Question about power transformer circuitry

[futuristicmonkey]

I opened up a power cube I have from an old cdplayer. The input is 120VAC, 60Hz, 8.6W. the output is 6VDC, 600mA. I've recently become interested in transformers (I took an electronics class in grade 10 - but it focused on DC circuits, I hope to take th enext class this year)..and so I disassembled this power cube.

Inside it, I found what I expected -> the transformer, the rectifier diodes, but I also found a capacitor, and a resistor.

Here's a link to a schematic diagram I made...circuit.jpg
EDIT: The capacitor actually has a value of 3300uf.

I understand how the diodes rectify the AC current into DC, and I understand the need for the resulting DC current to be filtered in order to "flatten out" the wave from the original AC current. But what I don't understand is: how are the capacitor and the resistor are able to do that?

 

[TuxDave]

So R1 is a 330 ohm resistor that directly goes across the positive and negative terminals of the DC output? Can you double check that? After you get back to me, I think I have a story why it's there.

 

[futuristicmonkey]

Originally posted by: TuxDave
So R1 is a 330 ohm resistor that directly goes across the positive and negative terminals of the DC output? Can you double check that? After you get back to me, I think I have a story why it's there.

It sure does. My original hypothesis was that it was there to prevent a short.

 

[Mark R]

The resistor is there to ensure that the capacitor discharges when mains power is cut off.

If it wasn't there, and there was no load - the capacitor would remain charged almost indefinitely. When connected to a device, the device would receive a short burst of power from the capacitor, which could potentially cause undesired operation.

 

[TuxDave]

Originally posted by: Mark R
The resistor is there to ensure that the capacitor discharges when mains power is cut off.

If it wasn't there, and there was no load - the capacitor would remain charged almost indefinitely. When connected to a device, the device would receive a short burst of power from the capacitor, which could potentially cause undesired operation.

That was my guess too. I was a little confused about the low resistance value but I saw how huge the capacitor was and I guess it make sense. Together they'll have an RC time constant of 0.1s.

To answer the OP's question, the diodes merely rectify the signal but it isn't flat. The capacitor stores the peak voltage and so when the signal returns to zero and the diodes turn back off, the output DC voltage will be held by the capacitor, hence the 'flattening' out part. Picture it as a peak detector which will hold the highest voltage that it sees. The resistor will discharge the capacitor but at a time constant that's 10 slower than what you're rectifying (I'm assuming a 60Hz) signal, you don't have to worry about the capacitor losing its charge too fast.

 

[futuristicmonkey]

Thanks for the great answers! Especially that last one - about the capacitors. It cleared up a few things for me - even a thing or two which I didn't ask.

 

[futuristicmonkey]

Originally posted by: TuxDave

Originally posted by: Mark R
The resistor is there to ensure that the capacitor discharges when mains power is cut off.

If it wasn't there, and there was no load - the capacitor would remain charged almost indefinitely. When connected to a device, the device would receive a short burst of power from the capacitor, which could potentially cause undesired operation.

That was my guess too. I was a little confused about the low resistance value but I saw how huge the capacitor was and I guess it make sense. Together they'll have an RC time constant of 0.1s.

Actually I just noticed that I made an error when I wrote down the value of the capacitor. It is actually 3300uf. Now - does that mean that if it was fully charged, and then disconnected from the wall - it would take the resistor 100 seconds to discharge the capacitor?

 

[TuxDave]

Originally posted by: futuristicmonkey

Originally posted by: TuxDave

Originally posted by: Mark R
The resistor is there to ensure that the capacitor discharges when mains power is cut off.

If it wasn't there, and there was no load - the capacitor would remain charged almost indefinitely. When connected to a device, the device would receive a short burst of power from the capacitor, which could potentially cause undesired operation.

That was my guess too. I was a little confused about the low resistance value but I saw how huge the capacitor was and I guess it make sense. Together they'll have an RC time constant of 0.1s.

Actually I just noticed that I made an error when I wrote down the value of the capacitor. It is actually 3300uf. Now - does that mean that if it was fully charged, and then disconnected from the wall - it would take the resistor 100 seconds to discharge the capacitor?

Take the capacitance value (3300*10^-6)and the resistance value (330)and multiply them together. That's the RC time constant in seconds. The percentage of voltage remaining after t seconds is e^(-t/RC). So when t goes to infinity, there is essentially no voltage left. For practical purposes to drop to below 1%, I usually use 5-7 time constants so for this case since RC = 1 second, it'll take roughly 5-7 seconds to fully discharge.

 

[Calin]

And it would take some 2 seconds to discharge it to 10% of its max voltage

 

[Torched]

Originally posted by: TuxDave

Originally posted by: Mark R

To answer the OP's question, the diodes merely rectify the signal but it isn't flat. The capacitor stores the peak voltage and so when the signal returns to zero and the diodes turn back off, the output DC voltage will be held by the capacitor, hence the 'flattening' out part. Picture it as a peak detector which will hold the highest voltage that it sees

Here's another way to put it. AC voltage has a rise and a fall time hence Alternating Current. Then to become DC current (or close to it i.e. ripple) it must be rectified. The rectifying diodes act to block the negative voltage see Here . The resulting output looks like this here.
The capacitor is needed to maintain the desired peak voltage. It act as a battery of sorts: on the rise of AC voltage the cap is charged, on the fall of AC voltage the cap discharges preventing the voltage from falling below a certain threshold(farad value would be the variable for that equation.) This charging and discharging looks like this. The difference would be quite obvious if you were to measure with a DMM or Oscilloscope: with the cap you would notice barely a ripple in the voltage. Without the cap the voltage alternation would be highly decernable.
Hope this helps.

 

[Dough1397]

67% will be charged or discharged in the first time constant... lol remember that from electronics class

 

[PowerEngineer]

The parallel resistor and capacitor form a low-pass filter that will partially block the higher frequency components of the rectified sine wave (think about the fourier series equivalent of that rectified sine wave). I'm too lazy to calculate the break frequency for the given componemt values, but I'm sure it's well below the 120 Hz noise inherent in rectifying the 60 Hz AC. As other have said, it helps smooth out the variations in the rectified wave to make it closer to the ideal DC output.