Hey guys,

Still in the same boat.. any other ideas?

Thanks!

 

What are the markings on the transistor? 2N3904? 2N3906? BC547? BC557?

 

Guru: It is an NPN, 2N3904 transistor.

 

Hey man,

Well I wired it up this way, and am now using a regular LED per your suggestion. The LED lights up and stays on when connected to the 4.5v batteries and is off when the batteries are removed from the circuit. So, I now have:

LED + --> 22 Ω --> +
LED - --> C of Transistor --> -
B of transistor --> 1k Ω --> +
E of transistor --> -

I need the IR LED to come on only when given a signal from TX. Am I missing something?

The reason the leds stay on is because your 1K is hooked to positive side battery.

Remove the 1K from the battery.

Then you input a signal to the base to turn it on.

Build your circuit just like the schematic in the first post.

 

It is difficult to see on your photo but I think you have the LED and its current-limiting resistor connected directly to the battery. The transistor has nothing to do with the LED and instead it simply shorts the battery and gets hot.

 

Thanks so much for the responses so far guys.

The reason the leds stay on is because your 1K is hooked to positive side battery.

Remove the 1K from the battery.

Then you input a signal to the base to turn it on.

Build your circuit just like the schematic in the first post.

Please pardon my ignorance. I am somewhat new at wiring, and I know this might be a noob question but I don't understand what the "TX" means. I wire things to + or -. I don't understand exactly what you mean when you say "input a signal to the base." I know you mean the base of the transistor, but I don't know what that means in terms of what it gets wired to.

Again, excuse my noob-ness. I would love to know how this "TX" port comes into play, and what that means in terms of wires, a transistor, and a breadboard.

Thanks again so much!!

-Mark

 

Hi,
Noob questions are always welcomed,
if a person has to apologize for asking newbie questions, then what's the use of this forum anyhow.

This forum is for just that, to help others in answering the best they can ,any question sincerely asked by another...

I don't know where you got your schematic from so I don't know what the TX stands for, but I do know what the schematic is showing, by the configuration of the circuit.

The circuit is designed as a switch, to turn the LED on when a positive voltage is applied to the resistor marked TX.

As TX is left open, as shown in the original schem. then no current flows thru the transistor, thereby the LED is off.

When you apply a positive signal voltage to the TX resistor the transistor will conduct current to flow thru the LED.

So TX could be driven by any circuit that has a positive voltage output,

Again TX is just the name this designer gave for that term.
as it refers to his overall circuit.

 

Hi,
Noob questions are always welcomed,
if a person has to apologize for asking newbie questions, then what's the use of this forum anyhow.

This forum is for just that, to help others in answering the best they can ,any question sincerely asked by another...

I don't know where you got your schematic from so I don't know what the TX stands for, but I do know what the schematic is showing, by the configuration of the circuit.

The circuit is designed as a switch, to turn the LED on when a positive voltage is applied to the resistor marked TX.

As TX is left open, as shown in the original schem. then no current flows thru the transistor, thereby the LED is off.

When you apply a positive signal voltage to the TX resistor the transistor will conduct current to flow thru the LED.

So TX could be driven by any circuit that has a positive voltage output,

Again TX is just the name this designer gave for that term.
as it refers to his overall circuit.

Hey,

That's why I posted this on here, I figured I would find a bunch of knowledgeable, helpful people who shared some of the same interests I do. I've learned a lot already from this forum and I've only created 1 thread!

Anyways, I drew up the schematic, per what my teacher told me. When I asked him what the "TX" meant he told me it was the "transmit" port...

I am just unsure as to how exactly that is wired.. if it is turned on when there is a signal then shouldn't it get wired to +? But then, it will always be connected to the batteries which are also wired to the + and - ...

Ughh I am so stuck on this! I appreciate all the help so far guys!

 

Hi Mark,

Please, give us as much background as possible concerning what your teacher wants you to do with this project.

What is the function of it, what device is going to be hooked up to it, so we have an idea of how to implement circuit values.

 

Hi Mark,

Please, give us as much background as possible concerning what your teacher wants you to do with this project.

What is the function of it, what device is going to be hooked up to it, so we have an idea of how to implement circuit values.

Hey,

This is the project:

I will be hooking up this infrared LED driver circuit to a cellular phone using the Android OS. It essentially turns the device into an infrared remote. I have been programming the application so the buttons play square waves, that, when this module is hooked up to the device, will convert into infrared signals for each of the functions on the remote. (Power, 0-9, Volume Up/Dn, Channel Up/Dn).

As of now, I have the MONO headphone jack hooked up to the breadboard, + and -. I have everything somewhat figured out except for this TX issue. I have the LED powering up when the batteries are connected, but the LED should ONLY get power when the signal is given from the cell phone to the headphone jack.

I should also mention that I am online (AIM) 24/7 haha. So if anyone feels inclined, I would love to talk to ya and get some live help! My sn is Icehock418.

Thanks so much guys!

-Mark

 

Hi,

alright, I'm assuming you are using a seperate supply for your circuit, here is how you combine 2 circuits together, when they each have there own seperate voltage supplies, and still be able to couple signals from one to another.


First make sure that your positive battery on your circuit is NOT connected to the pos. battery of the cell phone.

DO make sure that the NEGATIVE terminals are connected from your circuit to the cell phone.

That is for a signal ground. You have to have a common battery terminal between both batteries, and that usually is the neg. term.

Now put a 100uf cap. on the 1K resistor, leaving the other end of that same resistor connected to the base term.

Connect the other end of the capacitor to the output term. of your cellphone.

like this.



Depending on the frequency, output impedance, voltage output, as well as other parameters not yet investigated, this may or may not work.

However this is the proper way to connect 2 circuits together so a signal can be transmitted between them.

If it doesn't work,(which would not surprise me), then further tests need to be done on the cellphones output signal, to determine voltages and impedances, ect...

If the later is the case, you will need a multimeter, and some resistors. To do some preliminmary work, with the cellphone output signal.

If that don't work then try this one.



If the second one works a little, then we can modify it so it draws the least amount of current possible.

this will draw less current.

 

hobbyist: you are the MAN!!! You even went ahead and built 3 schematics for me dude that's awesome! Thanks!

Here's where I got with them:

I bought the capacitors from the shack. I got the 100uf ones like you said. There is a + and - side on these.. I believe I put the - at the resistor at the transistor's Base, and the + at the headphone jack's +. Unfortunately, no dice.

I just got the Android phone that I will be using for the project.. I bought a used htc Hero. Is there a way for me to test and find out the output impedance / voltage of the device? Or do I need to find a spec sheet? Or perhaps maybe I can inquire within htc.

I will try the other 2 schematics tonight.. although they look a bit more involved since there are more transistors

Again, You are a saint for taking the time out of your day to help me with this. Nobody else can seem to help me like you have been. My friend who is an electrical engineer was even like, "oh yeaaa man you do need a capacitor, my bad.."

I'm like dude! wtf!?

Haha I will post back tonight when I try the other 2 schematics. And again... THANK YOU.

-Mark

 

I didn't realize you had the availability of a output jack from your phone,
This makes it a lot easier to work with so then,
Instead of connecting the neg. terminals of both batteries, as I shown in the schems. it is better to connect your circuit neg. battery term. to the output jack ground term. and then connect the capacitor, to the signal term. of that output jack.

If you have a multimeter, with a continuity tester, you can determine what the ground side is on your headphone jack, by touching a probe to one of the jack pins, (if you have acces to it, and to the ground terminal on the phone device.

That;s the usual way of connecting 2 circuits together with a common ground, when and OUTPUT JACK is available.

That way there is no direct soldering wires to the phone batteries and all direct conections are made at the jack only.

That's how I coupled my portable CD player to my led display (primitive volume meter) circuit.

I used the output speaker jack and connected my circuit neg. to the ground term. of the CD jack, and then coupled the signal pin of the CD jack to a capacitor to the input of my circuit.

Like this.




I just built that 3rd circuit on the earlier post.

I used my osciloscope to check the frequencies of a square wave.
I set my power supply to a min. value of 4v.DC
I set my frequency generator to a squatrewave from 100Hz. to 100Khz.
I used a visible LED (green), it's voltage drop showed 1.8v. to full lighting.

for every freq. from 100hz up, the input voltage of the quarewave, had to be at around 2.8v.peek. in order for the LED to turn on.

anything lower than 2.8v, coming from the generator the LED remained off.

So depending on the vout from your phone signal will determine if the circuit will work for this application.

Otherwise different values for resistors and even power supply will need to be implemented.

 

I didn't realize you had the availability of a output jack from your phone,
This makes it a lot easier to work with so then,
Instead of connecting the neg. terminals of both batteries, as I shown in the schems. it is better to connect your circuit neg. battery term. to the output jack ground term. and then connect the capacitor, to the signal term. of that output jack.

If you have a multimeter, with a continuity tester, you can determine what the ground side is on your headphone jack, by touching a probe to one of the jack pins, (if you have acces to it, and to the ground terminal on the phone device.

That;s the usual way of connecting 2 circuits together with a common ground, when and OUTPUT JACK is available.

That way there is no direct soldering wires to the phone batteries and all direct conections are made at the jack only.

That's how I coupled my portable CD player to my led display (primitive volume meter) circuit.

I used the output speaker jack and connected my circuit neg. to the ground term. of the CD jack, and then coupled the signal pin of the CD jack to a capacitor to the input of my circuit.

Like this.

View attachment 17600


I just built that 3rd circuit on the earlier post.

I used my osciloscope to check the frequencies of a square wave.
I set my power supply to a min. value of 4v.DC
I set my frequency generator to a squatrewave from 100Hz. to 100Khz.
I used a visible LED (green), it's voltage drop showed 1.8v. to full lighting.

for every freq. from 100hz up, the input voltage of the quarewave, had to be at around 2.8v.peek. in order for the LED to turn on.

anything lower than 2.8v, coming from the generator the LED remained off.

So depending on the vout from your phone signal will determine if the circuit will work for this application.

Otherwise different values for resistors and even power supply will need to be implemented.

OK, so Battery GND -> output jack GND
CAP -> output jack signal.. I am assuming this is the + terminal? My capacitor seems to be polarized. So I obviously wouldn't be connecting both the + and - of the capacitor to output jack +, right? Just unsure how capacitors work exactly so I wasn't sure exactly where each terminal would go. Do I need a non-polarized one? Like this:

http://www.amazon.com/Parts-Express-100uF-Non-Polarized-Capacitor/dp/B0002KR4BQ

Thanks as always,

 

I ususaly hook the neg. side of a capacitor to the jack, and put the positive side to the resistor connected to the base of the input transistor.

The whole reason for using this capacitor, is too block any kind of DC and pass the AC signal.

For example, lets say that the phone output jack had a steady voltage of 3v. pos.

And when it pulses the voltage increases to 6v. then drops back to 3v.

Without the capacitor, then the DC of 3v. will be inputed to your circuit at all times, thereby activating it,.

By putting this coupling capacitor in between the output of the phone and the input of your switching circuit, then the DC of 3v. will be blocked from entering your circuit input.

So now the 3v. will sit on the output jack of the phone, BUT on the OTHER side of the capacitor, where your circuit resides, it will see zero volts, with respect to ground.

Therefor when the phone signal increases from 3v. to a 6v. pulse, as before, the capacitor will will charge up to this new value and at the same time pass that difference in voltage to the output across the capacitor into the input of your circuit.

So your circuit will see a 0 to 3v. peek pulse, when the phone signal goes from 3v to 6v. peek pulse.

The AAC book has a lot to show about capacitors on this forum website.

 

Hi,

Circuit design is a matter of trial and error.
The first schems I gave were quick configurations.
However after building and testing them I found that the input voltage had to be around 2-3v. for it to work properly.

So I redesigned it with specific input voltages in mind, around 50 to 100mV. pk. And did proper calculations for voltages and currents ect...

This was designed using a signal generator and osc. scope, where the generator has an output impedance of 700 ohms.

This circuit will work for modulating a visible LED (green) from the output jack of a portable CD player.

 

Hey bud,

I definitely understand that circuits are the result of trial and error. Hell, I've been trying, and unfortunately mostly failing, since January. I would have still been at square one if it wasn't for your extremely helpful posts in this thread.

OK, so on to a couple questions I have about that schematic. Would that work the same for an infrared LED as it would for the green LED? In other words, could this potentially be a working schematic for my circuit?

Also, quick question:

I see by the resistors it says RCQ, or RBQ. What do those letters mean? Also, the sensitivity control, Are those resistors just soldered together? so a 500k resistor into a 1meg?

So my parts list would be as follows:

4v battery supply (have)
IR LED (have)
3 2N3904 transistors (have)
1 .47uf capacitor
1 100uf capacitor (have)
910k ohm resistor
160k ohm resistor
68k ohm resistor
1.5k ohm resistor
100 ohm resistor
51 ohm resistor
500k ohm resistor
1Meg resistor ??? -- never heard of 1meg before

The connectivity of everything else seems pretty clear, and I will try this out tomorrow, pending your response. You know I come home from work and check the thread and say "mann I hope hobbyist posted again!" haha

Thanks a million,

-Mark

 

About to head to Radio Shack to get all of these parts. I guess I will test this circuit today.

Does it have a chance of working with a 940nm IR LED?

Also, I wanted to ask if it was cool with you to give you credit for all your help with this project. In the application's about / credits screen I have included your username for the circuit schematic.

Let me know if you're cool with that!

-Mark

 

Hi, Mark

The 1 MEG stands for 1 megaohms which is 1 million ohms.

RCQ means colector resistor for the transistor in question.
RBQ is the base resistor.

They are just the names I gave them, for this design.

This last circuit was designed as a small signal amp, to possibly pick up any small signal required from the phone output jack.

It will probably work an infrared LED, however as was posted in another thread by another person, about infra red detection, the infra red LED may require more power output, than what a visible LED would need.

When I breadboarded this circuit, I used my 2 resistor sub boxes, to switch resistors in and out of the circuit at the sensitivity controle. That's how I got the values of 1megaohms and 500K ohms.

By using variable resistors will allow the circuit to be "matched to whatever signal input voltage, and supply voltage being used within reasonable tolerances of course".

If the IR LED does not have enough output power, than a IR LED driver stage would need to be added at the output of this amp circuit, or a whole new circuit needs to be designed.

If you would like, I can give you a step by step process of how I designed that last circuit, so you can look over it, and get a good understanding of it and then have a good knowledge of how to modify it to work for your application.

 

About to head to Radio Shack to get all of these parts. I guess I will test this circuit today.

Does it have a chance of working with a 940nm IR LED?

Also, I wanted to ask if it was cool with you to give you credit for all your help with this project. In the application's about / credits screen I have included your username for the circuit schematic.

Let me know if you're cool with that!

-Mark

Hi,

That's very considerate, I appreciate that,

but it would be best to use the title of this forum as given credit to this build,

Because everybody on this forum as a whole should get the credit.

Thanks again.