# NE555 driving BC239C transistor

Discussion in 'General Electronics Chat' started by chaos51, May 8, 2012.

1. ### chaos51 Thread Starter Active Member

Jun 18, 2011
42
1
Hi,

I am trying to do a simple thing, and somewhere in my circuit my NE555 will drive a transistor of type BC239C. Bear in mind, I am quite a beginner, so my questions may seem silly.

While doing this, I found I came to some questions.

-How to calculate the base resistor. I can't seem to find the output impendance of the NE555 anywhere on the web. I have a collector current of about 15mA (I believe), and I have looked in the datasheet, and the hfe seems to be around 400 or so.... So I am getting to (with a bit of cheating to make the numbers round) around 5k. I found some circuits on the net, and there these base resistors seem to be just 1k. Meaning the NE555 has an output impedance of about 49k???? But I thought the output of the NE555 was very low, so I am confused.

-I am just driving two leds, so I could also directly source them from the NE555. But I am also driving other logic chips from the output of the NE555. I am wondering if putting the leds as a "real" load directly here, would interfere with the voltage levels....

Does anyone have any opinions on these?

Regards
DaC

2. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
A general rule of thumb for BJT transistors is the base current should be 1/10 the collector current. This is because saturation changes some basic assumptions, but the 1/10 current is guaranteed to saturate the transistor.

3. ### chaos51 Thread Starter Active Member

Jun 18, 2011
42
1
Hi,

Thanks.... But still, how to calculate the base resistor, if I don't know the output impedance of the NE555?

R = U / I -> U=5 and I will then be 1.5mA,
but R is made up out of Rb and R-ne555-outputimpedance, no?

Regards
DaC

4. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
There is no output impedance for a 555, not like you are thinking at least. It is a digital chip as seen from pin three. I have several articles on the 555. While the chip can output up to 200ma, 100ma is a safe number for output current.

Bill's Index

The 555 Projects

Lets use this schematic as an example.

You start with the needed collector current. Lets say it is 100ma. The base needs 10ma.

A 555 drops around 1.3V from Vcc (for an explanation of why this is so, see 555 Schmitt Trigger). If your power supply was 5V, the resistor R2 would need to be:

(Vcc-1.3VDC-Vbe) / I or (5V - 1.3V - .6V) / 0.01 = 3.1V / 0.01 = 310Ω ≈ 300Ω

I have a group of circuits and a LED tutorial you might find useful at LEDs, 555s, Flashers, and Light Chasers.

There are also a couple of cheat sheets I've made here...

5. ### chaos51 Thread Starter Active Member

Jun 18, 2011
42
1
Thanks.... That is something to get me started ).

6. ### chaos51 Thread Starter Active Member

Jun 18, 2011
42
1
BTW, just for curiosities sake. Maybe I already mentioned it, but first I forgot to put in R2 at all.... So, why did that not burn anything, the base current should go up all the way to 200mA, without resistor?

Or perhaps does that have anything to do with the limit I put on the collector current, by putting resistors to limit the emitter (and implicitly the collector) current perhaps?

Building on that (maybe false) assumption, could you in certain scenarios completely do without any base resistor then?

I hope my question was clear.

Regards
DaC

7. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
Nope, you are likely using a CMOS 555, such as a TLC555 or a 7555.

That or you have your load on the emitter. With no current limiting and a standard 555, the 555 or the transistor (or both) will burn up.

You haven't shown a schematic, so I can only speculate.