How to Calculate Transformerless Power Supplies


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 µF, 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.


Transformerless Power Supply Calculation

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 - 1.5)/0.02 = 11K,

Therefore the value of the resistor for controlling one red LED safely would be 11K.
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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11 comments

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October 26, 2012 at 10:23 PM delete

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

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Anonymous
January 29, 2013 at 4:05 PM delete

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

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January 29, 2013 at 4:07 PM delete

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

the zener diode won't be required.

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Anonymous
August 28, 2013 at 10:54 AM delete

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

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August 28, 2013 at 7:27 PM delete

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.

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July 25, 2015 at 1:58 PM delete

hi 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

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July 26, 2015 at 9:49 AM delete

Remove the zener and the capacitor and then check

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January 28, 2016 at 12:42 PM delete

Thank you for mentioning that you've assumed and deduced the "Transformerless Power supply theory".

You started with calculations converting the RMS voltage to a peak value, to calculate your maximum current. Unfortunately, the 70ma value is roughly the value obtained using the RMS voltage instead of the peak one. The real value is closer to 100ma (97.74 ma) if you deal with the peak voltage.

Later in you article, you talked about a theory of a linear voltage drop based on the load current. You're using the peak voltage and the 70ma value (obtained using the RMS voltage) to compute your 4.4v/ma voltage drop rate. How valid can this theory be if the values used for the computation aren't valid to start with ?

Afterwards you came up with the 20ma voltage drop example. You estimated it to 88 volts, using the RMS voltage, but you subtracted 62.8V from the input peak voltage to come up with a 220V after the bridge out of nowhere.

Globally you're simply shooting in the dark with your deduced and assumed theory. What logic is driving your assumptions ?

When you say "We all know how Ohm’s law works and how to use it for finding the unknown parameter when the other two are known.", I strongly believe you're not part of that group. I'm explaining.

Applying the equation "R = (supply voltage VS – LED forward voltage VF) / LED current IL, = (220 - 1.5)/0.02 = 11K" while the required voltage drop is so huge, is completely absurd. Did you take few seconds to evaluate the power dissipation required for that puppy ? We're talking of 4.4 watts. Pure non-sense. What an obvious risk of fire this is !!!

In conclusion, submitting material like this one based on assumptions and deductions, is putting the other hobbyists at risk. Keep in mind that some newbies don't know the rules when dealing with main voltages, care more or less about RMS, peak, average values, as long as the light turns on, or at least until something pop up, burns, etc...

Somebody will either get shocked or burned, specially when wrong advices are given to start with. There is a note in red indicating that the construction of the circuit should be initiated only if the concept is perfectly understood. What if the theory isn't valid to start with ?

Ideally, circuits should have been tested before submission, otherwise this will lead to failures, frustrations and accidents.

Here and on many other sites, some self-proclaimed electronics knowledgeable people are creating these useless circuit drawings. I wonder who got the editor's job !

Hoping to see better material in the future.

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January 28, 2016 at 2:58 PM delete

I have seen your previous comments and I am aware of the level of understanding that you have regarding electronics so I can only feel sorry for you.

In the above article I have tried explain the effect of a high voltage low current power supply such as a capacitive power supply on a load, may be with your lower level of understanding you missed the point.

The 70mA is correct, you are just embarrassing yourself by giving 100mA figure....please learn electronics instead of exposing your illiteracy in the filed.

Input AC is sinusoidal, so we ought to considering the average value of the voltage, current in the calculations….sorry it’s no use explaining to somebody who doesn’t even know the basics.
You can build a practical circuit and check the current……well only if you know how to do it.

I deleted your earlier comments just you save you from embarrassing yourself, this time I won't do that.

Even great people make mistakes and I am no exception, but discourteous people like you come here specifically to pick out the 1 negative out the 100 positives. I can understand your frustration since you are having a tough time digesting electronics,...but it requires brain for that.

You have already embarrassed yourself by showing your ignorance about IC 555, then you showed your frustration on the "earth" symbol in electronic circuits because you did not know it's a usual practice to use these symbol for the negative line.

You even could not build a simple LED fader circuit?
You better learn electronics first before saying all those rubbish….

Do your homework properly before criticizing somebody, especially somebody who has contributed so much for the newcomers.
Get well soon!

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January 28, 2016 at 3:09 PM delete

gosh! this is even more ridiculous:

"Applying the equation "R = (supply voltage VS – LED forward voltage VF) / LED current IL, = (220 - 1.5)/0.02 = 11K" while the required voltage drop is so huge, is completely absurd. Did you take few seconds to evaluate the power dissipation required for that puppy ? We're talking of 4.4 watts. Pure non-sense. What an obvious risk of fire this is !!!"


Do you understand anything, or have you gone nuts.

The calculation is not for a direct 220V input it's supposed to be after the capacitive reactance.

As far as risk is concerned, all mains operated non-isolated circuit are dangerous and I have appropriately provided the warnings in those articles.


It's just a waste of time......

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