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voltmeter on bridge rectifier

Farticus

Jun 21, 2015
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Question from "Noob".
I have a circuit using 24V DC generated from a 24V AC supply via a "Bridge Rectifier".

I want to instal a 30V digital voltmeter in the circuit but I cannot get it to work. It just flashes the voltage reading for a split second then reads "0" Volts.

The Voltmeter works just fine on a battery popwered 24V DC source.

What am I missing????

PS I am a REAL NOOB so not to high tech please

Thanks all
 

davenn

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Sep 5, 2009
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hi there

mite help if you told us what sort of voltmeter ??
give a link to a datasheet please

Dave
 

Minder

Apr 24, 2015
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Possibly does not like the 120Hz 100% ripple if no smoothing capacitor used?
Max.
 

AnalogKid

Jun 10, 2015
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When 24 Vac is full-wave rectified by a diode bridge, the result is not 24 Vdc - it is closer to 32 V which is an overrange for a 29.99V DC meter. 32 V comes from the secondary AC voltage times the square root of 2, minus the forward voltage drop across two power diodes.

Of course, since we have no schematic or part numbers, this is all guesswork.

ak
 

Farticus

Jun 21, 2015
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I have some Cheap Chinese "Specials" Voltmeters. Only cost about $4 each so I know that they are crap but can't understand why they work on batteries but not Bridge Rectified.If it is the "ripple" any suggestion on the smoothing capacitor I should use?. Circuit has to handle up to 3 A.
 

Arouse1973

Adam
Dec 18, 2013
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I have some Cheap Chinese "Specials" Voltmeters. Only cost about $4 each so I know that they are crap but can't understand why they work on batteries but not Bridge Rectified.If it is the "ripple" any suggestion on the smoothing capacitor I should use?. Circuit has to handle up to 3 A.

Look up one post, you have your answer I think. :)
Adam
 

Minder

Apr 24, 2015
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Without the Capacitor, the size of which is DC load dependant and % of ripple desired at full load, the voltage is slightly less than AC input, with cap it will be = 24 x 1.414 (minus a couple of volt drop across the diodes).
~32vDC.
M.
 

Farticus

Jun 21, 2015
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Tx Analogkid I understand your maths but... Why is it so???



Why is it the Multimeter tests the output as 24V AC input 21.64V DC output after bridge rectification. This would seem to validate the maths as 24V DC output less 2.36V allowance for the 2 diodes.The Multimeter gives accurate readings on Dry cell batteries. EG 9.4V 18.8V 28.2V DC

Why does the multimeter not read about 32V DC.

Would appreciate your comments to help me understand.

Thanks
 

davenn

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Sep 5, 2009
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Why is it the Multimeter tests the output as 24V AC input 21.64V DC output after bridge rectification. This would seem to validate the maths as 24V DC output less 2.36V allowance for the 2 diodes.The Multimeter gives accurate readings on Dry cell batteries. EG 9.4V 18.8V 28.2V DC

Why does the multimeter not read about 32V DC.

Would appreciate your comments to help me understand.


Because Analogkid wasn't clear with his/her explanation.
he/she neglected to state if they were talking about RMS or peak values.
RMS is usually used as most multimeters will give you the RMS voltage reading, hence you see the 21.64VAC and not ~32VAC

Also, don't panic too much with the difference caused by the voltage drop across the diodes 0.7V is a typical value, it can vary a bit.
Also you need to take into account that the open circuit voltage out of the transformer will be a bit higher than a loaded output of the bridge rectifier. so that 24VAC may drop a volt or 2 with the bridge connected


Dave
 

duke37

Jan 9, 2011
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Without a capacitor, the meter will read the average voltage which is Vrms*0.9
With a capacitor the meter will read the peak voltage, Vrms*sqrt(2).
 

Minder

Apr 24, 2015
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Further the RMS value equals the resistive or heating value/quantity of a AC power supply that would be equivalent to the DC voltage value that would produce the same effect.
M.
 

Farticus

Jun 21, 2015
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Thank you for your replies.
That does seem to clear up a few issues I had.
Seeing that a multimeter reads the RMS value I assume you would have to use a capicitor in the circuit to get the multimeter to read the peak voltage.
If this is so Is there a formula I can use to calculate the value of the capicator required to to get the multimeter to read the peak voltage?
 
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