How much capacitance do I need in the input of LM78L05

Question

I want to use LM78L05 as a fixed output voltage regulator as shown in the datasheet, as I cited below. My regulator is 2 meters away from the power supply filter. My input voltage is 24V, however this is an automotive application, so I have a transient voltage suppressor of 33V in the input too.

Edit:

I thought I wrote about the current, but I was wrong. I am going to pull at most 5 mA of current. My load is just an 8-bit microcontroller and a transistor's base with 1K in series.

I know that I need at least a 10nF capacitor load in the output to limit high frequency noise, as per Note 4 in the electrical characteristics table. However, I am confused if I need a low value (220nF to 1\$\mu\$F) capacitor in the input for decoupling (or bypassing?) purposes, or for just bulk purposes. I saw only those values in the datasheet, not higher.

This stated, can I use only a 10\$\mu\$F electrolytic capacitor in the input and a 150nF polyester capacitor on the output as shown? If I do, will I encounter any problems?

Answer 1

The 33V TVS isn't good enough. Rated reverse standoff voltage is always lower than breakdown voltage. For instance the Littelfuse 1.5KE39A is rated at 33V, but breakdown voltage can be as high as 41V. The LM78L05's absolute maximum input voltage is only 35V. A TVS is still a good idea though, since you're working in an automotive environment. I'll get back to it in a minute.

Russell suggests using a series resistor, and I concur. The resistor will drop the input voltage and form a low-pass filter with the LM78L05's input capacitor. I would even go a step further, and also place a zener diode on the input, so that you get a shunt pre-regulator. This will cause a little higher current consumption, though, but with a 1k\$\Omega\$ series resistor and a 12V zener this will only be 6mA.
You can then safely use a 1.5KE20A TVS, rated at 17V and with the maximum breakdown voltage of 21V your regulator will be safe.

Then the capacitors. If TI says a 330nF is required at the input, for pete's sake, put it there! I would also add a 10 to 33\$\mu\$F electrolytic; the zener is a voltage regulator, and all regulated voltages need a buffer capacitor.

PS: unless the datasheet mentions a minimum dV/dt for the input voltage (it doesn't) there's no upper limit for the input capacitor. Go for that terafarad cap if you feel like it.

further reading
Application Notes for Transient Voltage Suppressors from On Semiconductors.

Answer 2

Please confirm < 1 mA max load current. This makes an immense difference.

Simple shunt regulator with zener diode preregulator to greatly reduce spikes :

Two resistors, a zener diode and a TL431 or similar shunt regulator and a few capacitors would be VERY adequate in many cases.
Resistors in series drop the 24 V to 5V. A zener at the resistor midpoint clamps this to say 12V or some convenient value and greatly reduces transients. A capacitor across the zener further increases this effect. TL431 dissipates unwanted current.

Example - Say you want <= 1 mA, Vin min = 20V, Vin max = 30V, Vout = 5V.
Rmax = Vdrop/I max = (20-5) / 0.001A = 15k.
Now design for 5 mA max for lots of headroom = 3k max for 5 mA so a bit higher OK.
Split !~= 3k by ~= 2:1 to be say 2k2 and 1k.
2k2 will drop ~10V at Vinmin=20V, and 1k will drop about 5V.
Place a ~+ 9V Zener at midpoint of two R's. Say 8V2.
Program TL431 to 5V with say 2 x 10k resistors.

Check:

Vin = 20V.
I zener unloaded = V/R = (20-8.2)/2k2 =~ 11.8/2.2 = 5.4 mA.
I in 1k for 5V out = V/R = (8.2-5)/1k = 3.2 mA.
OK as we want <= 1 mA - could decrease R.1k slightly

Dissipation in 2k2 = V^2/R = (20-8.2)^2 / 2200 = 63 mW.

Now try Vinmax = 30V.
Current in unloaded zener = V/R = (30-8.2)/2200 =~ 10 mA.
Situation for 1k does not change as 8V2 is the same.
In practice Vzener will rise very slightly.
Dissipation in 2k2 = V^2/R = (30-8.2)^2/2200 = 0.22 Watt.
Use 1/2 Watt resistor OR reduce current somewhat.

Add a say 10 uF and 0.1 uF ceramic across zener and result should be good in many cases. Time constant of 2k2/1k + 10 uF =~ 150 Hz. About 25 Hz corner frequency. Use a 100 uF electrolytic for more notional rejection. TL431 will add extra noise rejection.

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**Comment on LM29xx family regulators:

Many of these are are specially designed for an automotive environment (check spec sheet in each case) and will survive in places where an LM7805 will be destroyed or will protect your 5V powered circuit in situations where an LM7805 would allow it to be destroyed.

LM2940 data sheet here.
This is a 5V. 1A regulator. $1.65/1 at Digikey and $1.15/100 and about $0.60 in very large volumes - for TO220 package in each case.

They say:

• The LM2940/LM2940C positive voltage regulator features the ability to source 1A of output current with a dropout voltage of typically 0.5V and a maximum of 1V over the entire temperature range. Furthermore, a quiescent current reduction circuit has been included which reduces the ground current when the differential between the input voltage and the output voltage exceeds approximately 3V. The quiescent current with 1A of output current and an input-output differential of 5V is therefore only 30 mA. Higher quiescent currents only exist when the regulator is in the dropout mode (VIN − VOUT ≤ 3V).

• Designed also for vehicular applications,

  The LM2940 / LM2940C and all regulated circuitry are protected from
  reverse battery installations or 2-battery jumps [related to conditions experienced in cars].

  During line transients, such as load dump when the input voltage can momentarily exceed the specified maximum operating voltage, the regulator will automatically shut down to protect both the internal circuits and the load.

  The LM2940/LM2940C cannot be harmed by temporary mirror-image insertion.

  Familiar regulator features such as short circuit and thermal overload protection are also provided.

Features

• Dropout voltage typically 0.5V @Iout = 1 Amp

• Output current in excess of 1A

• Output voltage trimmed before assembly

• Reverse battery protection

• Internal short circuit current limit

• Mirror image insertion protection

Answer 3

The reason for this "required if more than xxx distance" requirement is that one of the things that the capacitor does is act as a short circuit for alternating current signals that appear on the power rails (and are unwanted).

If there is a capacitor there, but is far away, it means that there is a resistance in series with that capacitor. Some of these unwanted voltages will appear across that resistance.

So the capacitor is useful even if the DC coming in over those long wires is as flat as a pancake.

These "bypass capacitors", as a rule of thumb, have to be located close to the devices being powered.

Bypass caps are often put individually across the power supply legs of individual integrated circuits or other components to try to trap these stray signals.

(You may also see several different capacitances used, because large caps are good for going down to the low frequencies, but become inductive at high frequencies. That's where the smaller capacitors take over and provide bypass.)