# Another question Base Resistor

Discussion in 'General Electronics Chat' started by BReeves, Nov 7, 2013.

1. ### BReeves Thread Starter Member

Nov 24, 2012
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64
Same schematic as my question on the logic of leaving the opto LED on.

Q2 is a BD137, hfe ~ 40, the counter has a 92 Ohm 12 volt coil. When I calculate the base resistor for Q2 I come up with 3.3K. Whoever designed the original control board felt 10K would put Q2 in saturation. What am I missing, big difference between 3.3 K and 10K?

2. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,320
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What you are missing is how much collector current is flowing. You need to know that to pick the base resistor.

That, and the number you keep in your back pocket. Most people keep a "10" there to give the base 1/10 the collector current. I have a "20" there for small signal stuff.

It's not a hard and fast thing.

3. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
{12-Vcesat) = ~11V

Icoil = 11/92 = 120mA

Ib = Ic/10 = 12mA

Rb = (12-0.6)/12m = 0.95K, call it 1K

4. ### BReeves Thread Starter Member

Nov 24, 2012
412
64
I did neglect the voltage drop across the transistor and used the transistors gain

3K is allot closer to 1K than 10K but still leaves me wondering how the original designer would come up with 10K

Just did a quick lets-see, if the transistor had a current gain of 120, 10K would be about right. We know 10K works as the machine is working with the old controller. This means in reality anything from 1K to 10K would work. At least till someone replaced the transistor with one that actually only had an hfe of 20.

Last edited: Nov 7, 2013
5. ### SgtWookie Expert

Jul 17, 2007
22,182
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When you're using a transistor as a saturated switch, you really do want to use the Ib=10 figure. However, in this case you're up against a troublesome design; if you DO decrease the 10k base resistor for Q2, it'll wind up with increased power dissipation. A 1k resistor will have around 144mW power dissipation when the optocoupler output is conducting, requiring a 1/4W resistor. There's also the limitation of the optocouplers' output to sink that much current; depending on the model you might be limited to somewhere around 3mA; and as the optocoupler ages, that will get worse. This is most likely the reason the base resistor chosen is so large. I suggest that the design you have presented is somewhat dubious. Datasheet for the BD135:
http://www.fairchildsemi.com/ds/BD/BD135.pdf
See Figure 1 and Figure 2 on page 3; I think it's likely they looked at those curves and then fudged it due to the optocoupler's low current sinking capacity.

6. ### BReeves Thread Starter Member

Nov 24, 2012
412
64
Thanks, that makes sense..

I am also controlling it with an opto but the one I am using has an output transistor rated at 50mA. 3.3K will result in ~3.64 mA, well within spec. A 1K would be ~12mA still within spec.

I may just split the difference and use a 2.2 K

I probably should add that I will not use the reverse logic of the original design. I will simply turn on the opto for ~50ms when it needs to count.

Last edited: Nov 7, 2013
7. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
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Lots of experience? Most of those small transistors will run an actual beta easily over 100 when driving a 120mA load. So a 10k Rb is low enough.

Also keep in mind that there are different levels of "saturation" and if you can tolerate (say) 1v drop across the transistor Vce when on, you can achieve that low-ish level of saturation at a beta of maybe 200-300. That still leaves 11v for the relay coil, plenty enough to ensure pull-in, AND reduces coil current and coil heat by a small amount too.

If in doubt, build it and test it.