5
\$\begingroup\$

LM7812 voltage regulator is recommended to be used with decoupling capacitors at its input and output as in the below diagram:

enter image description here

And here is the part from the manufacturer Fairchild(see recommended C1 is 0.33uF):

enter image description here

But another manufacturer Texas Instruments have different C1 recommendation on their datasheet which is 0.22uF.

I have two questions:

1-) I don't have 0.33uF cap: but I have 0.22uF, 0.1uF and 0.01uF caps instead. For C1, can I use these three caps in parallel as their sum is 0.33 instead of a single 0.33uF cap?

2-) Why those two manufacturers recommend different C1 values for the same component(LM7812)?

\$\endgroup\$
13
  • 4
    \$\begingroup\$ I'm not sure those are exactly recommended values. Rather those are what THEY had on hand to build the test circuit with and tabulate the numbers. \$\endgroup\$
    – Trevor_G
    Commented May 31, 2017 at 19:58
  • \$\begingroup\$ @Trevor I put 0.22uF, 0.1uF and 0.01uF ceramic caps in parallel to obtain 0.33. But what if I use a single 0.47u instead? Isnt 4.7u a superset of 3.3u. I mean 4.7u does the decoupling job even better than 0.33u right? Im confused. Another thing I never understood sometimes they put decoupling caps for power rails in parallel like 1u and 100n ect. combinations. But 1u already does the job of 100n. Isnt it. Do you have any idea about it? \$\endgroup\$
    – floppy380
    Commented May 31, 2017 at 20:07
  • 1
    \$\begingroup\$ First, there's a difference between what's specced as a test circuit and what's a real "recommended value" -- second, generic caps like that are often toleranced at about 20%. There's no sense in trying to get closer than that. \$\endgroup\$
    – Scott Seidman
    Commented May 31, 2017 at 20:10
  • 2
    \$\begingroup\$ It really does not matter much. Those values are really the recommended minimums. Bigger is better. You see them in parallel because different cap types have different high frequency characteristics. Some low ESR caps in parallel with some bulk 10uF caps won't hurt. Paralelling is a little less common these days since you can get some pretty large value ceramics nowadays. \$\endgroup\$
    – Trevor_G
    Commented May 31, 2017 at 20:16
  • 1
    \$\begingroup\$ @Trevor Oh never thought of it. Do you mean small capacitors' small ESRs lower the equivalent ESR? So lets say if I have a decoupling cap at supply rail as 1uF and its ESR is 10 Ohm, by adding a 10nF in parallel which has 1 Ohm ESR, the equivalent ESR drops to 0.9 Ohm and it means less parasitics. Did I get you correct? \$\endgroup\$
    – floppy380
    Commented May 31, 2017 at 20:20

2 Answers 2

Reset to default
5
\$\begingroup\$

From Texas Instruments datasheet:

9 Power Supply Recommendations

If the device is more than six inches from the input filter capacitors, an input bypass capacitor, 0.1 μF or greater, of any type is needed for stability.

From Fairchild datasheet:

CI is required if regulator is located an appreciable distance from the power supply filter.

Although all the circuits in Fairchild data sheet show 0.33μF and the TI datasheet 0.22μF, they do NOT specifically say a size. As the TI datasheet says 0.1 μF or greater, so 0.47μF is fine. Better than 3 capacitors in parallel.

8.2 Typical Applications

8.2.1 Fixed Output Voltage Regulator *Although no output capacitor is needed for stability, it does help transient response. (If needed, use 0.1-μF, ceramic disc).

On the output, TI states:

8.1.1 Shorting the Regulator Input

When using large capacitors at the output of these regulators, a protection diode connected input to output (Figure 15) may be required if the input is shorted to ground. Without the protection diode, an input short causes the input to rapidly approach ground potential, while the output remains near the initial VOUT because of the stored charge in the large output capacitor. The capacitor will then discharge through a large internal input to output diode and parasitic transistors. If the energy released by the capacitor is large enough, this diode, low current metal, and the regulator are destroyed. The fast diode in Figure 15 shunts most of the capacitors discharge current around the regulator. Generally no protection diode is required for values of output capacitance ≤ 10 μF.

So you are free to pick appropriate capacitors for input and output to reduce ripple, noise, etc.

I'd go with the application information as indicated by the TI datasheet. The Fairchild datasheet is limited to just data, with minimal application guidance.

\$\endgroup\$
4
\$\begingroup\$

Ci can be as large as you want. A 0 impedance input voltage isn't going to hurt the regulator. In fact, the lower the better. I'd probably use a 10 µF ceramic, more if the input voltage is of questionable impedance and stability.

What you quote doesn't say what Co needs to be, only what value the rest of the specs are for. You have to look elsewhere in the datasheet to find the min and max allowed values. If I remember right (your job to check), the 78xx series regulators are stable with 0 ESR output caps. The only real limitation is therefore the initial surge current as the output cap is charged up. Since these regulators protect themselves from output shorts, there is effectively no upper limit from the regulator's point of view. I'd probably use a 1-10 µF ceramic, depending on what I'm already using in that design.

\$\endgroup\$
2
  • \$\begingroup\$ 0 ESR output caps? You mean tantalum type? \$\endgroup\$
    – floppy380
    Commented May 31, 2017 at 20:28
  • 2
    \$\begingroup\$ @don: No, I mean caps with 0 ESR. Of course there is no such thing, but some technologies, like ceramic, have low ESR with large enough values to be useful in this application. \$\endgroup\$
    – Olin Lathrop
    Commented May 31, 2017 at 20:30

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.