1
\$\begingroup\$

Noob question inbound:

I'm getting into PIC programming, and I'm using a kit to build a microcontroller setup on a breadboard. Following the instructions here:

http://www.bypic.co.uk/index.php/Part_2_Serial_connection_and_construction

The images below are of the schematic for the circuit, and a photo of an actual wired breadboard. I want to understand how the schematic relates to the breadboard photo before I start wiring my own. I think it's making sense to me with one exception:

How is the 3v3 current being applied to pin 28?

I can see how power is being routed to the right rail, and the bridge to pin 13. The ground connections make sense. However, I cannot for the life of me see how power is being applied to pin 28. Am I missing something, or is there a missing wire in the breadboard photo?

enter image description here enter image description here

\$\endgroup\$
0
1
\$\begingroup\$

It is hard to make out your breadboard photo, but it doesn't seem like you have connected pin 28 to your 3.3V rail (right hand vertical rail on breadboard).

You need to add a link from your 3.3V rail, to pin 28, just like you have jumped that vertical rail to pin 13.

\$\endgroup\$
1
  • \$\begingroup\$ To be clear, that photo is from the guide on their website. I wanted to make sure it was their mistake before I wired up my own, differently from their photo. \$\endgroup\$ – Dave Swersky Jun 8 '16 at 16:05
1
\$\begingroup\$

enter image description here

Figure 1. Closeup of breadboard power.

I think you've missed something.

  • There's a link above the black plug. This must go to the red '+' strip.
  • There's what looks like a white component between the link and pin 28 as far as I can see. It's a little lost in the shadow and the blue line.

I've no idea what the white component is. Fuse? - unlikely. Zero-ohm resistor? - maybe but why when everything else is a piece of wire. Diode? - unlikely due to voltage drop.

\$\endgroup\$
3
  • \$\begingroup\$ This is the correct answer. The detail photo clearly shows an uninsulated wire from 28. I think the white blob may be a splitter to also send the 3.3V to the USB cable. \$\endgroup\$ – usajnf Jun 13 '16 at 22:18
  • \$\begingroup\$ Why would anyone use a splitter on a breadboard when a second link would suffice? Thanks for the vote of confidence. \$\endgroup\$ – Transistor Jun 13 '16 at 22:24
  • \$\begingroup\$ Well I absolutely agree with you that I have never used a splitter before but the schematic shows 3.3 to the USB. You already ruled out real components. It could be a cookie crumb dropped on the board before the photo! \$\endgroup\$ – usajnf Jun 13 '16 at 22:29
0
\$\begingroup\$

Missing wire. There is no connection of power to pin 28. As this is the AVCC pin, the issue would be that the analog section of the PIC wouldn't work right. If used.

In either case, it would be wise of you to connect it.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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