8
\$\begingroup\$

Please understand that i am actually a really beginner in electronics.

I found a really nice topic on how to create a 5V DC from a 9v power supply. All is ok. But in order to smooth out ripple, the author use two capacitors in order to smooth values before the voltage regulator, and add another after the voltage output pin.

What I don't understand is that the capacitor seems to be placed in parallel with the voltage regulator. Not in serial manner like I was expecting to see. So I really don't understand how you can use the smoothed output values, since in the circuit diagram, it seems that the direct output goes to the ground.

I know that when capacitors are in series, you add their values. But the input pin of the voltage regulator seems to be on one terminal while the capacitor is on another. How can the voltage regulator benefit of the capacitor?

I know that what I say is plenty wrong, and the circuit diagram is correct. But I can figure out how this circuit is working?

Here is the schematic.

schematic

By the way, do you know where can I find tutorial explaining how to read schematics? There is lot of topic explaining electronics but I haven't found any valuable link for electronic circuit explanation.

\$\endgroup\$
  • \$\begingroup\$ It's a schematic, not a scheme. \$\endgroup\$ – Adam Lawrence Dec 9 '13 at 17:01
12
\$\begingroup\$

I'm afraid you need to review capacitors.

I know that when capacitors are in serial, you add their values.

When capacitors are in parallel, their values add

Not in serial manner like I was expecting to see.

Loosely speaking, a capacitor has "infinite" impedance at DC. So, if the capacitor were in series with the regulator output, there could only be AC current through. Thus, the load would not have a DC voltage across, only an AC voltage. This is just the opposite of what we want.

When the capacitor is placed across (in parallel with) the regulator output and ground, the capacitor presents a (hopefully) low impedance for AC current through the capacitor and ground, "shunting" the ripple current around the load thus reducing the AC voltage across the load.

But, for DC, the capacitor is effectively open so the full DC voltage appears across the load. This is just what we want.

\$\endgroup\$
  • \$\begingroup\$ I was just about to press post on my answer, yours is extremely similar, and much more eloquent. \$\endgroup\$ – Jeff Langemeier Dec 9 '13 at 14:21
  • \$\begingroup\$ Hi Alfred, thank u for your response. If i understand clearly, a capacitor blocks DC current when placed in serial (this is something you just remember me). But if a capacitor is on one branch, and i test the voltage into the other branch (like with the regulator), then it will be smoothed anyway? I thought the smoothed output will be at the end of the branches node. \$\endgroup\$ – Mr Bonjour Dec 9 '13 at 15:05
5
\$\begingroup\$

First of all, these capacitors aren't there for smoothing the ripple, but to maintain stability of the regulator. You say you're a beginner to electronics, so (for now), just take it as a fact, that they need to be there. :-)

The 78xx regulator works roughly this way. There is a bipolar transistor placed between IN and OUT pins in the regulator, you can imagine that as a variable resistor. You could just place a fixed resistor there instead (leaving GND pin open) and calculate its resistance as R = (VIn-VOut)/IOut. The pity is that you generally know neither IOut nor VIn, as both may vary as the circuits works. So you need a mechanism that would set the resistance according to changes of these variables. This mechanism is called negative voltage feedback. There is a complex circuitry in the regulator IC that measures output voltage (voltage between OUT and GND pins) and compares it to an internal stable voltage source (again, for now, don't care where this voltage comes from). If the regulator detects a voltage drop on the output (i. e. you connect another LED on the output), it opens the transistor more, lowering its resistance and delivers more current to the load. When you put the additional load away, the voltage would rise and the regulator closes the transistor, cutting the overvoltage away.

An ideal regulator wouldn't require any of the capacitors, but there are some properties of the real circuit design that make it unstable (oscillations of voltage would appear on the output). That's why you need to place a correct cap on both input and output; just follow the datasheet and (important!) place the capacitors as close to the IC as you can.

Hope this helps. :)

\$\endgroup\$
  • \$\begingroup\$ these capacitors aren't there for smoothing the ripple, but to maintain stability of the regulator Isn't that the same thing, especially in this case? \$\endgroup\$ – Passerby Dec 9 '13 at 17:10
  • \$\begingroup\$ I believe it is not. On "smoothing the ripple" I imagine the work the capacitor does in an ordinary 50-Hz rectifier with diodes (holding the output voltage while input voltage is less or even negative). This is IMO the purpose the OA actually means by this term. On the other hand, especially the output capacitor of 7805 is more related to the frequency and phase properties of the feedback loop. The 7805 is not meant to produce AC on the output, as contrasted to a transformer. But maybe it is just a terminology issue. \$\endgroup\$ – Ladislav Dec 9 '13 at 17:30
  • \$\begingroup\$ Thanks for the reply. I really see my lacks on electronic! Look like i will need more than 10 years in order to understand all those components work... ;) \$\endgroup\$ – Mr Bonjour Dec 10 '13 at 10:37
1
\$\begingroup\$

Ladislav had it right. There are situations where a regulator may oscillate, and it has less to do with coupled noise (ex. from long wires) and more to do with the control loop stability inside of the regulator.

If you use a regulator that uses a NPN pass element (older/classical style) then you can probably get away with alot of things since it is a pretty stable topology. However, low-dropout (LDO) linear regulators are becoming more popular for good reason and these use a PNP pass element. The topology with a PNP or PMOS pass element requires more compensation to make it stable. You will most likely see oscillations with an LDO regulator if you aren't careful.

Here is a great application note: http://www.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=snoa842

\$\endgroup\$
-3
\$\begingroup\$

I find the above answers not appropriate.

Here's what I think:

  1. The Capacitor on the input : It is used to eliminate the electrical noise in case the sub unit containing this regulator IC is at a distance from the main transformer in the system. The length of the wire acting like an antenna attracts switching noise, motor noise etc. This input capacitance helps eliminate that.

  2. The capacitor on the output side: It is used to eliminate the transients caused by the switching on the Totem Pole outputs on the digital ICs that may be connected on the output. Know that if in a Totem pole output, if both the transistors in sequence on the totem pole are ON simultaneously even for a small instant, it creates a momentary short acting like a negative going ( but not negative ) transient pulse effectively pulling the regulator voltage to zero. Suppose that many such ICS are connectected to the o/p. In such a case, there is a transient propagation of this undesirable signal. The capacitor on the output side is used to eliminate the transients caused by the switching on the Totem Pole outputs on the digital ICs

\$\endgroup\$
  • \$\begingroup\$ The other answers are perfectly correct. Without the output capacitor, the regulator could oscillate even if no transients were introduced by the load. Load transients are better handled by bypass capacitors near the loads that generate them. Also, you fail to address OP's misconception about parallel and series capacitors. \$\endgroup\$ – The Photon May 17 '14 at 5:07
  • \$\begingroup\$ I would appreciate if you please tell me why the regulator oscillates ? \$\endgroup\$ – Manjunath Pai H May 17 '14 at 5:19
  • \$\begingroup\$ Read Ladislav's answer, he explains the control loop well. Like any control loop, if the phase goes to 180 degrees before the open loop gain goes below 1, it will oscillate. Many linear regulators are designed so that an external output capacitor is required to avoid this condition. \$\endgroup\$ – The Photon May 17 '14 at 5:32
  • \$\begingroup\$ Sorry to say that your answer is unclear \$\endgroup\$ – Manjunath Pai H May 17 '14 at 5:45
  • \$\begingroup\$ Try it for yourself then. Go make a linear regulator in your favorite simulator and check the stability conditions. Then add the output capacitor and see how they change. \$\endgroup\$ – The Photon May 17 '14 at 5:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.