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need help to design transformerless power supply circit

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aruna1

Member
hi,I'm trying to run 20 series LEDs using 230V main AC supply.assuming one LED require 20mA and has 2V voltage drop across one LED it re quires 40V 20mA supply to turn on all LEDs.

I calculated resistor value to drop rest of voltage.

R=(230-40)/20 = 9.5KΩ 3.8W

total power dissipation is 3.8W+0.8W(for bulbs) = 4.6W

due to resistor nearly 82% power wastes.

my question is is there anyway to get 40V 20mA transformer less supply with less power waste? :confused:

Thanks
 

audioguru

Well-Known Member
Most Helpful Member
A high voltage capacitor connected in series with the LEDs will use its reactance to limit the current with no heating loss. A small resistor should be used in series to limit the current surge when the power is first turned on. A high resistance should be connected in parallel to the capacitor to discharge it when the power is turned off.
Everything in the circuit is a severe shock hazzard.
 

aruna1

Member
you mean circuit like this?
can you help me to calculate capacitor values and resistor values please?
it would be great if you can give me some equations to calculate capacitor and resistor values so i can calculate values so i can design circuit to get other voltage outputs and current outputs.

in here i need 40V and 20mA
and are there any ratings for capacitors and resistors?
thanks
 

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audioguru

Well-Known Member
Most Helpful Member
I forgot to add a rectifier in series. If it is a half-wave rectifier then the LEDs will blink quickly at the mains frequency. If it is a full-wave rectifier then the LEDs will blink at double the mains frequency. Add a filter capacitor to reduce the blinking.

I would never make such a dangerous circuit.
 

aruna1

Member
any idea how to calculate values?

what do you mean by dangerous circuit? is there any non-dangerous circuit?
 

millwood

Banned
A high voltage capacitor connected in series with the LEDs will use its reactance to limit the current with no heating loss.

after you have gone through the analysis to determine the value of that capacitor for this application, you will find it impractical or uneconomical to go down that route.

I do not know if a simple yet efficient and safe solution exists for this particular ask.
 

Hero999

Banned
I forgot to add a rectifier in series.
You mean parallel.


any idea how to calculate values?
1) work out the required reactance of the capacitor, for example the LED current is 10mA.

For a start, this is a half wave rectifier so we need to double the current at the LED is only conducting for half a cycle.

In this case you can ignore the LED voltage drop of the value of the series and parallel resistor because they'll be insignificant compared to the mains voltage and capacitor reactance.

I = 20mA

[latex]X_C =\frac{V}{I} = \frac{230}{0.01} = 11.5k \Omega[/latex]

Now calculate the capacitor value:

[latex]C = \frac{1}{2 \pi \times 50 \times 11500} = 277nF[/latex]

Just use a 220nF capcitor.
 

MrAl

Well-Known Member
Most Helpful Member
Hello,

Wouldnt the required value be more like 0.33uf though?

The voltage across the cap is 230v-40v, not 230v, right?

The other problem though is that if there is a diode in series (and another
in the opposite direction to ground) that means only 1/2 the power gets
to the LEDs, so in that configuration we would need a 0.66uf cap.

On the other hand, if he uses a full wave bridge the cap value can be 0.33uf.

Maybe a 470 ohm resistor in series to lower the turn on surge too.


Double check my values...
 
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