Voltage divider for ac

Status
Not open for further replies.
P

PRIYADHARSHINI PALANISAMY

Member level 2
Joined
Dec 23, 2013
Messages
52
Helped
2
Reputation
4
Reaction score
2
Trophy points
8
Visit site
Activity points
371
I am using 15v/1A transformer for my circuit.To reduce this voltage to 5v i am using voltage Divider circuit .(i.e) R1=2k,R2=1k.But the output of the circuit seems that the negative peak of the sine wave having reduced amplitude when compared to positive peak.
what is the reason for this problem?
and how can i solve this?
 

It will be better if you can post your input and output waveform.As of what i understood from problem description i think you want to reduce the AC voltage,well in that case Use Electrolytic Capacitors instead of Resistors for dividing voltage.
 

vicky001.iith said:
It will be better if you can post your input and output waveform.As of what i understood from problem description i think you want to reduce the AC voltage,well in that case Use Electrolytic Capacitors instead of Resistors for dividing voltage.
Click to expand...


What u have understood is correct.i want to reduce my ac voltage from 15v to 5v using resistor divider.but the amplitude of the positive and negative peaks are not same.negative peak value is reduced when compared to positive.
1.what is the difference between capacitive and resistor divider network?
 

This is an AC question so be careful using electrolytic capacitors, you must use non-polarized types.

However, the issue is more likely to be due to another factor - the load on the circuit. Please tell us how you are measuring these peaks, are you looking at the waveform on an oscilloscope and if so, where do you have the probes connected. With relatively high value resistors the output waveform will easily be distorted by current drawn from it. A schematic would help.

Brian.
 
Surely the voltage divider will not treat positive and negative half cycle differently.

I rather guess you didn't describe your setup well and have additional components connected to the voltage divider output. If you e.g. connect it to a 0-5 V ADC input, the negative half cycle will be cut because it exceeds the permitted voltage range.
 

hi
your Voltage apply to your circuits as differential. positive pick divide by 1/2+1 and negative voltage divide by 2/3.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…