# Laser Tripwire Question

Discussion in 'General Electronics Chat' started by krow, Feb 17, 2011.

1. ### krow Thread Starter Member

May 25, 2010
49
0
I'm trying to build a simple laser tripwire circuit and there's something I don't get, I'm sure you guys can help me understand what's wrong.

Yesterday night, I managed to get the expected values but today I loaded my file in Electronics Workbench and there's something strange.

My question is simple, what do I not get 2.5V in the voltage divider? I used 2 1-K resistors just to make it easier for me to calculate the values (manually) and make sure I'm wiring things up right, however, I get 744.87 mV.

ps if the mistake I'm making is too stupid please don't judge me, I'm learning

Is it about the voltage sources? is it about how I connected the grounds?

Please see the picture.

Thanks a lot!

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Aug 27, 2009
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3. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
298
The transistor base draws a current of 3.426mA from the potential divider output. This current reduces the divider output voltage.

We can even check that this works out: The potential divider output equivalent resistance is 1kΩ in parallel with 1kΩ, or 500Ω. The voltage drop ca usedby the base current is therefore 3.426mA*500Ω or 1.713V.

The potential divider output should therefore be 2.500V-1.713V = 0.787V (as expected)

4. ### krow Thread Starter Member

May 25, 2010
49
0
Thanks for the reply but I have one more question now, how do I calculate the base current manually? I just want to be able to do all the calculations without any programs or simulators.

5. ### hobbyist Distinguished Member

Aug 10, 2008
764
56
Your voltage divider is across the supply of 5v.

According to voltage divider this should put your base voltage at 2.5v.

However the base emitter junction is across the bottom resistor, and it acts as a voltage drop of around 0.7v. due to the base emitter diode.

To get the base voltage to be at 2.5v. you need to place a emitter resistor in the loop.

Here is a equation to use for solving for a emitter resistor.

RE (emitter resistor) = [(VB - Vbe) / IC]

Your IC value is 55mA.
so RE = [(2.5v. - 0.7v.) / 55mA.] ~= 33 ohms.

Your emitter voltage is now (VB - Vbe) = 1.8v.
and (1.8v. / 33 ohms) ~= 54.5mA. = IC.

You may have to juggle the resistor value down a little to get a higher collector current.

I had to lower it from 33 ohms to 27 ohms to get back the collector current of 54mA.
However that lower value resistor will now load the voltage divider causing it to decrease down to 2.2v.

But thats how it works with the bjt. as current operated devices.

Now back to your question on calculating values.

Using your values for collector current and supply voltages.

VCC = 9v.
IC = 55mA.

Using single battery bias is a more economical way of designing many circuits.
This being one of them.

Keeping your resitor of 100 ohms for the LED.
makes the voltage drop across this resistor = to (IC x 100ohms) = 5.5v.

Assuming a 2v. drop across the LED makes a total so far of 7.5v.
(9v. - 7.5v) = 1.5v. is whats left to work with which is the voltage across the series resistance of the transistor and its emitter resistor.

So as to not saturate the transistor, make the voltage across the transistor to be 1v.
This leaves 0.5v. (VE) to be across the emitter resistor.

So now (VE / IC) = RE = (0.5v / 55mA.) = 9.1ohms = (R1) on bottom schematic.
VB now = (VE + Vbe) = (0.5v. + 0.7v) = 1.2v.

Make base to ground resistor (R4) be 10 times R1 which = 91 ohms.
Now the current (IB4) through this resistor is (VB / R4) = (1.2v / 91 ohms) = 13mA.
Therefore supply to base resistor (R3) = [(VCC - VB) / IB4] = [(9v. - 1.2v.) / 13mA.] ~= 560 to 620 ohms. I'll choose 560 ohms.

Now prototype the circuit and adjust values until yuou get your desired results.

Last edited: Feb 19, 2011
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