Hi, Mark
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.

Man that would be awesome!!! I mean, I understand it for the most part.. but that sensitivity control I am still a little unsure exactly how that is wired.

Should I try this circuit out? I have a spare breadboard ready to go. I was gonna head out to Radio Shack in a bit, I really hope they have all of these parts!

I am definitely getting a MUCH better understanding for how current travels and how wiring works just from your posts and a few other threads like the one you mentioned. Hopefully I can figure out the solution on my own if this schematic doesn't work for me!

-Mark

 

I definitely understand that circuits are the result of trial and error.

I have designed circuits for 48 years using parts spec's from their datasheet and simple arithmatic. Every circuit has worked perfectly for years. No trial and error.

Would that work the same for an infrared LED as it would for the green LED?

LEDs with different colours have different forward voltages that affect the calculation of the value for the current-limiting resistor.

I see by the resistors it says RCQ, or RBQ. What do those letters mean?

the designer should know.

Also, the sensitivity control, Are those resistors just soldered together? so a 500k resistor into a 1meg?

A 500k potentiometer and a 1M potentiometer. I think the 500k should be a 100k or a 25k pot to be used for fine tuning.

 

The potentiometers were drawn in the schematic, I used my resistor substitution box, because my resistors only go as high as 910K ohms.

There is no fine tuning with the second pot.

The values I was getting was 1.3 megaohms for a 4v. supply.
and around 720K ohms when I used a 6v. supply.

Mark,
I am ONLY a HOBBYIST in this field,
there are experts in this field, on this forum who can give you the exact circuit you need,
that's why with ME it is trial and error, because I too am just learnming more about this as I participate in this FORUM.

This forum has taken me from just putting resistors in and out until something works, to doing actual calculations, and purposefully calculating voltages and currents and placing a value component in when it is at my knowledge to do so.

When I said trial and error, I was referring to when I gave you those quick schematics without any calculations at all.

That's why when I breadboarded it I found that it took heavy signal voltages to get it to work properly.
that's when I did it over again with calculations.

Here it is in steps that I wrote up for you.

Chose 10mA. to flow thru LED.
Chose 4v. supply for all calculations.
2N3904 (Data sheet shows Beta min. 100 @ 10mA.)

Step 1:
LED driver stage
With 2v. across the LED leaves remaining 2v. across the rest of the circuit.
Of that remaining 2v. I chose to put 1v. across the RCQ1.
RCQ1 = (1v. / 10mA) = 100 ohms.

I chose to use a emitter resistor, (REQ1) and put 0.5v. across it.

So REQ1 = (0.5v. / 10mA) = 50 ohms. (standard value is 51 ohms)

Now Vbe of 2N3904 is around 0.7v. So I used 0.7v. for calculation purposes.
VEQ1 was chosen in the last step to be 0.5v.

So the base voltage of Q1= (VEQ1 + Vbe)= (0.5v + 0.7v)= 1.2v = VBQ1
Now with a Beta of 100 for Q1 @ 10mA. makes the base current for Q1 be (IC / beta min)
which is (10mA. / 100) = 100uA. = IBQ1

Now base resitor for Q1 to give this 100uA. base current is {(VCC - VBQ1) / IBQ1}
which is {(4v. - 1.2v.) / 100uA} = 28000, standard value use 27K ohms.

so for the first stage:

RCQ1 = 100 ohms
REQ1= 51 ohms
RBQ1= 27K ohms.
ICQ1= 10mA.
IBQ1= 100uA.

VCQ1= 1v.
VEQ1= 0.5v
VBQ1= 1.2v.

Now breadboard it and tested the voltages:
and the performance of the stage.
LED is on. with 2v @ ~ 12mA.
---------------------------------------------------
Step 2.
Inverting amplifier.

Now to turn the LED off with this next stage.
Chose to bring the VBQ1 down from 1.2v to 0.5v. so Q1 will be turned off and the LED is off.

If 0.5v is at the base of Q1 that means than that 3.5v remains across RBQ1.
therefore the new current flowing thru RBQ1 will be (3.5v / 27K ohms) = ~ 130uA.
Q2 will now supply 130uA to the base of Q1 to cause the VBQ1 to drop to the new value of 0.5v.

Now I chose to make the emitter voltage of Q2 be 0.2v.= VEQ2.

So now the emitter resistor of Q2 which is called REQ2= (0.2v / 130uA) = 1.5K ohms.

The base voltage of Q2 = (VEQ2 + vbe)= (0.2v + 0.7v) = 0.9v = VBQ2

In order for Q2 to supply this current the base current for Q2 needs to be calculated.
Fron the DATA sheet, it shows for 100uA of collector current, beta min. = 40.
So the base current for Q2 called IBQ2 will be (ICQ2 / beta min) = (130uA / 40)= 3.2uA.

so now RBQ2={ (VCC - VBQ2) / IBQ2} = 1million ohms = 1MEG.

Then built and test both stages for performance and the LED shut off.
Worked properly.

Step 3:
This is the signal amp stage.

This was designed with values that when there is no signal at the base of Q3 then it's collector voltage remained at around 1v. thereby keeping the Q2 stage on.

The base voltage of Q3 was chosen thru trial and error to be just at the point of conduction, so that a small millivolt signal of 100mV. would turn it on.

This required some elaborate calculations and test measurements beyond the scope of whats been explained with the other 2 stages.

But If you study this carefully you can get an understanding of how the first 2 stages were designed (resistor values chosen), and then you can use this information to design a curcuit for your application.
--------------------------------------------------------------------------------------------------------------------------------
NOW here is my actual recordings that I did during this build.


Data:
Green LED (ON) = 2V. @ 10mA.
Green LED (OFF) = 1.6v. @ 3uA.
2N3904 = B=~ 90 @ 10mA IC. (Data sheet), NOTE Data sheet shows 100 @ 10mA.
I chose 90 because I know the IC will be a little different, than 10mA.

Design:

STEP 1:
VCC =4v.
LED has 2v. @ 10mA.
choose 1v. across RCQ1 and 1v. for VCQ1,
so 1v. / 10mA= 100 ohms for RCQ1.
choose 0.5v. across REQ1 so REQ1= 51 ohms.
VEQ1= 0.5v.
VBQ1 = 1.2v.
B ~=90
IBQ1 = (10mA / 90) = 111uA.
RBQ1 = {(4v - 1.2v) / 111uA} = 25.2K make it 27K ohms.
-------------------------------------------------------------------------------
TEST 1:
LED has 2v. @ ~ 12mA.
VBQ1 = 1.3v.
VCQ1 ~ 0.8v.
----------------------------
STEP 2:
Calculate temporary base to ground bleeder resistor needed to bring VBQ1 to 0.5v. thereby turning LED off.
VBQ1 to be 0.5v.
so Vdrop RBQ1 = (4v. - 0.5v) = 3.5v.
current thru RBQ1 under this condition = (3.5v / 27K) = 129uA.
Therefore Resistance at base of Q1 to ground = (0.5v. / 129uA) = 3.9K ohms.
Now this temporary resistor will be replaced by a transistor to supply this current.
With Q2 saturated the VCE will be ~0.3v. so VEQ2 =0.2v.
REQ2 = (0.2v. / 129uA) = 1.5K
Data sheet shows B ~40 @ 100uA.
So (129uA / 40) = 3uA for IBQ2
VBQ2 = {(0.2v + 0.7v) = 0.9v.
and RBQ2 = {(4v - 0.9v) / 3uA} = ~ 1MEG.
-----------------------------------------------------------------
STEP3:
Now I want to bring the base of Q2 to 0.5v. so the LED comes on again.
Vdrop across RBQ2 will NOW be (4v. - 0.5v) = 3.5v.
and current thru RBQ2 = ( 3.5v / 1Meg) = 3.5uA
So Q3 will supply 3.5uA collector current to drop the VBQ2 to 0.5v.
Again assuming saturastion of Q3 to be 0.3v. makes VEQ3 = 0.2v.
and (0.2v / 3uA) = 68K

TEST 2:
Input sinewave = 50mV. peek
input squarewave = 25mV. peek
LED lights up with 1.3v. across it.


Thanks.

 

Hey,

Haha, I suppose "trial and error" was the wrong thing to say. Maybe for newbies like me that is true, but I know circuits are made using math for precise voltage calculations. Unfortunately, I never was good at math

But, I consider myself a pretty quick learner. I would like to be able to calculate voltages and build my own circuits. (Sucks being colorblind and not being able to determine resistor values though).

Thanks for writing that for me! I am getting an idea of how to make calculations for voltages and current.

So this worked properly to turn off your green LED, but Audioguru says that an infrared LED would have a different forward voltage, effecting the value of the current-limiting resistor. Is that ALL of the resistors? Or just the one coming off the LED? I'm assuming all of them since one effected value would throw all of the calculations off?

Question about this:

"The potentiometers were drawn in the schematic, I used my resistor substitution box, because my resistors only go as high as 910K ohms.

There is no fine tuning with the second pot."

So the resistor substitution box is for determining the right values for the circuit. Am I going to need a potentiometer in my circuit (where that sensitivity control is)? Or, are you talking about the potentiometers on the resistor sub box itself?

The remote application on the output device (cell phone) will be outputting square waves to the audio jack around 36kHz. Do I need to account for this as well?

On the IR LED package it says:

Forward Voltage: 1.2v
Forward Current: 100mA

So the forward current on your LED was 10mA? Would I be substituting 100mA into the math then?

If you could steer me in the right direction to get this to work with an IR LED, that would be amazing. In the meantime I am going to see if I can get this circuit built, the only thing that is cloudy to me is what exactly the sensitivity control is. Is it a potentiometer? Is it a 1meg resistor and a 500k resistor soldered together?

Thanks so much, everyone, for all of your responses. I definitely have gotten very interested in electrical engineering just from this thread! And sorry if this post seemed all over the place. I was replying, then reading again, replying, and thinking of new things as I went along haha.

-Mark

 

Hi mark,

The sub box, is just a box with switches to switch individual resistors in and out.

The schematic pots, were drawn as 2 potentiometers, because off hand I didn't know what values above 1 meg. are available.

First the sensitivity controle question.

I had to use around 1.3 megaohms for a battery supply of 4v.
Then I had to switch it down to around 720K ohms when I raised the voltage to 6v.

A potentiometer in the range of 2 to 5 megaohms if they have them would probably suffice,

However pottentiometers (pots), are either linear taper or logarithmic, which probably for this application wouldn't matter which kind you use.

If you have a visible LED, please build the circuit using that first, so you have a means of detecting if the circuit is working at all.

With your value of 1.2v. @ 100ma. would require some new resistor values for sure.

You could try the same design calculations for this 100mA diode, and see if it will work,

Here is a starting point:

It is important to get a data sheet for the transistor your using

You can do a google search for 2N3904 transistor data sheet,
and look under :ON Characteristics, on the sheet, and it will show the beta of the transistor for different collector currents.

2N3904
DC current gain (beta) min.

IC=0.1mA @ beta 40
" 1mA @ " 70
" 10mA @ " 100
" 50mA @ " 60
" 100mA @ " 30

Absolute rating IC = 200mA ,continuous current.
-------------------------------------------------
LED= 1.2v. @ 100mA.
VCC = 4v. min.


Try to do calculations with this new value, 100mA. and I will see if I can use a dummy 12 ohm resistor to drop 1.2v. @ 100mA. since I don't have a IR LED with those specs.

and I will calculate new values and breadboard it and see if I can get it working, using my multimeter to check for performance voltages at key points.

I don't know how long it will take but whatever the results, I'll post back and let you know.

If you can get a package of different values resistors, from the shack, it would be better than to buy individual values, if you do, then may I suggest building the circuit with a visible LED, doesn't have to be green, any color should work for this particular circuit, and just see if you can get a circuit working ahead of time before moving to the higher current non visible LED's.


One more thing:
There are numerous ways of designing for this application, as is with almost every circuit, I am going to stick with this particular circuit configuration,, and recalculate, new values if it doesn't work, then a whole new circuit design may be required.

thanks..

 

I used a visible LED but got peek voltages at around 1.2v.
This may work for your aplication.
The variable resistor could be a 50K potentiometer. For the sensitivity.



VCC=4v.
IC=100mA
LED=1.2v @ 100mA. (12 ohm resistor dummy load)
Freq=36 khz.

Design:
Stage Q1:
VCE=1v.
VCQ1=2v.
VEQ1=1v.
Vdrop,RCQ1=0.8v.
Bmin=30 (data sheet)
ICQ1=100mA.
IBQ1=3.33mA.
RCQ1=7.5 ohms
REQ1=10 ohms.
RBQ1=680 ohms
---------------------------------
Test 1:
VCQ1=1.6v.
VEQ1=1.25
VBQ1=2.1v.
-------------------------------
Stage Q2:
VBQ1=VCQ2=0.5v.
ICQ2={(4v. - 0.5v.) / 680}= 5.14mA.
Bmin=85
IBQ2=60.5uA
VEQ2=0.2v.
REQ2=39 ohms.
VBQ2=0.9v
RBQ2=51K ohms
---------------------------------
Test 2:
VCQ2=VBQ1=0.36v.
VEQ2=0.22v.
VBQ2=0.93v.
VCQ1=4v.
Vdrop load (12 ohm dummy resistor) = 0v.
------------------------------------------------------------
Stage Q3:
VCQ3=VBQ2=0.5v.
ICQ3=69uA.
Bmin=40
IBQ3=1.7uA
VEQ3=0.2v.
REQ3=2.7K
VBQ3=0.9v
RB1Q3=180K
Now the output is high (LED on)
To turn it off:
VBQ3=0.5v.
Vdrop RB1Q3= (4v. - 0.5v.) = 3.5v.
current thru RB1Q3= (3.5v. / 180K) = 19.4uA.
bleeder resistor to make this drop in voltage = (0.5v. / 19.4uA}= 24K
RB2Q3= variable resistance of around 24K ohms.





See if this works.




If it still has not enough output power, than 2 things can be done.
couple the output of this circuit thru a opto isolator into a higher current transistor.

OR

Drive the IR LED from the battery thru an appropriate resistor, and use a inverter stage to turn it off, as of right now the transistor is driving the LED, but if the transistor was used to turn it off, then when this transistor is turned off, then full power will be delivered to the LED thru its resistor and battery.

that would require another design modification.