Showing posts with label transformerless. Show all posts
Showing posts with label transformerless. Show all posts
Sunday, July 7, 2013
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.
Copyright: Elektor Electronics Magazine
Author: Srdjan Jankovic & Branko Milovanovic
Author: Srdjan Jankovic & Branko Milovanovic
Sunday, March 31, 2013
How to Calculate and Deduce Current Voltage Parameters in Transformerless Power Supplies
After carefully studying the relevant patterns, I devised a simple and effective way of solving the above issues, especially when the power supply used is a transformerless one or incorporates PPC capacitors or reactance for controlling current.
Typically, a transformerless power supply will produce an output with very low values but with voltages equal to the applied AC mains (until it’s loaded).
For example, a 1 µF, 400 V (breakdown voltage) when connected to a 220 V mains supply will produce a maximum of 70 mA of current and an initial voltage reading of 220 Volts.
However this voltage will show a very linear drop as the output gets loaded and current is drawn from the “70 mA” reservoir.
In case the load consumes the whole 70 mA would mean the voltage dropping to almost zero.
Now since this drop is linear, we can simply divide the initial output voltage with the max current to find the voltage drops that would occur for different magnitudes of load currents.
Therefore dividing 220 volts by 70 mA gives 3.14. This is the rate at which the voltage will drop for every 1 mA of current added with the load.
That means if the load consumes 20 mA of current, the drop in voltage will be 20 × 3.14 = 62.8 volts, so the output now will show a voltage of 220 – 62.8 = 157.2 volts (see figure).
However the voltage across the LED terminals will show a voltage equal to forward voltage drop of the particular LED, because the above concept again becomes applicable for the section which comes after the resistor and across the LED.
Here, the resistor further controls the current and by dividing 157 by 4700 we get 0.033 or a max of 33 mA of current and a voltage of 157, which drops linearly, so dividing 157 by 33 gives 4.75, which is the rate of voltage drop for each mA rise in the load current. The connected LED draws probably the whole 33 mA, which deducts around 33 × 4.75 = 156.75 Volts from the resistor’s output 157 Volts.
That leaves the volts across the LED to the relevant values of around a couple volts or to be precise, the forward voltage drop of the particular type of LED.
Conclusion: From the above discussion and analysis, it becomes clear that voltage in any power supply unit is immaterial if the current delivering capability of the power supply is "relatively" low.
For example if we consider an LED, it can withstand 30 to 40 mA current at voltages close to its "forward voltage drop", however at higher voltages this current can become dangerous for the LED, so its all about keeping the maximum current equal to the maximum safe tolerable limit of the load.
While calculating series resistor values with LEDs, instead of using the standard LED formula directly, we can use the above rule first.
That means either we choose a capacitor whose reactance value only allows the maximum tolerable current to the LED, in which case a resistor can be totally avoided.
If the capacitor value is large with higher current outputs, then probably as discussed above we can incorporate a resistor to reduce the current to tolerable limits.
Example: In the shown diagram, the value of the capacitor produces 70 mA of max. current which is quite high for any LED to withstand. Using the standard LED/resistor formula:
R = (supply voltage VS – LED forward voltage VF) / LED current IL,
= (220 - 1.5)/0.02 = 11K,
Therefore the value of the resistor for controlling one red LED safely would be 11K.
The above theory has been assumed and deduced by me, I am not very sure about its feasibility, though
Typically, a transformerless power supply will produce an output with very low values but with voltages equal to the applied AC mains (until it’s loaded).
For example, a 1 µF, 400 V (breakdown voltage) when connected to a 220 V mains supply will produce a maximum of 70 mA of current and an initial voltage reading of 220 Volts.
However this voltage will show a very linear drop as the output gets loaded and current is drawn from the “70 mA” reservoir.
In case the load consumes the whole 70 mA would mean the voltage dropping to almost zero.
Now since this drop is linear, we can simply divide the initial output voltage with the max current to find the voltage drops that would occur for different magnitudes of load currents.
Therefore dividing 220 volts by 70 mA gives 3.14. This is the rate at which the voltage will drop for every 1 mA of current added with the load.
That means if the load consumes 20 mA of current, the drop in voltage will be 20 × 3.14 = 62.8 volts, so the output now will show a voltage of 220 – 62.8 = 157.2 volts (see figure).
However the voltage across the LED terminals will show a voltage equal to forward voltage drop of the particular LED, because the above concept again becomes applicable for the section which comes after the resistor and across the LED.
Here, the resistor further controls the current and by dividing 157 by 4700 we get 0.033 or a max of 33 mA of current and a voltage of 157, which drops linearly, so dividing 157 by 33 gives 4.75, which is the rate of voltage drop for each mA rise in the load current. The connected LED draws probably the whole 33 mA, which deducts around 33 × 4.75 = 156.75 Volts from the resistor’s output 157 Volts.
That leaves the volts across the LED to the relevant values of around a couple volts or to be precise, the forward voltage drop of the particular type of LED.
Conclusion: From the above discussion and analysis, it becomes clear that voltage in any power supply unit is immaterial if the current delivering capability of the power supply is "relatively" low.
For example if we consider an LED, it can withstand 30 to 40 mA current at voltages close to its "forward voltage drop", however at higher voltages this current can become dangerous for the LED, so its all about keeping the maximum current equal to the maximum safe tolerable limit of the load.
While calculating series resistor values with LEDs, instead of using the standard LED formula directly, we can use the above rule first.
That means either we choose a capacitor whose reactance value only allows the maximum tolerable current to the LED, in which case a resistor can be totally avoided.
If the capacitor value is large with higher current outputs, then probably as discussed above we can incorporate a resistor to reduce the current to tolerable limits.
Example: In the shown diagram, the value of the capacitor produces 70 mA of max. current which is quite high for any LED to withstand. Using the standard LED/resistor formula:
R = (supply voltage VS – LED forward voltage VF) / LED current IL,
= (220 - 1.5)/0.02 = 11K,
Therefore the value of the resistor for controlling one red LED safely would be 11K.
The above theory has been assumed and deduced by me, I am not very sure about its feasibility, though
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