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Salazar

Member Since: February 19, 2011

Country: United States

Profile

Organizations

Carnegie Mellon University; BAE Systems

Programming Languages

C, C++, Java, Verilog, x86, x86-64, Ada, Fortran,

Universities

Carnegie Mellon University

Expertise

Embedded Systems Design, Biomedical Engineering

  • NOTE: You CAN use 3.3V on the base input, but you need 5V at the JP2-pin3 to supply the relay. For this: you do NOT need to change the resistor resistor network feeding the base of the transistor. If you look at the schematic, this uses a 2N3904 bjt which has a turn-on voltage of 0.7v, and Beta=100. That means that the base-current, as it stands, is ~2.53mA (with 1k/10k combo) when 3.3V is applied to the input (3.3-0.7=2.6V across R2). This means that the base is allowing up to 253mA to flow through the collector (Ic=Beta*Ib). BUT, since the coil in this relay has R=27Ohm…even if you had a full 5V drop across your coil, your max current will be 185.2mA (e.g. the relay turn-on current, per the datasheet). So basically, 3.3V on the input is STILL enough for the base…it’s the Vcc(“RAW” on schematic) that matters. ———————-> 1.) Hook up a separate 5V supply (maybe strip a 5V usb wall plug/knob or use arduino 5V source) to JP2-pin3 ———————-> 2.) Hook your 3.3V arduino digital output to JP2-pin2. Hookup the grounds, and you should be good. The flyback diode will protect your 5V supply. See for yourself: SIMULATOR

  • NOTE: You do NOT need to change the resistor resistor network feeding the base of the transistor. If you look at the schematic, this uses a 2N3904 bjt which has a turn-on voltage of 0.7v, and Beta=100. That means that the base-current, as it stands, is ~2.53mA (with 1k/10k combo) when 3.3V is applied to the input (3.3-0.7=2.6V across R2). This means that the base is allowing up to 253mA to flow through the collector (Ic=Beta*Ib). BUT, since the coil in this relay has R=27Ohm…even if you had a full 5V drop across your coil, your max current will be 185.2mA (e.g. the relay turn-on current, per the datasheet). So basically, 3.3V on the input is STILL enough for the base…it’s the Vcc(“RAW” on schematic) that matters. ———————-> 1.) Hook up a separate 5V supply (maybe strip a 5V usb wall plug/knob or use arduino 5V source) to JP2-pin3 ———————-> 2.) Hook your 3.3V arduino digital output to JP2-pin2. Hookup the grounds, and you should be good. The flyback diode will protect your 5V supply. See for yourself: SIMULATOR

  • NOTE: You do NOT need to change the resistor resistor network feeding the base of the transistor. If you look at the schematic, this uses a 2N3904 bjt which has a turn-on voltage of 0.7v, and Beta=100. That means that the base-current, as it stands, is ~2.53mA (with 1k/10k combo) when 3.3V is applied to the input (3.3-0.7=2.6V across R2). This means that the base is allowing up to 253mA to flow through the collector (Ic=Beta*Ib). BUT, since the coil in this relay has R=27Ohm…even if you had a full 5V drop across your coil, your max current will be 185.2mA (e.g. the relay turn-on current, per the datasheet). So basically, 3.3V on the input is STILL enough for the base…it’s the Vcc(“RAW” on schematic) that matters. ———————-> 1.) Hook up a separate 5V supply (maybe strip a 5V usb wall plug/knob or use arduino 5V source) to JP2-pin3 ———————-> 2.) Hook your 3.3V arduino digital output to JP2-pin2. Hookup the grounds, and you should be good. The flyback diode will protect your 5V supply. See for yourself: SIMULATOR

  • …and extra blades for lots of cutting

    https://www.sparkfun.com/products/10325

  • HERE IS HOW TO CUT THE TRACES PROPERLY: simple hobby knife (link below). Check the connections w/voltmeter on Ohm-setting. Only thing that worked easily for me. You don’t need to go rambo, just a few runs of the blade, and an ohm-meter check.

    https://www.sparkfun.com/products/9200

No public wish lists :(