How to figure transistor base resistor?

[Glenn Ashmore]

This must be very basic but I just can't get a handle on it. Everything
I can find on transistors is either simple stuff about the silicon or so
complex I need to be a Greek major. I need a cookbook methodology for
figuring out the proper values.

Say you have a relay coil that draws 45mA at 12V and you want to hold it
on with a 2N2222 transistor. The transistor is held on by the 12V
supply and turned off by draining the base through another transistor.

First, I keep hearing to "force the beta to 10" but the data sheet says
the gain ranges from 35 to 100. Do I use 45/10, 45/30 or something
else to find the current?

What voltage do I need to saturate the base and how do I get it from the
12V supply? What do I need to drain off to turn the transistor off?

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[John Larkin]

This must be very basic but I just can't get a handle on it. Everything
I can find on transistors is either simple stuff about the silicon or so
complex I need to be a Greek major. I need a cookbook methodology for
figuring out the proper values.

Say you have a relay coil that draws 45mA at 12V and you want to hold it
on with a 2N2222 transistor. The transistor is held on by the 12V
supply and turned off by draining the base through another transistor.

Presumably there is a resistor from +12 to the base to define the "on"
base current. You can't just apply +12 to the base or it will fry.

First, I keep hearing to "force the beta to 10" but the data sheet says
the gain ranges from 35 to 100. Do I use 45/10, 45/30 or something
else to find the current?

OK, technically if the min beta is 35, you'd need at least 45ma/35 =
1.3 ma of base current to pull your collector load down to ground,
which is all you really want to do to operate a relay. If you use more
base current "just for luck", the actual ratio of load current to base
current is called "forced beta" because you are force-feeding the poor
thing. That's a good idea, since it guarantees a good low collector
saturation voltage and provides lots of margin for error. So to
"force" beta to 10, you'd need 4.5 ma of base current, or maybe a 2.2k
resistor from +12 to the base.

What voltage do I need to saturate the base and how do I get it from the
12V supply? What do I need to drain off to turn the transistor off?

See above. Base voltage will be about +0.7 above emitter voltage when
the transistor is on; but design to force *current* into the base, not
voltage. Base current is an extreme (exponential) function of voltage,
and varies a lot with temperature and between parts, so it's not
practical to use base voltage calculations to get consistant results.
So run a resistor from +12 to the base as above.

Then base current Ib = (12-0.7) / Rb

Actually, it's the collector you saturate, not the base. The
transistor is "saturated" when the collector-emitter voltage is very
low, below the base-emitter voltage at least.

To turn it off, make the base current zero. You could do that by
opening the resistor that's supplying base drive from +12, or you
could short the base to the emitter (ground here, I guess) with a
switch or with the collector of another transistor.


John

 

[Glenn Ashmore]

Presumably there is a resistor from +12 to the base to define the "on"
base current. You can't just apply +12 to the base or it will fry.

That is correct.

OK, technically if the min beta is 35, you'd need at least 45ma/35 =
1.3 ma of base current to pull your collector load down to ground,
which is all you really want to do to operate a relay. If you use more
base current "just for luck", the actual ratio of load current to base
current is called "forced beta" because you are force-feeding the poor
thing. That's a good idea, since it guarantees a good low collector
saturation voltage and provides lots of margin for error. So to
"force" beta to 10, you'd need 4.5 ma of base current, or maybe a 2.2k
resistor from +12 to the base.

That is kind of what I figured but where did you get the 2.2K? Figuring
backwards: 2.2K*4.5mA = 9.9V so we get 2.1V into the base?

Base voltage will be about +0.7 above emitter voltage when
the transistor is on; but design to force *current* into the base, not
voltage. Base current is an extreme (exponential) function of voltage,
and varies a lot with temperature and between parts, so it's not
practical to use base voltage calculations to get consistant results.
So run a resistor from +12 to the base as above.

Then base current Ib = (12-0.7) / Rb

Not sure I follow. Lets see: 11.3V/2.2K = 5.1mA A little off but if
we had a 2.5K resistor it would be right on. OK so we are talking
strictly current. That is where my brain got tangled up.

Actually, it's the collector you saturate, not the base. The
transistor is "saturated" when the collector-emitter voltage is very
low, below the base-emitter voltage at least.

To turn it off, make the base current zero. You could do that by
opening the resistor that's supplying base drive from +12, or you
could short the base to the emitter (ground here, I guess) with a
switch or with the collector of another transistor.

That's what I was planning. So using the same process the controlling
transistors need to drain that 4.5 to 5 mA and will need about 450 to
500uA into its base.?

John

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[John Larkin]

That is kind of what I figured but where did you get the 2.2K? Figuring
backwards: 2.2K*4.5mA = 9.9V so we get 2.1V into the base?

The more-correct value is R = (12-0.7) / 0.0045 = 2511 ohms. I just
sort of automatically pick the next lowest standard resistor value.

No, using 2.2K will *not* increase the base voltage very much.
Instead, base voltage will stay close to 0.7, and using a lower
resistor will increase base *current*. To maybe 5 ma, in this case. No
big deal.

There are nice equations for Ib vs Vbe, but you don't need to be that
precise for simple switching stuff like this. Just assume you'll have
*about* +0.7 on the base when there's a useful amount of base current
flowong.

Not sure I follow. Lets see: 11.3V/2.2K = 5.1mA A little off but if
we had a 2.5K resistor it would be right on. OK so we are talking
strictly current. That is where my brain got tangled up.


That's what I was planning. So using the same process the controlling
transistors need to drain that 4.5 to 5 mA and will need about 450 to
500uA into its base.?

Yup. Since you're working in a boat, and they tend to get wet, as I
understand it, try to avoid circuits that work with very small
currents. If a transistor would turn on with a tiny amount of wet-pcb
surface leakage current, you can add a base-emitter resistor to steal
some of the base current and make it less sensitive. If you put, say,
a 1K b-e resistor, it would swipe 0.7 ma of base current when the
transistor was trying to turn on, so that would prevent small leakages
from turning it on. Some people, myself included, never trust a
transistor with a floating base to be "off", and always have a b-e
resistor or equivalent; legacy of old leaky germanium transistors, I
guess. Your relay driver stage is fine: a resistor pulls the base up
hard, and a transistor yanks it down. The lower-level stage needs to
be similarly robust.

Salt water is *very* conductive stuff!

John