ac 220v measurement using transformer less circuit

is this possible to measure 220v ac voltage using transformer less circuit . i-e first step down the voltage using resisters divider then through adc of any microcontroller measure the voltage . .

like

https://www.google.com.pk/imgres?im...ct=rc&uact=3&dur=1064&page=5&start=80&ndsp=19

aqildad said:

is this possible to measure 220v ac voltage using transformer less circuit . i-e first step down the voltage using resisters divider then through adc of any microcontroller measure the voltage . .

like

http://www.google.com.pk/imgres?img...ct=rc&uact=3&dur=1064&page=5&start=80&ndsp=19

Click to expand...

yes infact adapters do that as it is shown in image.
there should be a bridge rectifier but do not use capacitor else it will be a dc voltage & then give to micrcontroller via adc to measure ac voltage.

I think you better use opto-opamp line AD202. the picture which is shown has not measuring AC voltage capability.

Also, whatever you do, please consider electrical safety.

It is in general easy to find transformers specified to withstand mains voltage safely. With a capacitive connection you have to use capacitors classified as "Y" capacitors, if they are between mains and user accessible circuits. With resistors one has to have sufficient voltage ratings, and usually several in series.

Using optical isolation may be one of most elegant, but it is often not very simple. And you have to use optoisolators fulfilling the safety requirements.

he is actually using the ckt only to step down the voltage & not to measure. Later after stepping down it will be given to adc and then voltage can be measured using microcontoller.

Hi,

the circuit shown is not developed to measure mains voltage. It is a transformerless supply.

But the circuit has some mistakes in development.
For example there is a fuse rated 100mA. Usually 100mA is the hold current.
The same 100mA flowing through FR. This may be a fusable resistor. With 100mA there is 1W of dissipated power. This is a lot of heat.
Even worse. Again the 100mA are flowing through R1 rated at 1kOhms and 2W. But dissipated power is 10W.
So either the fuse is useless or the Rs have wrong rated power. R1 will blow before the fuse blows.

The main current limit is by C1. with 50Hz it is 6770 Ohms. limiting current to about 32mA.

*******
For AC measurement !!! without !!! electrical isolation one can use a voltage divider, a capacitor and a half_vref_biased ADC input.

Another solution could be to switch a optocoupler at a desired voltage level (maybe at half of estimated peak voltage). the result is a pwm signal (with duty cycle depending on input voltage) that can easily be calculated back to the input voltage by a ucontroller.

Klaus

KlausST said:

Hi,

the circuit shown is not developed to measure mains voltage. It is a transformerless supply.

But the circuit has some mistakes in development.
For example there is a fuse rated 100mA. Usually 100mA is the hold current.
The same 100mA flowing through FR. This may be a fusable resistor. With 100mA there is 1W of dissipated power. This is a lot of heat.
Even worse. Again the 100mA are flowing through R1 rated at 1kOhms and 2W. But dissipated power is 10W.
So either the fuse is useless or the Rs have wrong rated power. R1 will blow before the fuse blows.
Klaus

Click to expand...

lets us calculate:
let fs1(fuse resistance) value be negligible. And then the zener voltage is of 16v rating. As seen from diagram the voltage across two conducting diodes+load resistance together which are in parallel with the zener diode have voltage across them=16v.
now,
calculating the total resistance value before the zener i.e 100R (here R=2.125 for 110v so 4.12(approx) for 230v)
100R=100*4.25=425ohm
Xc1=1/(2*pi*50*.47u)=6772.55077ohm.
R1=1K.
total resistance=1K+6772.55+425=8190.55Ohm.
so current flowing will be (230-16)/total resistance
current=26.1053mA.
1)fuse rating is around 100ma & as we calculated current is 26.1053ma hence safe.
2)the power dissipation in R1=(26.1053mA)^2*(1K)=.6814watt and rated is 2w hence this is also safe.
how did you calculate?

Hi,

Why do one use a fuse?
It s not meant for normal operation. Ith normal operation one can calculate with about 30 mA. Everything is safe.
It is used to break the circuit if a device fails ant to prevent from overheat and fire.

Therefore the normal operation is meaningless. One must calculate for worst case.
To blow a fuse rated to hold 100mA you need at least 130...150mA.

I bet the 1kohms 2W resistor blows before the fuse. Therefore i wrote:

So either the fuse is useless or the Rs have wrong rated power.

Click to expand...

Klaus

i think we are deviating from the question that he posted that is: He just wants to step down 230v ac using the circuit(as per link) & wants to measure the ac voltage may be by using uc, in this we aren't loading the circuit. As the value of current is calculated to be at the max 30mA there are no changes in the load conditions(infact there is no load as such just a stepping down voltage circuit) then why will the current change when none of the parameters are changing? And hence why do we need to think about current going beyond the limit of that of fuse when that condition will never take place for the supplied voltage kept fixed (and the circuit parameters @ constant) . And as you quoted yes it is correct that using fuse is useless in normal operation. But it doesn't harm if it is there.

KlausST said:

I bet the 1kohms 2W resistor blows before the fuse.
Klaus

Click to expand...

sorry i didn't get this on what assumptions did you tell this.