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Finding voltages across resistors

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love_electronic

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Hi guys
I want to find out the voltages across RB1K, RB2K, RB3K and R4K.
How to do that?
at I/P-L and I/P-N 220v AC supply is connected.
IMG_20170919_141301.jpg
 

Re: finding voltages across resistors

1) Use a volt meter.
2) Use Kirchoffs laws. To make things easier, just consider one polarity, i.e., I/P-L is 220 volts (RMS) higher than I/P-N so that only two diodes are conducting. Then you've got three loops: one through the four resistors, two conducting diodes and R5BK. The second loop is through R5BK,R7Bk, ZD1BK and R6BK. The third loop is R6BK and the optoisolator diode.

I'll leave it to you to do the math.
 

Re: finding voltages across resistors

For a quick estimation of resistor voltage and power rating, consider that the capacitor is initially discharged, so each resistor is loaded with maximal 55 V AC.
 

Re: finding voltages across resistors

Are the 3 loops equation like that

Vr3bk + Vr4bk + Vd2bk + Vd4bk + Vr2bk + Vr1bk = 0
Vr5bk = Vr7bk + Vzd1bk + Vr6bk
Vr6bk = Vopto-diode
 

Re: finding voltages across resistors

Vin = Vr3bk + Vr4bk + Vd2bk + Vd4bk + Vr2bk + Vr1bk + Vr5bk
Vr5bk = Vr7bk + Vzd1bk + Vr6bk
Vr6bk = Vopto-diode

in the datasheet of opto-coupler the led forward voltage is 1.2v
so,
Vr6bk = 1.2v

Vr5bk = Vr7bk + 30v + 1.2v
Vr5bk = Vr7bk + 31.2v ------eq.A

Vin = Vr3bk + Vr4bk + 0.7 + 0.7 + Vr2bk + Vr1bk + Vr5bk
Vin - 1.4v = Vr3bk + Vr4bk + Vr2bk + Vr1bk + Vr5bk
Vin - 1.4v = i(178k +178k +178k +178k +402k)
Vin - 1.4v = i(1114000)
i = (Vin - 1.4)/1114000

Are my calculations right?
What value of Vin should i use, 220v RMS or 155V peak?
 

Re: finding voltages across resistors

You have neglected the capacitor, which is going to have an effect. As far as whether you should use RMS or peak, well, that depends on whether you want know the RMS or peak voltages, doesnt it? Also, 220 volts RMS is not 155 peak, it's 311.
 

Re: finding voltages across resistors

There is considerable source impedance; the input to the diodes 0.7M does not let the capacitor charge fully during each cycle.

The capacitor does not have high capacitance (is it 10uF or 0.1 uF?) to hold the voltage const because the output impedance is not high enough- 102K in parallel with the capacitor.

The zener and the IR LED cannot be considered as simple load because they are voltage dependent.

Just to get some idea, consider the RC time constant for 102K and 10uF - about a second. Hence without the other load, the capacitor is fine and will hold the charge for some time...

Look at the other side: the same cap with 5K will cause the voltage to drop fast and the LED will flash once and the cap voltage will drop close to 30V.

I guess the LED will flash at the line frequency. The voltages and currents need to be calculated graphically.
 

It is very simple if you can take some approximations.

Same current passes through R1, R2 and R3, R4. So the voltage drop across them will be same. You need to calculate the supply current.

Approx 1: Assume steady state; draw the sine voltages (after rectification)

Calculate the capacitive reactance at 2f: calculate the current.

Include the resistor: calculate the current.

Include the LED now: calculate the current. Do a graphical analysis. If include the zener and see the result.
 

Use a voltmeter if the physical circuit is at disposal. Since theoretically solving it will be a complex task (In term of calculations).
For theoretical calculations the diodes are serving a full bridge rectifier, they are converting to dc, then comes the capacitor (Check how it behaves for dc). check the components on right side oc circuit and their specs as well. Use the forum for hints and comment of love_electrical as well (I hope he did all clcs well).
 

Since theoretically solving it will be a complex task...

I agree but such problems are not at all uncommon. Many of us will go for a simulation but I love the good old fashioned graphical solution. It is simple and gives a very open view into the operation.

I can draw but I am lazy! So I shall just outline the steps:

1. R1 and R2 are in series; they can be combined. Same is true for R3 and R4.

2. If you look hard, you can see that the same current must always flow via these two pairs: hence they can be combined further: we replace R1, R2, R3 and R4 with one resistor 712K...

3. The input 220V AC (assume RMS: peak 310V) is rectified by the bridge of four diodes. Ignore diode drops and plot the result. You will get the following graph:
View attachment case1.pdf

4. Now add the R5 as load; the curve will not change but the peak will come down by some: 310*102/(102+712)=39V this will be the max voltage once we put the 102K load. The graph will remain same in form but the peak will become 39V.

5. Now put the capacitor; the capacitor will charge max to 39V (actually less) but the Xc for the cap has to be calculated at 100Hz. You will be charging the cap at const curr approx 310/712mA=0.4mA

6. The sine curve will be distorted (look more like a ramp) and it goes upto 39V.

7. Next branch is left as an exercise for the student. You have to divide the voltage 0V to 30V and 30V to up (two steps)
 

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