Display cabinet - PIR motion sensor & LED lights

[Ronnie_Space]

I'm coming in a bit late on this one, but a good way to connect devices like this PIR to your breadboard are cables like this.

No real functional difference, it just makes the process even more "solderless".

My final circuit will be soldered and likely on flying leads rather than breadboard. However, these leads would be very handy for prototyping and possibly the final build. I didn't know they existed. Thanks for your help.

Ronnie

 

[Ronnie_Space]

Thanks for the reply kpatz.

Please see video below measuring the current draw of the circuit when idle and illuminated. I current have only x6 LED's in series, the final circuit will have x10. The multi-meter reads 0A current draw when idle and 80mA(?) when illuminated.

Ronnie

Would I be correct to assume that based on AA batteries having a capacity of 2000mAh, that this current prototype circuit would last 2000/80 = 25 hrs, running continuously?

No. of 15 second Activation's:

25 hours = 1500 mins

1500 mins = 90000 seconds

90000/15 = 6000 activations

I calculated the resistor value at 150ohms to give 3.2V @ 20mA. I measure 3.2V across the LED, however, I was expecting to see the current draw be 120mA (6 x 20mA), not 80mA. Any reason for this?

Thanks

Ronnie

 

[BobK]

The listing you pointed to said the Vf was 3.2V to 3.8V. So you are using the minimum, which pretty much guarantees that you will not get the desired current. You will have to raise the voltage to get the full current.

Bob

 

[Ronnie_Space]

The listing you pointed to said the Vf was 3.2V to 3.8V. So you are using the minimum, which pretty much guarantees that you will not get the desired current. You will have to raise the voltage to get the full current.

Bob

Hi Bob,

Thanks, I don't understand why that is the case, I guess that is the difference between real life and theory, accounting for the efficiency of the LED?

I used V=IR to calculate the resistor:

Input voltage: 6V
Output voltage: 3.2V
Current draw: 20mA

Vdrop / I = R

6-3.2 / 0.02 = 140 ohms

Nearest resistor value 150 ohms

Does the transistor use a certain amount of voltage/current?

Ronnie

 

[gorgon]

The normal Vce voltage drop over a single transistor will be in the order of 0.2- 0.4V, and this is easy to measure. I would think that using 120 ohm resistors will bring you closer to the goal of 20mA, depending on how the Vf is reacting to a greater current. How much more light you get remains to be seen. I think I would have added a couple of LEDs, instead of increasing the power on the few you have. It all comes down to what, and how big your exhibition is, and how your lighting is placed.

 

[BobK]

Hi Bob,

Thanks, I don't understand why that is the case, I guess that is the difference between real life and theory, accounting for the efficiency of the LED?

The range means that the forward voltage of any particular LED at 20mA will be somewhere in that range. Each one might be different. The large range is probably due to the fact that they have several sources for the LEDs. And even LEDs from the same source will vary from batch to batch and even between different LEDs in the same batch.

Since you did your calculation based on the low end of the range, it is very likely that you would not get 20mA at that voltage since most will require a higher Vf to pass 20mA.

Bob

 

[gorgon]

As long as he measure 3.2V over the LED, there is no reason he can't get 20mA through it, IF the available voltage is present.

The problem has to be the total available voltage. Is the 6V supply 6V? How much does the transistor drop?
With 80mA for 6 LEDs, the average current is 13.33mA that is equal to a 2V drop over the resistor, leaving 0.8V missing somewhere. This is divided between low supply voltage and Vce of the transistor.

Should be no problem to measure where it goes. If you have a large Vce, I would have tried a fresh transistor, I'm not sure how well it takes to be reversed.

 

[BobK]

As long as he measure 3.2V over the LED, there is no reason he can't get 20mA through it, IF the available voltage is present.

The problem has to be the total available voltage. Is the 6V supply 6V? How much does the transistor drop?
With 80mA for 6 LEDs, the average current is 13.33mA that is equal to a 2V drop over the resistor, leaving 0.8V missing somewhere. This is divided between low supply voltage and Vce of the transistor.

Should be no problem to measure where it goes. If you have a large Vce, I would have tried a fresh transistor, I'm not sure how well it takes to be reversed.

Not true. The voltage required to get 20mA might be anywhere between 3.2 and 3.8V.

Bob

 

[gorgon]

A single normal LED will not jump around like that, if you have 3.2V @ 13mA, you will not have 3.8V @ 20mA. The 3.2 - 3.8V is the range that you can get the Vf in for the LEDs as a group, not as a single LED.

The max Vf you will see for the 3.2V unit will be around 3.3V close to max current, if it is that much. It just the same for a LED with Vf = 3.8 @ 20mA, It will definitly not be 3.2V @13mA

 

[BobK]

If you have 13mA at 3.2V, how do propose to get 20mA at 3.2V? You will get 20mA at some higher voltage, unknown until you actually test it.

What I am saying is the if you use the minimum voltage as Vf in the calculation, you will only get 20mA for an LED that happens to be at the extreme of the range.

Bob

 

[gorgon]

It is a LED - Light Emitting Diode we talk about, not a resistor?

And the OP measured it to 3.2V @13mA

 

[Ronnie_Space]

Just a quick post to say thanks all for your help, this project was a success..!

Final assembly used x9 LED's and note the PIR sensor in the front of the lid, activated for 10 seconds.

Will post final circuit diagram.

Best Regards

Ronnie

 

[Ronnie_Space]