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# Transformerless power suply problem

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#### Cubyte

##### Newbie level 2
Hi
Just found this forum today and it looks like you guys know your stuff.
im using the circuit below on 220v
View attachment TRANSFORMERLESS-POWER-SUPPLY.bmp
which is a basic resistor capacitor psu which works fine on 220 volts A/c
however if i try to run it off a phase cut dimmer the safety resistor gets very hot.
my question is can a x2 capacitor run directly off 220v A/C ( being switched on at peak voltage all the time or simply putting it a 220 volt square wave)?

#### andre_luis

##### Super Moderator
Staff member
Cubyte,

Once the RMS value of the peak is higher than mean values, a pulsed haveform tends to geherate more heating than an square waveform ( Note that RMS calculations don´t assigns aritmetric equal weights to each point, but powered of 2 )

The efficiency of this kind of AC/DC power supply, is strongly decreased according load current consumption decreases ( At first sight, it seems to be a paradox ) due working is based on resistive divider.
So, despite the integrated current value be low, the peak current value at resistence is hight, performing a higher heating.

+++

#### KerimF

Hi,

I am not sure if I understood well your problem but I usually use a (high voltage) capacitor to supply a low current circuit on 220Vac 50Hz.
The role of the capacitor (as 0.94µF in your example) is to limit the supply current, based on the formula (RMS value and sinusoidal waveform):
I = V / Z = V * w * C = V * 2 * PI * F * C
so for V=220Vac, F=50Hz and C=0.94µ
I_cap = 66 mA (RMS)
Actually it will be a bit less than this value since we neglected the other small drop voltages in series with the capacitor.
The role of the resistor Rs (in series with the capacitor) is to to limit the inrush current till the capacitor reaches its steady state. It will dissipate P_Rs because of I_cap:
P_Rs = I_cap * I_cap * Rs
P_Rs = 0.066 * 0.066 * 10 = 44 mW (a relatively low value)

Now what happens if the input voltage is a ±220V square wave.
The formula above V*wC is no more valid :-(
And the time constant RC of the circuit is equal to 0.94µ * 10 = 9.4µsec
The time period of the input voltage is equal to 1/2F = 1/100 = 10ms
So in each half cycle, the capacitor will be charged by Rs from +/- 220V to -/+ 220V. The voltage step on the capacitor is therefore 440V. The repeated inrush current is equal:
I_inrush = V / Rs = 440 / 10 = 44A.
Approximately let us we assume this current as constant during the time constant RC and zero elsewhere. In this case the power dissipation on Rs:
P_dis = I_inrush * I_inrush * Rs * RC * (2 * F )
P_dis = 44 * 44 * 10 * 9.4e-6 * (2 * 50 ) = 18 Watt (relatively too high)

Hope this helps.

Kerim

Last edited:

##### Super Moderator
Staff member
Whoever thought up the idea of using capacitor reactance to drop AC volts, it was an ingenious idea.

My simulator shows your circuit will work for the most part. It will only succeed on AC sine. You'll need over 10 W rating for your safety resistor.

Is there a reason you're using two capacitors in parallel? Are you basing this on typical LED driver circuits which use .22 uF to drop house current to power an LED under 20 mA? I notice you want 35 mA.

The reactance concept works for sine waves of course. This is why you get current surges through your safety resistor when you run it on a lamp dimmer. Sudden pulses get through relatively unhindered in both directions.

You ask about using square waves. For similar reasons you'll get massive current surges through the diodes. You'll see voltage swings of several times the zener value across the load.

Even as it stands you're getting several amps through your circuit through each cycle. This is a lot of overhead for needing only a few milliamps. You may be better off using an ordinary diode bridge. Then to get 18V you would put a single 18v zener in parallel with load and filter cap (a customary method for low current regulation).

#### Cubyte

##### Newbie level 2
i'm trying to use the circuit on a dimmer to generate a 12volt and a 5v rail to run a pic12f615 and a driver for a mosfet.The pic looks at the phase angle to calculate the pwm to the mosfet.
is there away of getting this supply to work off a dimmer without using high power resistor?
i'm trying to get around 350mA so the capacitors in the circuit have been increased to 4.7mfd.

#### andre_luis

##### Super Moderator
Staff member
...is there away of getting this supply to work off a dimmer without using high power resistor?
i'm trying to get around 350mA...

Code:
P_resistor = ( 220V * 1.42 ) * ( 350mA ) ~ 107 W !!

That kind of power supply is indeed to use at low consumption circuit.

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#### KerimF

Code:
P_resistor = ( 220V * 1.42 ) * ( 350mA ) ~ 107 W !!

That kind of power supply is indeed to use at low consumption circuit.

+++

I fully agree with you.

In fact, the higgest value of the capacitor I may use in a direct power supply from 220Vac, is 2µ2 if half wave and 1µF if fullwave. The first can provide about 56mA @180Vac and in the second case we get about 51mA.

Note: Most of the time I need to use the half wave configuration (2 diodes only) since the powered circuits have to drive triacs. So one power terminal should be connected directly to the triac node M2 (or A2, it can be seen as a thyristor cathode).

Kerim

##### Super Moderator
Staff member
i'm trying to get around 350mA so the capacitors in the circuit have been increased to 4.7mfd.

Not recommended. An explosion may occur. Your capacitor and all diodes will have several amps going through them. The capacitor will overheat and blow, if the diodes don't blow first.

When you want 350 mA it's time to use a transformer.

#### kalyanasv

##### Full Member level 3
This may sound rather vague:But could be worth an attempt.
Essentially you have a lot of power to dissipate and want to do it without a coupled transformer I suppose you want to avoid the bulk transformer stuck on the driver next to the LED.
using Tesla's wireless power idea:
Though you will need a transformer type coil to get the EM waves you do not need the bulk of laminates etc.
i.e. You have a supply and power the coil ..it works for a couple of feet distance with very low efficiency.
But serves your purpose of not needing a bulky transformer nor a power resistor.

#### mvs sarma

leave alone the other parameters, for life risk( many times others use what we design and many are not careful)
we should avoid using non-isolated power supplies. What ever alarm/caution you may give, it may legally support the designer, but life, if gone, cant be restored.

Thus I seriously discourage transformer less/ non isolated power supplies.

#### KerimF

Thus I seriously discourage transformer less/ non isolated power supplies.