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What is the purpose of R1 (120 Ω, 2W)?

  1. When the voltage drops from 24 V to 5 V, a lot of heat is generated in the 78L05.
  2. Put a resister on the 78L05's input. Say we need 5 V, 100 mA output from the 78L05.
  3. What is the resistor for?
    • limit current
    • limit voltage
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    \$\begingroup\$ Looks like this resistor is only here to drop an excessive amount of voltage before we apply it to the voltage regulator. In simple words, to reduce the power dissipation in the voltage regulator. Because 24V-5V = 19V So the power dissipation is 19V * 0.1A = 2W. This is a lot for such a small package. This is why R1 was added. This is my guess. \$\endgroup\$
    – G36
    Commented Dec 28, 2022 at 7:56
  • \$\begingroup\$ (C. reduce power dissipated in 78L05) (The "+24V" annotation in the schematic is funny.) (The 24v tag is deprecated.) \$\endgroup\$
    – greybeard
    Commented Dec 28, 2022 at 7:56
  • \$\begingroup\$ L7805 data sheet Page 27 \$\endgroup\$
    – StainlessSteelRat
    Commented Dec 28, 2022 at 8:02
  • \$\begingroup\$ 120 ohm is a tad bit high for a fusable resistor but perhaps several reasons, input impedance for the TVS, shift losses and fusable resistor? \$\endgroup\$
    – winny
    Commented Dec 28, 2022 at 10:59

1 Answer 1

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It performs two functions. If an input transient comes along it will be clamped by the TVS with excessive voltage dropped across the resistor.

It also reduces the maximum dissipation by dropping voltage.

However it should be noted that this is a rather marginal design. The TVS doesn't really start to clamp until the input voltage is well over 30V which is the stated maximum input and also the absolute maximum input voltage for the L78L05.

At the stated maximum output current of 100mA there will be about 105mA through the resistor and it will drop 12.6V and dissipate 1.3W (acceptably within the 2W rating for many purposes).

The regulator, with 30V in, will dissipate about 1.8W so the junction will be about 100°C hotter than ambient with 6cm^2 of copper heatsink. Chance are the junction will be hotter than absolute maximum even in a cool office environment, and if Ta gets much hotter than that, and there's an enclosure keeping in the 1.3W from the resistor and the 1.8W from the regulator the semiconductor will be well into danger territory.

So there's two ways for this circuit to die catastrophically.

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  • \$\begingroup\$ so this is very cheap&dangerous way right? \$\endgroup\$
    – KimJiHoon
    Commented Dec 28, 2022 at 9:44
  • \$\begingroup\$ i'd rather using step down regulator(such as a CN3903) thank you \$\endgroup\$
    – KimJiHoon
    Commented Dec 28, 2022 at 9:48
  • \$\begingroup\$ Yes, it will run cool. If the input is a regulated 24V then the input range might be safe enough. \$\endgroup\$
    – Spehro Pefhany
    Commented Dec 28, 2022 at 10:04

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