Need Circuit Explanation

Discussion in 'General Electronics Discussion' started by rdevendrakumar1989, Oct 22, 2012.

1. rdevendrakumar1989

29
0
Aug 9, 2012
Hai friends,

I have some doubts in the attached circuit..

1. How to calculate the output voltage of RC parallel circuit in input side.

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2. Harald KappModeratorModerator

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Nov 17, 2011
The impedance of a parallel circuit is
Xp=Xr*Xc/(Xr+Xc)

Xr=R
Xc=1/(2*pi*f*C) where f is the frequency of the signal.

You insert these values into the equation and get the complex impedance of the circuit. As a first approximation in this application you can use the absolute value of the RC-circuit's impedance as a series resistance to the rest of the circuit.

3. rdevendrakumar1989

29
0
Aug 9, 2012
can you please explain more.. what am understand from this circuit is..

Input voltage = 230v

After RC parallel circuit O/P = 230v . It restricts amount of current only..Because its doesn't restrict voltage..

after bridge o/p= 330v D.c

Here the capacitance value only 63v dc and LED took aroud 60 vdc.. what about the excess volt.. It doesn't affect the cicuit capacitor and LEDs.. This is my question.. am confused toooooo much .. please explain about the circuit..

4. Harald KappModeratorModerator

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Nov 17, 2011
The RC combination acts as a series resistor, limiting the current to the rest of the circuit. Limited current can only develop a limited voltage across the electrolytic capacitor. This voltage is used to drive a current through R2 to light up the LEDs.

Harald

5. duke37

5,364
772
Jan 9, 2011
The parallel circuit restrics the current and so the voltage is less. If you do not have a load to take this current, then as you say, the voltage on the 63V capacitor will be over 300V. That would be quite exciting.

I would think that the capacitor and resistor in series with the leds are not needed.

I hope that the entire circuit is placed where no fingers can touch

6. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,849
Jan 21, 2010
How can it restrict current but not voltage?

for the purpose of this circuit it acts as a resistor with the value given by the real part of the reactance.

If your input voltage is 220V, then do your calculations as if it were 320V.

You then have the reactance, a resistor, and a structure with a fixed voltage across it all in series.

if you remove the LEDs (15 * 3.4V) = 51V then you can model the rest as a voltage divider (for the capacitor voltage), and series resistance s for current.

You have the reactance and the resistance in series (with 320V - 50V = 270V across it) and ohms law will tell you the current. From that you can calculate the voltage across the resistor and thus (when you add 50V) the voltage across the capacitor.

For the purposes of these calculations, ignore the resistor across the capacitor on the mains side of the bridge rectifier and the bridge rectifier itself.

This gives steady state operating characteristics, it works slightly differently as the power is applied or when there's a spike

7. rdevendrakumar1989

29
0
Aug 9, 2012
Thanks for your valuable comments.. Thanks to Mr.Steve,harald and duke.. am not cleared in excess of voltage given to the electrolytic... The load only decides the current.. Then what about the volatge? As per steve Led consumes LEDs (15 * 3.4V) = 51V & Capacitor Voltage 63V. Totally 114V . But the output of rectifier around 326v .. what about Remaining (326-114=212) 212 V.. Please clarify

8. Harald KappModeratorModerator

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Nov 17, 2011
No!
The current through the load will lead to a voltage drop across the RC circuit in the input.
You can't add the voltage across the LEDs and the voltage rating of the electrolytic capacitor.
For one the 63V is the rating, meaning that the capacitor can be used with voltages up to 63 V. This doesn't mean the voltage is actually 63 V.
Second the capacitor and the LEDs are in parallel. voltages in parallel circuits don't add, they are the same (look up Kirchoffs laws).

I made a short non-representative simualtion and it turns out the voltage across the electrolytic capcitor is on the order of 45 V ... 50 V.

Harald

9. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
You completely fail to understand

A) the load has 51 V across it,
B) the 560 ohm resistor R2 has some voltage across it
C) the capacitor C1 has some voltage across it.
D) the Capacitor C2 ha the sum of (a) and (B) across it.
E) your mains voltage can be assumed to be 220*1.4142136, or about 320V
F) the voltage across (B) and (C) is (E) - (A) = 270V

You determine the reactance of C1 at your mains frequency -- this is a value that can be presumed to be a resistance.

Using ohms law you use the resistance of R2 and the reactance of C1 along with the voltage (F) to determine current.

There is more, but I'll just be repeating what I said above.

10. rdevendrakumar1989

29
0
Aug 9, 2012
Mains frequency is 50Hz. Then reactance is around 14.475K ohm. please tell me the reamaining .. My major doubt is,

1. Is the load side decides the input of bridge ..
2. Rectifier output??? after connecting load

11. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
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Jan 21, 2010
V = IR, I = V/R, R = V/I

OK, you now have sufficient information to calculate everything (A to F above).

Give it a go.