This post explains how to calculate transformerless power supply circuits using simple formulas like ohms law.

We all know how Ohm’s law works and how to use it for finding the unknown parameter when the other two are known. However, with a capacitive type of power supply having peculiar features and with LEDs connected to it, calculating current, voltage drop and LED resistor becomes a bit confusing.

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 ÂµF, 400 V (breakdown voltage) when connected to a 220 V x 1.4 = 308V (after bridge) mains supply will produce a maximum of 70 mA of current and an initial voltage reading of 308 Volts.

However this voltage will show a very linear drop as the output gets loaded and current is drawn from the “70 mA” reservoir.

We know that if 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 308 volts by 70 mA gives 4.4V. 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 × 4.4 = 88 volts, so the output now will show a voltage of 308 – 62.8 = 220 volts DC(after bridge).

For example with a 1 watt LED connected directly to this circuit without a resistor would show a voltage equal to forward voltage drop of the LED (3.3V), this is because the LED is sinking almost all the current available from the capacitor. However the voltage across the LED is not dropping to zero because the forward voltage is maximum specified voltage that can drop across it.

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 it's 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 - 3.3)/0.02 = 10.83K,

However the 10.83K value looks pretty huge, and would substantially drop the illumination on the LED....none-the-less the calculations look absolutely legitimate....so are we missing something here??

I think here the voltage "220" might not be correct because ultimately the LED would be requiring just 3.3V....so why not apply this value in the above formula and check the results? In case you have used a zener diode, then the zener value could be applied here instead.

Ok, here we go again.

R = 3.3/0.02 = 165 ohms

Now this looks much better.

In case you used, let's say a 12V zener diode before the LED, the formula could be calculated as given below:

R = (supply voltage VS – LED forward voltage VF) / LED current IL,

= (12 - 3.3)/0.02 = 435 Ohms,

Therefore the value of the resistor for controlling one red LED safely would be around 400 ohm.

The above theory of How to Calculate and Deduce Current, Voltage Parameters in Transformerless Power Supplies has been assumed and deduced by me, I am not very sure about its feasibility, though

We all know how Ohm’s law works and how to use it for finding the unknown parameter when the other two are known. However, with a capacitive type of power supply having peculiar features and with LEDs connected to it, calculating current, voltage drop and LED resistor becomes a bit confusing.

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 ÂµF, 400 V (breakdown voltage) when connected to a 220 V x 1.4 = 308V (after bridge) mains supply will produce a maximum of 70 mA of current and an initial voltage reading of 308 Volts.

However this voltage will show a very linear drop as the output gets loaded and current is drawn from the “70 mA” reservoir.

We know that if 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 308 volts by 70 mA gives 4.4V. 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 × 4.4 = 88 volts, so the output now will show a voltage of 308 – 62.8 = 220 volts DC(after bridge).

For example with a 1 watt LED connected directly to this circuit without a resistor would show a voltage equal to forward voltage drop of the LED (3.3V), this is because the LED is sinking almost all the current available from the capacitor. However the voltage across the LED is not dropping to zero because the forward voltage is maximum specified voltage that can drop across it.

**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 it's 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 - 3.3)/0.02 = 10.83K,

However the 10.83K value looks pretty huge, and would substantially drop the illumination on the LED....none-the-less the calculations look absolutely legitimate....so are we missing something here??

I think here the voltage "220" might not be correct because ultimately the LED would be requiring just 3.3V....so why not apply this value in the above formula and check the results? In case you have used a zener diode, then the zener value could be applied here instead.

Ok, here we go again.

R = 3.3/0.02 = 165 ohms

Now this looks much better.

In case you used, let's say a 12V zener diode before the LED, the formula could be calculated as given below:

R = (supply voltage VS – LED forward voltage VF) / LED current IL,

= (12 - 3.3)/0.02 = 435 Ohms,

Therefore the value of the resistor for controlling one red LED safely would be around 400 ohm.

The above theory of How to Calculate and Deduce Current, Voltage Parameters in Transformerless Power Supplies has been assumed and deduced by me, I am not very sure about its feasibility, though

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can we use this circuit for power supply for water level circuit and fluorescent tube circuit

ReplyDeleteyes can be used...

Deletehi, sir,

ReplyDeletecan we connect 3 blue led with resistor for this circuit, when 70 ma is the output current

hi, yes you may connect them, but use 0.33uF/400V capacitor instead of the shown 105/400V.

Deletethe zener diode won't be required.

Hello Dada

ReplyDeleteIf I want To create 5V 500mA output, can I connect 5 250v 225K capacitors in parallel because they have 100ma and 24V output voltage?

No you cannot do that, unless the output load operating voltage doesn't match the input mains level, adding caps won't work, rather would become dangerous for the LEDs.

Deletehi guys ,, i have done this circuit as you illustrated in the diagram.... but my output is just 1.5v after brige circuit... can u help me with this

ReplyDeleteRemove the zener and the capacitor and then check

DeleteSir,

ReplyDeletePlease explain what type calculation behind "220v X 1.4=308v",why we using and where we get 1.4 here to multiple, any standard or calculation.

Thanking you,

kumaran muthu

Kumaran,

DeletePlease research "RMS voltage" you will get answer

Hi sir!I'm a student.My project is transformerless power supply (220VAC-12VDC). I want to know about how to caculate C1 and R1 by using real formular. Can you help me?

ReplyDeleteHi sokleang, you can refer to this article for the details:

Deletehttp://www.homemade-circuits.com/2015/01/calculating-capacitor-current-in.html

R1 is not important, you can use any resistor above 330K and below 2M2 for this resistor

Hi sir.how to known value of C1 R1?

ReplyDeleteHi Sir,

ReplyDeletehow I will calculate the current limiting resistor of transformerless power supply. Please feedback my question

Hi Joni, It will need to be calculated as per the specifications of the load that you intend to connect at the output

Deleteyou can use Ohm's law for calculating the parameters

Hi there.. I need to get 5VDc 1A output from 220VAc 60hz.. Please give me the C and R value

ReplyDelete1 amp is too high and is not recommended from capacitive circuits.

Deletehello dear swagatam , can i use above ckt as power supply for PIR motion sensor ckt

ReplyDeleteHi Shankar, I won't recommend it, better to go for a SMPS adapter, your cellphone adapter will do the job nicely

DeleteHai sir, just tell me am i thinking right, i want to use lm2596 to produce 36volt and 3 amps, im planning to give transformerless ac to dc converter as a input of lm2596, i confusing that what is the voltage and current for the input and how do i get from dropping capacitor

ReplyDeleteHi, 3 amp is too high for a capacitive power supply...I would advise you to build an SMPS instead

Delete