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Filter capacitor in power supply is increasing the voltage

J

Jon Danniken

Jan 1, 1970
0
I'm building an unregulated 12V power supply, but I've run into an issue I
can't seem to resolve. The supply consists of a mains transformer that
converts 120VAC into 12VAC. The resulting 12VAC is fed into a full-wave
bridge rectifier, with the resulting output of ~12VDC.

Now obviously I need some filtering on that 12VDC, so I added a capacitor
between the (+) and the (-) output of the bridge rectifier. The resulting
output is filtered, but as a result the voltage at the capacitor is now
~17VDC.

I'm somewhat familiar with why this occuring, having built a voltage
multiplier in the past, but is there any "trick" to prevent it from
occuring, or possibly a way (other than a resistor network or a voltage
regulator) to reduce it afterwards?

Thanks for any suggestions,

Jon
 
J

Jag Man

Jan 1, 1970
0
Why not use something like a 7812 regulator IC?

Ed
 
J

john jardine

Jan 1, 1970
0
Jon Danniken said:
I'm building an unregulated 12V power supply, but I've run into an issue I
can't seem to resolve. The supply consists of a mains transformer that
converts 120VAC into 12VAC. The resulting 12VAC is fed into a full-wave
bridge rectifier, with the resulting output of ~12VDC.

Now obviously I need some filtering on that 12VDC, so I added a capacitor
between the (+) and the (-) output of the bridge rectifier. The resulting
output is filtered, but as a result the voltage at the capacitor is now
~17VDC.

I'm somewhat familiar with why this occuring, having built a voltage
multiplier in the past, but is there any "trick" to prevent it from
occuring, or possibly a way (other than a resistor network or a voltage
regulator) to reduce it afterwards?

Thanks for any suggestions,

Jon
Unloaded, the cap will always charge up to near the peak voltage of the
incoming 12Vac. (1.4 X 12Vac is 17Vdc).
Other than inductors or regulators then the best option is maybe just use a
9Vac transformer.
Bear in mind, the DC voltage can drop a few volts when you put load on the
power supply.
regards
john
 
J

Jon Danniken

Jan 1, 1970
0
john jardine" said:
Unloaded, the cap will always charge up to near the peak voltage of the
incoming 12Vac. (1.4 X 12Vac is 17Vdc).
Other than inductors or regulators then the best option is maybe just use a
9Vac transformer.
Bear in mind, the DC voltage can drop a few volts when you put load on the
power supply.

Thanks, John, the voltage did indeed drop down to ~12.5VDC when I applied a
5A load. I'm curious about your statement regarding the inclusion of an
inductor though. In what way would an inductor be applied to reduce the
voltage?

Thanks,

Jon
 
J

john jardine

Jan 1, 1970
0
Jon Danniken said:
john jardine" said:
"Jon Danniken" wrote:
[clip]

Thanks, John, the voltage did indeed drop down to ~12.5VDC when I applied a
5A load. I'm curious about your statement regarding the inclusion of an
inductor though. In what way would an inductor be applied to reduce the
voltage?

Thanks,

Jon

Used in this situation the inductors are usually called "chokes". The idea
is to connect a choke to the output of the main capacitor, then feed into
yet another capacitor that supplies the load. It doesn't 'drop' the voltage
like a switcher or regulator or resistor would. It simply acts with the
extra capacitor to 'reshape' or 'mould' the nasty incoming waveform to
something flatter and smoother. As ...

___ o/p V
+-+--+--+------UUU-----+------o to load
o-. ,--, A A | choke |
AC in )|( |-+ | --- ---
)|( -(-+ --- ---
o-' '--' A A | cap1 cap2 |
+-+-----+--------------+------o

The choke and Cap2 form a very low frequency low pass filter acting as a
'flywheel' energy store that can smooth out the effects of the peaky
voltages that are charging up Cap1. Nett result is that the load voltage is
a nice smooth voltage hovering somewhere near the average (not peak) of the
incoming voltage. I.e near 12V.
Problem is that to be effective (at mains frequency), the choke inductor
needs lots and lots of Henries and ends up as some humungous, expensive lump
of steel and copper that is similar in size and cost, to the power
transformer. Hence at the high currents and low voltages of modern
equipment, is a technique little used nowadays.
regards
john
 
J

Jon Danniken

Jan 1, 1970
0
john jardine said:
Jon Danniken said:
john jardine" said:
"Jon Danniken" wrote:
[clip]

Thanks, John, the voltage did indeed drop down to ~12.5VDC when I
applied
a
5A load. I'm curious about your statement regarding the inclusion of an
inductor though. In what way would an inductor be applied to reduce the
voltage?

Thanks,

Jon

Used in this situation the inductors are usually called "chokes". The idea
is to connect a choke to the output of the main capacitor, then feed into
yet another capacitor that supplies the load. It doesn't 'drop' the voltage
like a switcher or regulator or resistor would. It simply acts with the
extra capacitor to 'reshape' or 'mould' the nasty incoming waveform to
something flatter and smoother. As ...

___ o/p V
+-+--+--+------UUU-----+------o to load
o-. ,--, A A | choke |
AC in )|( |-+ | --- ---
)|( -(-+ --- ---
o-' '--' A A | cap1 cap2 |
+-+-----+--------------+------o

The choke and Cap2 form a very low frequency low pass filter acting as a
'flywheel' energy store that can smooth out the effects of the peaky
voltages that are charging up Cap1. Nett result is that the load voltage is
a nice smooth voltage hovering somewhere near the average (not peak) of the
incoming voltage. I.e near 12V.
Problem is that to be effective (at mains frequency), the choke inductor
needs lots and lots of Henries and ends up as some humungous, expensive lump
of steel and copper that is similar in size and cost, to the power
transformer. Hence at the high currents and low voltages of modern
equipment, is a technique little used nowadays.

Thanks again for your reply, John. I've got enough now to play around with
componenets and look for results.

Jon
 
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