0
\$\begingroup\$

I implemented a breadboard prototype based on the circuit below (image created by a LED circuit calculator here).

enter image description here

The LEDs are SFH4556P. Voltage for the breadboard is provided by a wall wart adapter (phone charger) and supplies 5V (1A max) via a USB cable salvaged from a PC mouse (has 5 wires, two for data, shield ground, plus and minus - of which only the latter two are used). I also added a 500mA fuse to the circuit, the resistor is rated for 2 Watts.

I hooked everything up and started my measurements and immediately hit an issue. The current draw of the circuit was below 30mA meaning my LEDs were way too dimm.

I replaced the resistor with a 1.5 Ohm one (same wattage rating) and current draw rose up to around 40mA, which is still too low.

I then I removed the resistor and voila - around 50mA.

I also found the culprit - the salvaged USB cable. It seems that due to thin wires used in it, each of the two used cables cause a resistance of around 5 Ohm so it causes the circuit to already contain 10 Ohms of resistance.

How bad of an idea would it be to just leave the circuit as it is (without a resistor)? Can I calculate if the wires in the cable are capable of sustaining the circuit without any resistor in it?

The circuit is part of a DIY TrackIR project and that is also its intended usage.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ That USB cable sounds suspicious; I would not trust it. Suggest trying a new cable. Might find that two parallel sets of two LEDs in series, or three sets of parallel single LED's might work better with such a low supply voltage (5v.) \$\endgroup\$
    – rdtsc
    Dec 17, 2019 at 20:20
  • 2
    \$\begingroup\$ Relying on the cable itself as a resistor is not good practice. Just think about how the circuit were to change if you were to use a different cable that uses a different gauge wire. I agree with the statement that you are probably using too many LEDs in series. You need to have sufficient voltage overhead to reliably bias your LEDs. Try going with 2 LEDs in series max, and play with the numbers to see how supply voltage variation or cable resistance affects the current. \$\endgroup\$
    – BEE
    Dec 17, 2019 at 20:25
  • \$\begingroup\$ I ended up using another cable and changing the circuit so the three LEDs are in parallel. \$\endgroup\$
    – predi
    Dec 23, 2019 at 9:56

2 Answers 2

2
\$\begingroup\$

I suspect most of your resistance more in the 4 contacts than the wires. It's OK if they are predictable and stable over time. Copper wire is often used as a 1/8W 50mV current shunt.

If you are OK with an unregulated constant dim IR current, this will work, but if you want a little better performance, you will want to regulate current and aim ahead a bit to get a competitive edge and look ahead a bit to turn the motors to follow better. Optical path control is important.

SFH4556P Specs: 130 deg (discontinued July '19)

1.7V nom @ 100mA pulsed 2.0 V max
2.5V nom @ 450mA pulsed
3.6V nom @ 1000mA pulsed 4.6V max

Other

  • I hope you choose to regulate it with a low side 50mV shunt dual comparator and low Ron logic level NFET and pulse it.

  • Height variation must be eliminated for reflection path loss variation.

  • Beamwidth should match your reflection target for good edge detection.

I would use say 1~5% duty cycle pulsed at higher current using an active current limit at say >10kHz with a BPF and detect the wave precision envelope using Photo diodes with 100kHz TIA's which will be far matched than PT's and as accurate as your matched resistors. You may want to do auto calibration as well.

Above 100mA the bulk resistance dominates this diode curve.
Vf= 2.1 [Ω] * If + 1.51 V nominal pulsed voltage
Vf= 2.9 [Ω] * If + 1.7 V worst case ( process tolerances)

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ A TrackIR is a computer input device that consists of three IR LEDs mounted on a fixed frame, a camera with an IR band pass filter and a piece of software that tracks LED positions relative to the camera and translates this information into desired input. It doesn't take much to make the whole thing work, so unregulated current is okay. Anything else would probably be overkill. \$\endgroup\$
    – predi
    Dec 18, 2019 at 7:28
  • \$\begingroup\$ If you want to compete and prevent the robot from going off track at faster speeds , you must do some things better than everyone else who just use baseband detection which is proven lower SNR than AM carrier modulation. since position is the 2nd integration of acceleration and motor current , you need to understand Control Theory to see why my idea is better for faster servos. try driving a car looking down only. \$\endgroup\$ Dec 18, 2019 at 15:16
2
\$\begingroup\$

Copper and LEDs both have significant variation over temperature, as well as some over other conditions. If you just want the LEDs to light up and don't care exactly how bright or how long they last, this might be fine for you. If you want more reliability, you're better off with something like a TPS61261.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Copper resistance shifts by 4% for every 10 degrees C. That is more than most resistors, but it is not THAT bad. The thing is, what kind of crap USB cable has 10 Ohms of resistance? The whole thing is all wrong. OP should get a normal USB cable and fix the circuit. \$\endgroup\$
    – user57037
    Dec 17, 2019 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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