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walid

Transformerless power supply discussion!

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I wouldn't call it a power supply because its ouput current is nearly nothing.
It only puts about 1.6mA into the zener and output capacitor only half the time because it is half-wave. Therefore the 1.6mA will go into the zener and the output current and voltage will be very low.
R2 limits the charging current into C2 to 0.6A if it was connected to the mains at the moment it was at its peak of +311V.

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Hi Audioguru
First of all I thank you for your kind answer. you must complete the road with me to make me understand as you hope, so
(1) The 1.6mA, I think you depend on my calculations to get this number, that is 1.6mA approx = 220Vrms/130.5Kohm, is this true?
(2) If (1) is true, R2 must connected to GND, where is it?     
(3) I agree with you that it is a half wave, D2 passing the +ve half to zener, and D1 passing the -ve half to GND, the question is why D1, I think that D2 is sufficient to do the job!
(4) If I want this circuit to provide me a 12 VDC 50mA, I do the following:
        ->for 12VDC I use a 12V zener
        ->for 50 mA I reduce the Rtot to 220/50m = 4.4K ohm\
        -> calculate C2, R1, R2 as follows:
              C2 is the same 10nF 400V Ceramic or Polyester Capacitor
              R2 also the same = 470R 

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(1) The 1.6mA, I think you depend on my calculations to get this number, that is 1.6mA approx = 220Vrms/130.5Kohm, is this true?

220V - 12V / 130.5k = 1.6mA, but since it is half-wave it will be about half.

(2) If (1) is true, R2 must connected to GND, where is it?

The zener is connected to gnd.

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Hi audioguru
Thank you very much for your very good answer, especially that part about D1 which helps in discharging the cap C2.
Now, I use Electronic Workbench EDA software to simulate the circuit shown below as Fig.PU1. I noticed that the readings are dependant on R_load value. At low values of R_load zener OFF and I_z = 0 A.
I record some of these values as follow:

R_load I_c2 I_R1 Iz I_load V_load
10K 106.4 1.466 46.30 1.2 12.05
5K 105.8 1.457 44.24 2.4 12.05
1K 105.9 1.456 34.33 12.02 12.02
0.2K 104.4 1.452 0 46.25 9.25
0.01K 107.2 1.467 0 47.5 0.475

Can you please explain why this dependence?
thank you very much.
walid.

post-2833-14279142463354_thumb.jpg

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Guest Alun

Notice how the voltage doesn't change much until the load is under 1k? That's because (as audioguru said) it's not a proper power supply.  You'll find this is the same for any supply, try putting a 6 ohm load on a LM7812 power supply and see what happens to the load voltage. The maximum current you circuit can supply is about 10mA which is fine for small CMOS circuits but not much use for anything else, also it isn't isolated from the mains so the circuit running from it should be enclosed in an insulated box.

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hi

please look at the following circuit, it is a transformerless power supply:

AlimStr.GIF


this is a 12 volt and can supply a max current = 15 mA
I calc this current as:
Xc = 1/2 pi f C = 14.47 K ohm
I = (220 -12 - 1.4)/Xc = 15 mA
the 1.4 volt is the drop voltage on two diodes
Now, please tell me What are the disadvantages of this circuit, and
why the zener diode is Often burned
what if i connect 100 to 300 ohm resistor in series with that diode.

the last question: at the ollowing circuit, why using R3 =220 ohm
its value did not affect the the current at all
index.php?action=dlattach;topic=9093.0;a
guru said before that: "The circuit needs this resistor to limit the current in C1 if the circuit was connected to the mains at the moment it is at its peak of 311V"
I did not fully understand the intent, especially, I know that the freq of electricity = 50 that the voltage reaches 311 fifty times per second, how can this be?

thank you very much.

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Hi Walid,
Your mains is a sine-wave that goes from a peak voltage of 311V then through 0V to 311V with the other polarity and continues to alternate. If the circuit is connected to the mains at a moment when the voltage is at a 311V peak and there is not a resistor to limit the capacitor's charging current then the current will be huge and might cause damage.

I don't see anything in the circuit to cause the 1W zener diode to burn. Its max dissipation is 12V x 15mA= only 180mW.

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Hi guru

(1) the word "at the moment" is the problem
if we agree that the voltage reaches 311 volt 100 times per a second
that is every 1/100 sec the voltage value is 311v
in the human sense it is every time, every any bit of time..........

(2) in this circuit AlimStr.GIF
there is no resistor to limit the capacitor's charging current, why there is no problem?

(3)Also related to the above circuit; if i increase C1 to 900nF to get about 60mA, when there is no load, zener eats all this current
if it is, for discussion, 1/2 watt it will be burned, so if i don't want to use a 2watt zener, and i want to use a 300 ohm resistor in series with the 1/2 watt zener, would this protect the zener?
How much this resistor decrease the current?
thank you very much

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It is a dangerous and bad circuit. Of course there is a problem if nothing limits the capacitor's charging current. If the capacitor is not rated for high currents then it will fail.

The fuse limits the current like a resistor. A fuse is a resistor.

If a 300 ohm resistor is in series with the zener diode then it reduces the current only a tiny amount, because it is also in series with the much higher reactance of the current-limiting capacitor.

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hi guru

to the following circuit,
AlimStr.GIF
(1)if i changed it to be a half wave rectifier, is the current still 15 mA as with full wave
(2) what the difference of putting a 300 ohm resistor in place of the red line or blue line in the followin circuit.
series res.jpg

thanks

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With a half-wave rectifier the current will be about half and the ripple voltage will be doubled.

If a resistor is added in series with the zener diode then it is also in series with the much higher reactance of the current-limiting capacitor, so the zener diode current or the load current will be reduced only a tiny amount.

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Hi
I answer myself

look at the followig figure:
transformerlessps.jpg
At the +ve half of the main AC voltage (311 V peak) the upper zener acts as a diode and the lower zD2 acts as a zener regulating the voltage to 16.7v
At the -ve half of the main AC voltage (-311 V peak) the lower zener acts as a diode and the upper zD1 acts as a zener regulating the voltage to -16.7v

In my software there is no 16V zener, there is a 15 v zener so I use it to simulate this part of the circuit

THANKS TO MY TEATCHER AUDIOGURU WHO LEARN ME 99.9999% of what I know in ELECTRONICS

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