DIY IR Droid Module - 3.5mm to Infrared - Breadboard Prototype
Hey guys, I figured since I had grabbed so much great information from you all I might try my luck with a little more! ;-)
I have built this on my breadboard, powered with 6V 150mAh AC-DC converter. First connect saw at least one of the LEDs flicker dark orange and then go out. Assuming this means I burned it out, as afterwards I didn't notice any response.
To prevent burning IR LEDs, I'll use visible LEDs, but I'm still concerned that my breadboard has a mistake. I'm pretty much worthless (but willing to learn, at least outside of the classroom setting!) at Phys II, so the debugging is foreign to me. Pulling out the multimeter... What process should I be using to confirm that everything is powered correctly? The schematic is attached if you would like to customize your response to the hardware, but a general guide to debugging amplified LED circuits would be much appreciated as well!
Additionally, if amplification the best way to go? How about a relay? Does that not offer the same (not Phys II...) amplitude? Smaller the better!
Source List:
- Schematic - Attached PDF
- PCB Creation - https://www.instructables.com/id/Extreme-Surface-Mount-Soldering/
- Mounting plan - https://www.instructables.com/id/LEGO-Nikon-IR-Remote/
Thanks in advance!
Andrew
Oh, and I'll post a picture of it within the week to get another set of eyes. I must confess, it is a little ratty...
Discussions
8 years ago
Is it for Left and Right audio channels ?
Answer 8 years ago
I'm thinking it is using both, but outputting mono...
Answer 8 years ago
To use both you need to create a middle virtual audio ground.
Just one Op-Amp and two 47K 1/8 W resistors in a follower configuration..
A
Answer 3 years ago
Is anybody able to explain in detail what each resistor/capacitor in the schematic does?
Answer 3 years ago
Glad to. After you tell me your skill level.. Do you know how an Op-Amp works ? and Ohms law for resistors ?
Answer 3 years ago
Well, I know what is a voltage divider, what are inverting and non-inverting inputs of op-amps, I know that capacitors make only AC pass through but they stop DC... but I don't know hot to put all together in your circuit.
It would also be useful to figure out how to modify this minimal audio amplifier tpo fit out needs:
<img src="https://i0.wp.com/sci-toys.com/scitoys/scitoys/computers/solderless/small_1_watt_audio_amplifier.jpg">
Answer 3 years ago
Who ever put the L and R inputs makes this appear as an stereo input amp which it is Not !
There is no bias for AC input or provision for IR or any diode LED switch-over gap.
The R2/R1 input [11:1] ratio reduces a DC signal to the Op-Amp which without feedback will rail to full positive or ground if the input signal goes ever so slightly negative.
C4, R5 are audio frequency adjustments which do nothing for a rail to rail switcher.. C1, R3 are shunting high frequency out of the output but here limiting rise and fall.
C5 & C3 are meant to pass AC for the bidirectional LEDs but that cannot ever happen because there is No negative power supply and D2 will never light.
R4 limits the current pulse through C5 to D1
You should ask your own question so I can get a chance to be adorned with a Best-Answer.
This is a Biber toxic schematic.
Answer 3 years ago
The circuit is supposed to amplify a rectified signal received on L/R (see picture)
When you play a 19 kHz electric signal, you get a 38 kHz light-signal (tested: it works!)
Answer 3 years ago
image
8 years ago
Somebody forgot to mention that the Schematic was taken from http://www.irdroid.com which is a remote control module for Android wright?
Answer 8 years ago
Yes. Open-Hardware. Not trying to claim as my own. My apologies, just trying to get her running!
Man, electricity is awesome, but its not easy! Pretty sure I'm gonna have to put yours on my X-Max list! -- Oh, I should mention they are for sale now... after I go and learned how to make it myself... breadboard and all..
8 years ago
No current limiting resistor to LEDs = bad idea.
What, exactly, are you trying to do ?
Steve
Answer 8 years ago
LOL... Umm. Yea... Maybe with batteries it must play nice. I want power source flexibility! Let's add one. (Maybe 2? Most suggest using one for each in parallel.)
Resistance and power Resistor Values:
Resistance = (Vtotal - Vled forware) / Current
Power = Voltage * Current
I'll probably want to operate the LEDs underpowered, since 6V, 100mAh is giving me over 1/2 watt. on a SMD device, I want to keep it well below that to use 0805 parts.
RadioShack doesn't give real details, but 100mAh, Vled forward = 1.28V
So... since these are in inverse parallel I'm really confused about how this effects the voltage. (It is my understanding that this allows the pulses to add to each other, or something about additive frequencies, or amplitudes or something.
Making progress, at least I think so. Just running a few more sets of numbers... Thanks for the start!