# 2n3904 driving an NE555

## To put 2n3904 at the input stage ?

• ### No

• Total voters
2

#### engineertam

Joined Dec 4, 2011
6

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#### wayneh

Joined Sep 9, 2010
16,512
It turns on the 555, in response to a signal detected by the LDR, and the 555 produces a tone as an alarm.

#### engineertam

Joined Dec 4, 2011
6
Yes I am aware of that, but my dilemma is, is it necessary to drive a 555 using 2n3904? I used multisim to test if putting 1k and the ldr in voltage divider and connecting it to the pin 4 and 8 will still light up the led, and the LED still lit up

#### wayneh

Joined Sep 9, 2010
16,512
...is it necessary to drive a 555 using 2n3904?
Of course not. That arrangement was simply to control the power to the 555. If you can achieve the same thing some other way, that's fine.

You can probably get all the power you need for the 555 through your 1K resistor, but I'm a little surprised the LDR can shut it off reliably. It would have to pull the voltage down quite a bit? Seems unreliable as the power source ages (9v battery?). Could be that the transistor arrangement was used to increase the sensitivity of detecting light at the LDR.

#### engineertam

Joined Dec 4, 2011
6
Yes the source is a 9v battery, I don't think that the transistor is arrange to adjust the sensitivity of the LDR, the 1k and LDR forms a voltage divider towards 2n3904. The configuration is common collector amp, so it is a voltage buffer, a good impedance matcher, has a voltage gain of 1, high current gain, and has a power gain. Then what? I am sorry, I am just a beginner in applications of theory. I just want to know the purpose of having a 2n3904 as the input stage.

Because I almost get the same values for voltages at the load, the circuit w/o a transistor has even a higher voltage of about .10mv than the circuit w/ a transistor

#### engineertam

Joined Dec 4, 2011
6
I just want to know the sole purpose and importance of placing a 2n3904 at the input stage

Joined Dec 26, 2010
2,148
The output obtained directly from the LDR may not be able to pass enough current to switch the power input to the 555 directly. Depending on the type of LDR used, its resistance may be too big, even when strongly illuminated.

The transistor provides current gain so that more current is available to power the 555.

#### engineertam

Joined Dec 4, 2011
6
Thanks, I'm not that sure if the current is not enough, but I'll go with it.

Joined Dec 26, 2010
2,148
It depends on what LDR you have, and how brightly it will be illuminated. Some are many kΩ even in bright light, others are much lower. This site may be of interest: http://www.ladyada.net/learn/sensors/cds.html

The single emitter-follower circuit actually gives only moderate current gain and no voltage gain, hence rather soft switching as the light increases. Other transistor arrangements might work better, but a voltage comparator IC would be far more usual for this.

#### bloguetronica

Joined Apr 27, 2007
1,424
I would use the 2N3904. LDRs per se don't let pass enough current to control anything.