1
\$\begingroup\$

To get, say 5VDC, from the AC mains, after stepping down the high AC voltage using a transformer, we have to rectify it and regulate the output DC voltage. As I have seen, there seems to be two prevalent methods of voltage regulation on the internet. One is using a zener diode in reverse breakdown region and the other is to use LM7805 regulator IC. But I do not understand what are the pros and cons of using one or the other.

One more thing, if LM78XX IC is itself regulating the voltage what is the need of using capacitors in parralel with it? Can anyone please help me clear this doubt!

enter image description here

Thanks in advance!

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Please compare the price for each and the stand-by losses (no or very light load) and report back! \$\endgroup\$
    – winny
    Commented Aug 21, 2019 at 17:11
  • 2
    \$\begingroup\$ One very important detail would be the required current compliance for the regulated output. Another would be how precise the output voltage needs to be (and/or if you can accept the addition of an expensive potentiometer.) Another might be how stable, over operating temperatures, you want the output voltage to remain. Etc. A lot of different specs would go into making some choice. There are still a few situations where a zener and BJT might make sense. Fewer still where just a zener without a BJT may. \$\endgroup\$
    – jonk
    Commented Aug 21, 2019 at 17:14
  • 2
    \$\begingroup\$ You have asked two distinct questions. Please ask the second one in a separate thread. Thanks. \$\endgroup\$
    – TimWescott
    Commented Aug 21, 2019 at 18:12
  • \$\begingroup\$ [Following from the comment above] - But before you ask a new question about linear regulators (like the 7805) and its associated capacitors, please search for previous questions on that topic, and make sure that you are not going to ask a duplicate question. There have been many previous questions, where the purpose (and the need, or not) of those capacitors has been explained. The datasheets for those regulators usually have some explanation too. Thanks. \$\endgroup\$
    – SamGibson
    Commented Aug 21, 2019 at 18:27

3 Answers 3

7
\$\begingroup\$

Zener

You need to supply a standing current to the Zener that is greater than the largest load current. This current is always being consumed in the power supply. Moreover, a Zener diode doesn't have really good regulation (I'm going to accept @Justme's 4%, but it'll be in various data sheets), and a Zener diode is subject to temperature variations.

Linear Regulator

A linear regulator consumes the load current plus a "quiescent current", and that quiescent current is far lower than the standing current you need on a Zener (at least with modern regulators). So when the load current is zero, you're only consuming the regulator's quiescent current, not every bit of current you'll ever need, and more.

Linear regulators are far more accurate than Zeners, and can be easily trimmed if necessary for even more accuracy (or, these days, you can just spend extra $$ and get them pre-trimmed if need be).

Linear regulators are internally temperature compensated.

There's more advantages than that -- but it's been a long, long time since regulators commonly used Zeners as a reference, much less as the main regulating element.

\$\endgroup\$
6
\$\begingroup\$

7805 is a complex IC that can regulate the output fairly well with about 4% tolerance and is in series with the load. Just like other ICs, it needs the capacitors to keep voltages on input and output stable during load transients to operate properly, or it might oscillate so it would be a very poor regulator. A zener must shunt current and is much worse as a regulator. Switching mode regulators start to be just as common, so these may not be the two prevalent ways to regulate voltage.

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

A Zener shunt regulator will consume a lot of power because it's carrying the max current, less the load current. To work at all such a solution needs a series resistor to keep from frying the diode, which also will have loss. Zeners also very poor at regulating voltage over a wide current range.

A linear regulator will have almost no losses at minimum load (only quiescent current - the power needed to run the regulator’s circuitry.) Once you start to draw current, loss is (Vin-Vout)*Iout. Linear regulators have fairly good line and load regulation due to internal feedback gain and a stable internal bandgap reference.

One drawback of a linear regulator is that it requires an overhead voltage to work. For the 7805 that's about 2V. This sets your minimum loss at 2V * Iout. There are improved, Low Dropout (LDO) regulators that require less overhead, as little as 50mV in some cases, so lower losses are possible.

Linear regulators benefit from bypass capacitors to improve transient response and stability. That said, there are NMOS-based LDO's that claim not to need bypassing (e.g., TI 'cap-free' NMOS).

Have you considered a switcher? TI and others make them as drop-in replacements for 7805 in TO-220 package.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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