Here I have a circuit for a full-wave bridge rectifier with capacitor filtering.
And here is another circuit with two filter capacitors. What effect does the second filter capacitor provide?
\$\begingroup\$
\$\endgroup\$
2
-
\$\begingroup\$ The top circuit is full wave rectified. For each cycle power is pushed into the capacitor two times. For the bottom circuit each output is half wave rectified, so power is pushed out only once per cycle. \$\endgroup\$– user338146Commented Sep 30, 2023 at 1:30
-
\$\begingroup\$ Were you trying to create a full wave split supply in the second schematic? For that a center tapped transformer will often be used, with only the transformer's center tap being grounded. \$\endgroup\$– NeddCommented Sep 30, 2023 at 6:23
Add a comment
|
3 Answers
\$\begingroup\$
\$\endgroup\$
2
And here is another circuit with two filter capacitors. What effect does the second filter capacitor provide?
The two filter capacitors do not interact.
It is not just another filter capacitor. The second circuit has two separate power supply outputs, one positive with respect to GND and one negative. Note that the anode (+ end) of C4 is connected to GND. This is a clear indication that the circuit output is negative with respect to GND.
In the second circuit, D2 and D5 do not do anything. If you delete them and redraw the circuit, how it operates might be more clear.
-
\$\begingroup\$ It's a voltage doubler circuit! \$\endgroup\$– vu2nanCommented Sep 30, 2023 at 10:39
-
\$\begingroup\$ The D2 and D5 do protect the caps from reverse cap voltage during startup and shutdown, whem there are loads between both +V and -V outputs. \$\endgroup\$– JustmeCommented Sep 30, 2023 at 18:56
\$\begingroup\$
\$\endgroup\$
7
The first circuit is that of a power supply with a full-wave bridge rectifier.
The second circuit is redrawn for easier understanding.
Diodes D2 & D5 serve no purpose and hence may be removed.
It's the circuit of a voltage doubler type of power supply.
-
1\$\begingroup\$ Not quite. You drew a voltage doubler. The original circuit has a center tapped ground so you get positive and negative outputs in regard to the center tap which is 0V ground. And the diodes you removed protects caps from reverse voltage. \$\endgroup\$– JustmeCommented Sep 30, 2023 at 8:54
-
\$\begingroup\$ My redrawn circuit is no different from the original but for the earth symbols. It has the configuration of a voltage doubler circuit. \$\endgroup\$– vu2nanCommented Sep 30, 2023 at 10:54
-
2\$\begingroup\$ Sorry, I was wrong about center tapping. The original circuit uses it as dual half rectifier voltage output to get positive and negative supplies. If you don't use the middle 0V point then you do have a single ended output from voltage doubler. \$\endgroup\$– JustmeCommented Sep 30, 2023 at 10:55
-
-
\$\begingroup\$ I was hoping the TS would do the re-draw and figure it out for himself. Also, C3 and C4 do not need reverse-polarity protection. Side question to vu2 - what package do you use for schematics? \$\endgroup\$ Commented Sep 30, 2023 at 14:24
\$\begingroup\$
\$\endgroup\$
If you are asking about the small voltage difference from the first circuit to the second, note that in the top full bridge configuration the supply path coming from the transformer passes through two diodes, resulting in two diode voltage drops, while in the second circuit the path contains only one diode voltage drop.
The second circuit being configured as a half wave rectified supply will actually produce more ripple voltage than the first. A larger filter capacitor will better average out that ripple, hence the negative supply using the larger value capacitor shows a slightly higher average DC voltage.
