value. Charge pump ICs can either invert or double an input voltage (for example, 3 V to –3 V or 3 V to 6 V). The charge pump itself does not regulate the output voltage and one running off 3 V is not normally capable of generating intermediate output voltage levels like 5 V. However, by adding a comparator and a reference device, you can create arbitrary output levels like 5 V and regulate them as well. Charge pump IC1 (a MAX660) has an internal oscillator whose 45 kHz operation transfers charge from C1 to C2, causing the regulated output to rise.

Circuit diagram:

Inductorless 3-to-5 Volts Converter circuit schematic

Inductorless 3-5 Volts Converter Circuit Diagram

When the feedback voltage (pin 3 of IC2) exceeds 1.18 V, the output of comparator IC2 (a MAX921) goes high, turning off the oscillator via T1. The comparator hysteresis (easily added on IC2) is zero here simply because no hysteresis is required in the control loop. The oscillator when enabled generates two cycles, which is sufficient to drive VOUT slightly above the desired level. Next, the feedback turns the oscillator off again. The resulting output ripple will depend mainly on the input voltage and the output load current. Output ripple may be reduced at the expense of circuit efficiency by adding a small resistor (say, 1 ?) in series with C1. You’ll find that ripple also depends on the value and ESR associated with C1 - smaller values of C1 transfer less charge to C2, producing smaller jumps in V OUT.

Author: D. Prabakaran - Copyright: Elektor July-August 2004

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Simple Transformerless 5 Volt Power Supply

Description

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 100 mA 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×v2) volts in a resistor. Even if the load would require only about 10 mA, the loss would be approximately 3 watts, so a significant heat dissipation would occur even for such a small power consumption.

At 100 mA, 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 overvoltage/surge protection. Voltage divider R1-R2 follows the rectified 230 V and, when it is high enough, T1 turns on and T3 cannot conduct.

Circuit diagram:

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 6 mA 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).

CAUTION!

The circuit is not galvanically isolated from the mains. Touching any part of the circuit (or any circuitry it supplies power to) while in operation, is dangerous and can result in an electric shock! This circuit should not be built or used by individuals without proper knowledge of mains voltage procedures.

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