This is the first time I'm drawing a schematic, thus why this question seeks to get clarification from you experienced engineers :-)

A little background information:

I'm trying to draw a circuit for an oximeter "breakout board" for my Arduino. The circuit contains respectively two 860nm LEDs and two 940nm LEDs. (I'm not sure which LEDs to buy yet, hence why the resistors are not given any values)

To measure irregularities in the emitted light from the diodes, I bought the FDS100 photodiode from Thorlabs - Which also is the component that confuse me upon drawing the schematic.

I'm planning to supply the board with 5V from my Arduino, and from the datasheet, it seems as if the FDS100 can handle much more - hence why I haven't added any resistors. The FDS100 has 3 legs, where one of them is a case(??) - I've read in another post, that it is good practice to ground the case, but not necessary. (I have done it anyway)

Furthermore, I understand that the FDS100 is "reverse biased", meaning that current flows in the opposite direction, hence why the analog port in my Arduino should be wired to the cathode of the FDS100, in order to be able to measure how much light that the FDS100 picks up.

Below you can see my preliminary Eagle schematic: Oximeter Circuit

Have I done this correctly, or am I missing some important details, i.e. resistors and other components concerning the FDS100?

  • \$\begingroup\$ You're confusing a schematic with a board layout. A schematic should be a clear logical (not physical) depiction of the circuit. Rearrange your components so that they're orderly and roughly vertical between Vcc and ground. \$\endgroup\$
    – Reinderien
    Nov 21, 2018 at 23:29
  • \$\begingroup\$ Thank you for your answer. I see! So basically i do the same, but rearrange between VCC and GND to enhance transparency/readability? \$\endgroup\$ Nov 21, 2018 at 23:37
  • \$\begingroup\$ To enhance legibility, yes. \$\endgroup\$
    – Reinderien
    Nov 21, 2018 at 23:38
  • \$\begingroup\$ @Reinderien I have just edited the schematic. \$\endgroup\$ Nov 21, 2018 at 23:59
  • \$\begingroup\$ Much better, but reorient your transistors so that the arrow faces down. \$\endgroup\$
    – Reinderien
    Nov 22, 2018 at 0:00

1 Answer 1


To draw a reasonably easy to understand schematic you typically have inputs on the left, outputs on the right. Wherever possible arrange voltage domains with low voltage on the bottom and high on the top (this can be hard to comply with at times).

I've re-drawn a schematic with a few changes you might consider:

  1. Don't drive LEDs with an Emitter follower since you needlessly drop the available voltage. In this case you only have 5V available, and the IR LEDs will be 1.4-2V Vf. I've shown a simple FET low side drive.
  2. If you don't have symbols for your device you can build your own, but you typically should not mix physical characteristics with schematic symbols. Here I've shown the reverse biased Photo diode with a case connection.


simulate this circuit – Schematic created using CircuitLab

Note: I added a sense load resistor (R5) which is shown in the FDS100 datasheet. Since the diode conducts a current relative to the light input, you need a path for this current.

  • \$\begingroup\$ You should also note the addition of R5 and explain it, since the OP clearly has no idea why it's needed. And yes, this will require rather more explanation than the rest of the circuit. \$\endgroup\$ Nov 22, 2018 at 2:43
  • \$\begingroup\$ @WhatRoughBeast This is clearly shown in the FDS100 datasheet, but I added it to the answer. \$\endgroup\$ Nov 22, 2018 at 6:12
  • \$\begingroup\$ Thank you very much for your answer @JackCreasey. So I think that I understand most of it. However, you say that I should use MOSFET transistors, instead of the NPN ones that I have used in my own schematic. - Does what you say mean, that the NPN transistors drop the voltage, and the MOSFET doesn't? - Furthermore, why is the capacitor C1 needed? Bear in mind, that I have no electrical engineering background whatsoever if you decide to answer :-) \$\endgroup\$ Nov 22, 2018 at 12:36
  • \$\begingroup\$ @JeppeChristensen - NPN transistors which are used as switches require that the input current (base current) be about 1/10 of the load current. MOSFETs don't have that requirement, although for higher currents it's a good idea to drive the gate with a bit more voltage. Since you have not specified the LED current, using FETs makes sense, since it's entirely possible that an NPN will require more base current than your Arduino can supply. \$\endgroup\$ Nov 22, 2018 at 14:08
  • \$\begingroup\$ Once again, thank you for your answer. :-) That makes good sense. - so for the MOSFETs, I don't use any current to open the gate, and therefore don't overload the MCU? - Why is C1 needed for my circuit? \$\endgroup\$ Nov 22, 2018 at 15:05

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