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Still learning about electronics. I want to build a soft starter circuit, I found this schematic.soft start schematic

enter image description here

I can understand the schematic, but there's a 7805 voltage regulator circuit at the top right. I'm confused why this part of circuit is here as its +5 vDC output doesn't connect anywhere else in the overall circuit. Am I missing something or is the circuit incomplete?

Also, do you people think this is a well designed soft start circuit?

I want to do this as a learning exercise as well as having a practical usage.

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I'm confused why this part of circuit is here as its +5 vDC output doesn't connect anywhere else in the overall circuit. Am I missing something or is the circuit incomplete?

The schematic does not show it, but U4 requires and uses the 5 V supplied by the 7805. The schematic is is incomplete, and its author expects its users to know that U4 requires a power connection.

Also, do you people think this is a well designed soft start circuit?

The circuit in the schematic is quite clearly not the exact circuit in the video. An obvious pointer to this fact is that the circuit in the video has a LED, which is not present in the schematic.

Additionally, the circuit in the video does not appear to soft start the lamp, but rather to start it at full intensity and then, after a short bit, to dim it somewhat. However, it is possible this apparent dimming may be an artifact of the video camera adjusting to the sudden brightness.

I would add that the author of the site has a very poor understanding of electronics. To motivate a soft starter, he writes:

This surge current is much higher than the normal load current. Especially for inductive loads. When an inductor is cold or not energized, it works like a short circuit.

That is hogwash. That is not how inductors work. When an inductor is not energized, it impedes (i.e. it resists, in the colloquial sense) a sudden inrush current.

I want to do this as a learning exercise as well as having a practical usage.

Be aware that you would need to write code for, and program U4, the PIC12F675.

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  • \$\begingroup\$ If the schematic is correct the AC source is directly connected across the transformer, the triac control is only in series with the lamp L1. \$\endgroup\$ Commented Mar 31, 2024 at 11:13
  • \$\begingroup\$ @Nedd if the circuit is to work, the AC source needs to be directly connected across the transformer. The triac should only be in series with the lamp. \$\endgroup\$ Commented Mar 31, 2024 at 11:16
  • \$\begingroup\$ @MathKeepsMeBusy - The symbol marked "L1" is the output socket and is fed via the triac. The power input is directly across the transformer as required. \$\endgroup\$ Commented Mar 31, 2024 at 17:00
  • \$\begingroup\$ @KevinWhite good catch. Thank-you. \$\endgroup\$ Commented Mar 31, 2024 at 17:25
  • \$\begingroup\$ @MathKeepsMeBusy - It is not 'Hogwash' that iron-cored transformers can have very large inrush currents at switch on even when they are not loaded. Depending upon the phase at which power is applied it can even cause saturation where the currents increase dramatically. electrical4u.com/… \$\endgroup\$ Commented Mar 31, 2024 at 17:54
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The circuit is incomplete, but not wrong.

The microcontroller (U4) needs to be supplied with a supply voltage before it can perform its task of controlling the triac for soft start. This has to be a fixed voltage and this voltage is provided by the 7805 regulator (+5V) in the upper right of the schematic.

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  • \$\begingroup\$ The power circuit is in the upper right - the connections to the processor are implied and not directly shown. \$\endgroup\$ Commented Mar 31, 2024 at 17:55

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