IR Beacon

vla7

I succesfully built an IR beacon using a 555 timer and following the instructions on "IR Remote for the Boe Bot" in appendix C.· I would like to increase the intensity of the beacon by adding more than one infared LED to it.· Can I accomplish this by simply connecting more IR LEDs in parallel to the circuit?· Would doing so change·the frequency setting so that I would have to recalibrate?· Please refer to my schematic.

Mike Green

If you just put LEDs in parallel, you won't get more light. The LEDs will share the amount of current allowed by the 220 Ohm resistor (about 20mA). You need to have a separate resistor for each LED and the total amount of current can't exceed the amount available from the 555 timer (about 200mA total). Figure on at most 8 LED/resistor sets in parallel. As long as your power source is stable at that kind of current drain, the 555's calibration should remain stable. It might change a little with heating from the current load, but the 555 is supposed to be tolerant of that.

The output from the LEDs will tend to drop the more LEDs you add because the output voltage from the 555 will drop with increased current drain. The LM555 datasheet goes into details on that. In any event, you'll have a sort of diminishing returns behavior. If you increase the supply voltage to compensate, the heat dissipation will increase and you'll have to adjust the resistor values as well.

I'd probably use no more than 4 LED/resistor sets in parallel and I'd run the whole thing off a 6V battery pack, maybe using 270 Ohm resistors for the LEDs

Phil Pilgrim (PhiPi)

Most IREDs have a low enough forward voltage that you can put two in series, assuming a 5V supply. You will need to reduce the size of your series resistor accordingly to attain the same current. That way you could attach double the number of IREDs to any one output as you could by just paralleling singlets. You still need a separate resistor for each series-connected doublet.

-Phil

vla7

Would this work?· Notice that I plan to use a 9 volt battery.· Mainly because of size and convenience.

kwinn

Yes. that would work, but you are wasting a lot of power across the resistors. As Phil said earlier it would be better to put 2 (or even 3 with 9V) leds in series and reduce the series resistor. You get 2 (or 3) times the light for the same led current.

vla7

OK thanks

GWJax

I just plugged your circuit into the computer and it does not work at least with my program EWB8. The IR-LED's stay on weather or not your signal from the Stamp is moving from High to Low state. I will be working on the schematics for you so you can have at lease 8 IR LED's to run off the 9V battery. I'll get back to ya soon.

Jax

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If a robot has a screw then it must be romoved and hacked into..

kwinn

vla7, what I said previously about the circuit working is correct, but I did not realize that you were using P6 and P7 to connect to a microcontroller pin. To do this with a 9V supply will require level shifting on those signal lines. A transistor with 2 resistors for P6 and 2 resistors for P7 should be enough.

vla7

kwinn,
I will be using the basic stamp with the circuit to be able to obtain a frequency that can be transmitted at 38khz, following the direction on"IR Remote for the Boe Bot" in appendix C. Once I get the value of resistance of the potentiometer that gets me this value I will then not use the stamp and use a fixed resistors instead of the potentiometer and will use a 9 volt supply instead of the 5v that is regulated by the homework board. I should not have written 9v on the schematic, it should really say 5v since the power is being supplied by the homework board. I want to use 9 volts after I build the circuit seperate from the stamp using the 555 timer, a capacitor and resistors. Basically an IR beacon. My goal is to maximize the distance the beacon can be detected by increasing the number of ir emmitters.

GWJax

You can not switch from 5V to 9V in this circuit and keep the same values for the caps and resistors once calibrated. Also when using the VR once you measure the resistance you'll find that getting a resistor at that OHM reading will be very hard to do. I do suggest that you keep the VR and tack it down with hot glue or silicone so it will stay in-place at the given value. I will finish up the circuit that you need tonight an post it for you using the 9V battery, once you have it set you'll need to lock down the signal line with another small circuit that I will provide for you as well..

Jax

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If a robot has a screw then it must be romoved and hacked into..

GWJax

vla7 said...
Would this work? Notice that I plan to use a 9 volt battery. Mainly because of size and convenience.

OK Try this one, It has 7 IR LED's. Use the 9 volt power supply that your going to use and tie the grounds together to your homework board. Attach TP1 to pin 7 of your stamp and then adjust the VR to get at around 38Mhz once this is set then disconnect the grounds and your ready. Let me know how this works for you please..

Jax

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If a robot has a screw then it must be romoved and hacked into..

kwinn

Again, you are putting the leds in parallel and wasting a lot of power which will mean a short battery life.

Lets make the following assumptions (they should be close):

1-With a 9V battery the output from the 555 is 8.2V.
2-The IR leds have a forward drop of 2.2 volts
3-The duty cycle is 40%

---

6 leds in parallel
resistor ohms

---

220
led voltage 2.2
res voltage 8.2-2.2=6.0
current (mA) 6.0/220 x 6 = 164
power .164x9x0.4= 0.59W

---

2 strings of 3 leds in series
resistor ohms

---

56
led voltage

---

2.2x3=6.6
res voltage

---

8.2-6.6=1.6
current (mA)

---

1.6/56 x 2 = 57.2
power

---

.0572x9x0.4= 0.206W


This gives slightly brighter output with three times the battery life. You will have to measure the utput voltage of the 555 with the 9 volt battery and the forward voltage of the leds and adjust the resistor value accordingly to make sure it will works properly.

I would also suggest using the ir sensor you plan to use with the stamp to adjust the frequency.

Post Edited (kwinn) : 4/3/2009 4:53:36 AM GMT

vla7

Thanks kwinn, I appreciate all your help. I will let you know how it turns out.

vla7

Thanks for the schematic Jax. I will give it a try.

GWJax

kwinn said...
Again, you are putting the leds in parallel and wasting a lot of power which will mean a short battery life.

Lets make the following assumptions (they should be close):

1-With a 9V battery the output from the 555 is 8.2V.
2-The IR leds have a forward drop of 2.2 volts
3-The duty cycle is 40%

---

6 leds in parallel
resistor ohms

---

220
led voltage 2.2
res voltage 8.2-2.2=6.0
current (mA) 6.0/220 x 6 = 164
power .164x9x0.4= 0.59W

---

2 strings of 3 leds in series
resistor ohms

---

56
led voltage

---

2.2x3=6.6
res voltage

---

8.2-6.6=1.6
current (mA)

---

1.6/56 x 2 = 57.2
power

---

.0572x9x0.4= 0.206W


This gives slightly brighter output with three times the battery life. You will have to measure the output voltage of the 555 with the 9 volt battery and the forward voltage of the leds and adjust the resistor value accordingly to make sure it will works properly.

I would also suggest using the ir sensor you plan to use with the stamp to adjust the frequency.

@Kwinn If you change your resistance to 56 OHMS on the IR LED the circuit will provide 7.34 volts to each LED " Can you Say POP there goes the Magic Smoke'? also the current and power will have a spike on them as well along with using more Current and Power as well. Please see the attached screen shots for these readings I provided for both 200 OHM set up with 14 IR LEDs and the 56 OHM setup as you instructed. This is one reason I use the computer simulations because then I can design circuits and test them before building and not only tat but to see how they preform as well. Also the Power is calculated by the following P=IR and not by the 40% duty cycle.. This may be where you got confused.



Jax

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If a robot has a screw then it must be romoved and hacked into..

Post Edited (GWJax) : 4/4/2009 4:34:45 AM GMT

GWJax

kwinn said...
Yes. that would work, but you are wasting a lot of power across the resistors. As Phil said earlier it would be better to put 2 (or even 3 with 9V) leds in series and reduce the series resistor. You get 2 (or 3) times the light for the same led current.

Also here is the comparison running 6 IR LED's in series.. As you see the 200 Ohms has Just a little bit more Power consumption and Current consumption than a 56 OHM resistor, not only that but the Voltage it too high on them and will over heat the IR's thus giving you less usage on them including the spikes that occur with this value . See the attached images to compare.

Jax

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If a robot has a screw then it must be romoved and hacked into..

Post Edited (GWJax) : 4/4/2009 5:04:42 AM GMT

kwinn

GWJax, I think you missed or misunderstood the part about putting 3 leds in series and using 2 strings of 3 leds in series. The 3 leds would have a total forward voltage drop of approximately 6.6V (2.2V x 3). The 555 timer output will also have a small voltage drop between the battery input and the output pin which I am estimating to be 0.8V. These voltages are estimates and will vary slightly depending on the specific leds and timer chip, but are close enough to estimate the voltage across the current limiting resistor ( 9-(6.6+0.8)=1.6V) A 56 ohm resistor with 1.6 volts across it will have a current of 1.6/56=0.028Amps or 28mA. Since the 555 is turning the led on and off at 38KHz it will be on for less than 50% of the time (555 can not get to 50% duty cycle) at most the average current will be 14mA. As long as you do not exceed the maximum power dissipation and peak current of the leds this should work just fine, and most ir leds have ratings higher than this

It would be a good idea to temporarily put a higher value resistor (say 100 ohms) in series with the leds, measure the actual voltage across it, and use that to calculate the value of the permanent resistor needed to limit the current to the desired value.

GWJax

kwinn said...
GWJax, I think you missed or misunderstood the part about putting 3 leds in series and using 2 strings of 3 leds in series. The 3 leds would have a total forward voltage drop of approximately 6.6V (2.2V x 3). The 555 timer output will also have a small voltage drop between the battery input and the output pin which I am estimating to be 0.8V. These voltages are estimates and will vary slightly depending on the specific leds and timer chip, but are close enough to estimate the voltage across the current limiting resistor ( 9-(6.6+0.8)=1.6V) A 56 ohm resistor with 1.6 volts across it will have a current of 1.6/56=0.028Amps or 28mA. Since the 555 is turning the led on and off at 38KHz it will be on for less than 50% of the time (555 can not get to 50% duty cycle) at most the average current will be 14mA. As long as you do not exceed the maximum power dissipation and peak current of the leds this should work just fine, and most ir leds have ratings higher than this

It would be a good idea to temporarily put a higher value resistor (say 100 ohms) in series with the leds, measure the actual voltage across it, and use that to calculate the value of the permanent resistor needed to limit the current to the desired value.

OK kwinn, rebuilding this circuit again the way you instructed now shows that your design is using alot more Current and more Power with the 56 OHM resistor than the 220 OHM resistor setup. Also you forgot about what happens when you add resistors or LED in series that the voltage will drop when you go down the line even though the Power and Current remains the same. With this said the first LED in the series will be brighter and when you go down the row they will get dimmer. This is why in my first post I used a pull-up resistor bank to keep the voltage the same on all LEDs. As of the output from the 555 timer this will remain the same until the battery charge starts to drain past the threshold. Also this is another reason why I use the computer to do the math and testing because even though we know the math how it should work we are all human and make small mistakes while building electronics.

Jax

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If a robot has a screw then it must be romoved and hacked into..

Phil Pilgrim (PhiPi)

GWJax said...
With this said the first LED in the series will be brighter and when you go down the row they will get dimmer.

That is simply not true. The current through every LED in a string of series-connected LEDs is the same. The voltage drop across each LED will also be equal if the LEDs are of the same type. Therefore the power consumed by each one will be equal, as will the light output.

With a nine-volt supply, say, and three LEDs in series, you will get three times the light output per watt as you would with a single LED and a larger series resistor operating at the same current.

-Phil

GWJax

Phil Pilgrim (PhiPi) said...

GWJax said...
With this said the first LED in the series will be brighter and when you go down the row they will get dimmer.

That is simply not true. The current through every LED in a string of series-connected LEDs is the same. The voltage drop across each LED will also be equal if the LEDs are of the same type. Therefore the power consumed by each one will be equal, as will the light output.

With a nine-volt supply, say, and three LEDs in series, you will get three times the light output per watt as you would with a single LED and a larger series resistor operating at the same current.

-Phil

@ Phil,

OK now I have to try to prove you wrong with a photo of 9 LED in series. In theory this is correct but each LED has a tolerance in which the manufacture must make them by so finding all LEDs with the same resistance other than military specs which are at a 1% tolerance and cost alot more than what a hobbiest is willing to pay, you'll have to test alot of them to get just the right ones. I'll post a picture of the LED's in series from the same spool and we'll compare them in a little bit and if I'm wrong then I'll retract my statement.

Jax

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If a robot has a screw then it must be romoved and hacked into..

Phil Pilgrim (PhiPi)

LEDs of a given type are typically binned by the manufacturer, and the light output within a bin and for a given current can vary by as much as 100% in some cases. This is not my quibble, though. Rather, it's the assertion that light ouput will diminish as a funciton of position in a chain of series-connected LEDs.

-Phil

GWJax

Correct Phil, I just ran 8 Green LED's on my test bench and had to remove the resistor and applied 15V to them just to turn them on, The light that comes out of them are NOT even but the camera does not pick up this very well, As I move down the row the last 3 are some what less dim but without having the light frequency chip that Parallax sell's I cannot prove this theory with a photo. As you stated above the LED's will vary in it's light output. So with that said I will have to stick with the known theory and retract my statement..

Jax

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If a robot has a screw then it must be romoved and hacked into..

kwinn

GW, if you put 15V on a string of leds and got them to light up without blowing any of them up you are very lucky.

I hope you will agree that if you connect several components such as diodes, resistors, and inductors in series and apply a voltage across them each component will have the same current flowing through it. The voltage and power will be different, but the current is the same.

I agree that individual leds will produce different intensities with the same current applied to them, but that is true for series or parallel connected leds.

Leds are diodes, so when the voltage increases to where the diode starts to conduct (forward biased) the current increases very rapidly with small increases in voltage unless there is a resistor in series to limit the current.

If you put several leds in series this is still true, but the voltage required to start them conducting will be the sum of the forward voltage drops of the leds. All of the leds will have the same current through them, so the intensity differences are strictly due to manufacturing tolerances, not to the location in the series chain.

GWJax

Kwinn, Now I only stated that I applied 15V, I did say what current it had. In fact I set my power station to only allow 300mA which is not enough to blow out the LED's. So I guess I'm not lucky after all, LOL, and just think I was going to buy Loto tickets tonight. I guess where I was mislead was when I looked at the voltage drop and over looked the same current flow as stated in my previous posts on your set up that you were talking about, my mind just did not click two to two together. But still you were wrong about the Power consumption with using the 56 Ohm resistor in series verses the 220 Ohm resistor. This was clearly documented via the simulations. As I said we humans make mistakes and over look some LAWs in electronics where a simulation program that has been proven time and time again is a very useful tool.

Jax

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If a robot has a screw then it must be romoved and hacked into..

kwinn

GW, if we make the assumptions I did in my earlier post as follows:

The ir led has a forward voltage drop of 2.2V
The battery provides 9V
There is a drop of 0.8V through the 555 when the leds are on

LEDS WITH 220 OHM CURRENT LIMITING RESISTORS X 6

One led with a 220 ohm resistor will have 8.2 volts applied by the output of the 555 when it is on

The led will have 2.2V across it leaving 6.0 volts to be dropped across the resistor

Since I=V/R hat means there must be 6.0/220=27.273mA through the 555/resistor/led

If you connect 6 leds and 220 ohm resistors to the output (in parallel as drawn in the schematic) of the 555 then you have 6x27.273=163.636mA or 0.163636 Amps in total

Since this is powered by a 9V battery the total power (VxI) is 9x0.163636=1.472727W

If the duty cycle of the 555 is 40% ( on 40% of the time ) then the average power is 1.472727x0.4=0.589091Watts

3 LEDS AND 56 OHM CURRENT LIMITING RESISTOR X 2

If you connect 3 leds and a 56 ohm resistor in series and apply the 8.2 volts from the output of the 555 each of the leds will have 2.2V across it for a total of 6.6 volts across the 3 leds, leaving 1.6 volts across the 56 ohm resistor

Since I=V/R there must be 1.6/56=28.571mA through the 555/resistor/3 leds

If you connect 2 sets of 3 leds and 56 ohm resistors to the output of the 555 then you will draw 2x28.571=57.143mA or 0.057143 Amps total

Since this is powered by a 9V battery the total power is 9x0.057143=0.514286 Watts

With a duty cycle of 40% that would be an average of 0.514286x0.4=0.205714 Watts

kwinn

After sending the last reply it occurred to me that we may have different ideas as to how the leds and resistors are connected in the series and parellel circuit so here is a schematic.

We make the following assumptions:
The current drawn by the 555 that is not going to the output pin to drive the leds is the same for both circuits and is low enough to be ignored.
We have picked resistors that provide the same current through the leds in both circuits (220 ohms for circuit A, 58.66667 ohms for circuit
The same leds are used in both circuits

Under those conditions circuit B will draw one third of the power drawn by circuit A and both circuits will provide the same ir output.

GWJax

kwinn said...
After sending the last reply it occurred to me that we may have different ideas as to how the leds and resistors are connected in the series and parellel circuit so here is a schematic.

We make the following assumptions:
The current drawn by the 555 that is not going to the output pin to drive the leds is the same for both circuits and is low enough to be ignored.
We have picked resistors that provide the same current through the leds in both circuits (220 ohms for circuit A, 58.66667 ohms for circuit
The same leds are used in both circuits

Under those conditions circuit B will draw one third of the power drawn by circuit A and both circuits will provide the same ir output.

This would be correct Kwinn,
Showing the results of both circuits on the simulator does prove this setup to work just fine and will save battery life. See attachments below.

Jax

PS. I'm glad we resolved this debate..

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If a robot has a screw then it must be romoved and hacked into..

kwinn

I am also happy to have it resolved, although I do feel a bit silly. If I had thought to send the schematic earlier instead of the long tedious explanations it would not have taken anywhere near as long. Oh well, from now on a schematic or block diagram first.

Proof of the old adage that a picture is worth a thousand words

GWJax

kwinn said...
I am also happy to have it resolved, although I do feel a bit silly. If I had thought to send the schematic earlier instead of the long tedious explanations it would not have taken anywhere near as long. Oh well, from now on a schematic or block diagram first.

Proof of the old adage that a picture is worth a thousand words

Ok now I did it, I just looked at the attached files and I sent the wrong ones, LMAO resending the correct screen shot photos..

Jax

Oh you have to love this, I just uploaded the wrong ones again!!! Not only that but I forgot to save the images to my HDD.. I am really having a hard time with this, LOL Sorry Guys for that

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If a robot has a screw then it must be romoved and hacked into..

Post Edited (GWJax) : 4/5/2009 11:48:25 PM GMT

vla7

Ok, I built the attached circuit.· I used·kwinn's·idea of 2 strings·of 3 leds in series and 47 ohm resistors.· (closest I had to 56 ohm).· I included actual voltage and current measurements on the schematic.· Remember that my goal is to maximize the range of detection for the IR beacon.· Please let me know if you have any other suggestions.

Phil Pilgrim (PhiPi)

Can I assume you measured the "voltage drop" with a meter and not with a scope? It's awfully low to be a peak reading, but if it's averaged over the duty cycle, it's probably about right.

-Phil