I am a newbie to electronics and currently working on my very first electronics project on Atmega16.I want to make sure that no matter how high current is supplied(by mistake), it shouldn't damage my laptop.So here's what I did till now,

enter image description here

I haven't soldered barrel jack near voltage regulator. Here I connected MISO, MOSI, RESET and SCK from the connector to IC as shown in the imageenter image description here.

I was warned that if the voltage regulator doesn't work as intended and large current enters the circuit it will not only damage my circuit but also laptop's port and fry the motherboard. So, in theory, I thought if I don't connect VCC coming from my laptop to USBASP then my laptop should be safe. So my question is, 1)is it a good idea to not connect VCC from the computer and completely be dependent on an external source for power? 2)if yes then will I face some issues while programming the device as I am not supplying the VCC from a computer? 3)Also is it ok to disconnect ground from the laptop also?

I am not well with designing tools like eagle or ORCAD so here is the hand-drawn circuit diagram,enter image description here

Also here is the back of the board.I want to remove the white and brown wire which are ground and VCC to ensure the safety of my laptop. Is it a good idea? And sorry for terrible soldering.It was my very first soldering experience. enter image description here

I know this is not a discussion forum but still, if anyone finds a mistake in the circuit and if this platform permits such discussion then I will be pleased to hear it as I can correct them and avoid such mistakes in future. If not then please ignore this part and let me know in comments I will remove this from question.

Thank you,

  • 1
    \$\begingroup\$ It's impossible to find anything with your circuit, right or wrong. You need to at the very least post schematics and how the prototype board is wired (physically, not how it's meant to be) \$\endgroup\$
    – MrGerber
    Nov 2, 2017 at 5:14
  • \$\begingroup\$ If you are suggesting not connecting VCC on the USBASP connector to protect your notebook, that will not work with most programmers. The programmer uses the VCC from the connector to know what voltage the target processor is running at so it can use the correct voltage on the IO pins. If you disconnect VCC the programmer will just think that the target it not powered up and it will do nothing. \$\endgroup\$ Nov 2, 2017 at 5:49
  • \$\begingroup\$ @MrGerber can you please review the question.I added the schematic and back of the circuit board.Also is it a good idea to disconnect the VCC? \$\endgroup\$ Nov 2, 2017 at 5:50
  • \$\begingroup\$ @DeanFranks then how can I ensure the safety of my laptop?Is it ok to first upload the code and then disconnect it from pc and the power using an external source to run the code?Also, I don't want my circuit to take too much current while programming it which might result frying my motherboard. \$\endgroup\$ Nov 2, 2017 at 5:51
  • \$\begingroup\$ @laptop2d provided some suggestions below, I would add: check the voltages with a multimeter before connecting your computer, use a USB hub rather than plugging the programmer directly into your computer [I use this one often], use a real Atmel programmer or a direct clone of one so the programmer has the necessary protection circuitry, use a USB charger adapter (old 500mA variety) for power, check the adapter with a multimeter first, invest in a used lab quality power supply with current limiting [highly recommended in the long run, can be >$USD100] \$\endgroup\$ Nov 2, 2017 at 10:44

1 Answer 1


No, it still will not protect against overvoltage on the serial ports. There are a few ways to do this here are two suggestions:

You still need to run the numbers on the current and size the resistor correctly. If the laptop port needs 1mA @ 5V to recognize a high or '1' condition, then you'd need 5V/0.001=5k max resistance, but you would also limit the current to the port and prevent a burn out. Diodes can also be used to clamp the voltage on the board to prevent the voltage getting higher or lower than VCC and ground.

You could also use a zener diode to keep the voltage below a certian value, in the other example, the zener will turn on if the voltage goes beyond 5V and shunt excessive current.

There are other ways these are just a few suggestions. Make sure your ground is connected to the laptops ground or you might have issues. If the VCC of your circuit is not of the potential of the port on the laptop (you haven't specified where you'll connect it with the laptop, then don't connect it. )


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ thanks for the reply, I currently have few 1N4007 diodes .will they work in place of 1N4148? \$\endgroup\$ Nov 2, 2017 at 7:36
  • \$\begingroup\$ Yeah, they have a forward voltage drop of 1.1V so they won't 'turn on' until 6.1V (if your rail is at 5V) whereas the 1N4148 will 'turn on' at 5.7V. So the 1N4007 offers less protection but its better than nothing. \$\endgroup\$
    – Voltage Spike
    Nov 2, 2017 at 15:16
  • \$\begingroup\$ ok, now I understood the logic how the diode was selected.Thank you for your help :) \$\endgroup\$ Nov 3, 2017 at 20:03

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