I've got a bunch of general purpose NPN transistors
No particular designations, then? Well, it's not terribly important
so long as they aren't very ancient.
and I want to turn on and off 15Vdc with 5Vdc
Since they are NPN, I assume that you mean to use them with emitter
connected to ground and the 15V load (whatever it is) between the
collector and the +15V supply side.
from logic chips like binary
counters or different logic gates (XOR, NAND,etc).
Hobby play?
From what I read about these transistors I guess I want them to go into
saturation which is maximum CE conduction.
Well, saturation in the case I mentioned above is just a matter of
forcing the collector voltage close enough to ground voltage (0V) so
that the base-to-collector junction begins to be forward-biased. And
the current through the collector needed to do that will depend on the
load you have connected, not upon the transistor. So which exact type
of transistor you use won't really matter on this narrow point. Do
you see why?
Is there a rule of thumb to turn a transistor all the way on
Yes. When the current passing by way of the transistor's collector
and through the load causes the load itself to drop almost all of your
supply voltage, then the transistor will be "all the way on." When
that happens, the load is taking up almost all of the voltage
available, leaving only a smidgen left to support the collector to
emitter voltage of the transistor. For design purposes, that
remaining smidgen is usually estimated to be just a few tenths of a
volt, perhaps somewhere between .1 and .4 or so.
So just estimate the collector current needed by taking your 15V,
subtracting a small "smidgen" from it for the transistor, then
dividing that by the load resistance of what you are connecting to the
collector and supply. Let's say it is 1k ohm load and you want to
estimate a saturation Vce of .2V, you'd calculate (15-.2)/1k as your
collector current needed for saturation. As you can see, with a 15V
supply in hand, changing your saturation estimate by a tenth or two,
one way or another, won't seriously impact your calculated saturation
current. The actual voltage for Vce that you will get will depend on
other particulars (such as the exact device used and the base
current), but it's not important at this stage to worry about the
transistor data sheet.
It's not uncommon to then blindly estimate the needed beta as 10, for
saturation. That's because most modern transistors are pretty much
going to be doing as much as you can expect from them as a switch when
you supply 1/10th of the collector current as base drive. There are a
few exceptions. Actually, it turns out that in many cases you can
drive them with much less base current than this rule of thumb
suggests. Maybe 1/30th or even 1/50th. But almost any transistor
will get there with 1/10th. So you might not even need to look at the
data sheet for even this value, if you are willing to accept 1/10th of
the collector current as base drive.
Also, it's not uncommon to estimate the base voltage as .7V for this
purpose. The exact base voltage will depend largely on the base
current used (not entirely, as the collector current does play a role,
but almost.) But since it varies up and down by roughly 60mV for a
10-fold change in base drive current, the exact value won't be all
that far away. If your collector current is large, then the base
drive current will be large too and the base voltage may be nearer
..8V, .85V, or even .9V. But for design purposes, being off by one
tenth of a volt won't kill your design. And .7 is a good place to be
for many cases.
So far, no need yet for a data sheet to make a reasonable design.
(1) Compute collector current from the supply voltage, the load
resistance, and a blind estimate of Vce, as (Supply - Vcesat) / load.
(2) Compute needed base drive current as 1/10th the value computed in
step (1).
(3) Use .7V as a blind estimate for the base voltage. You can modify
this per Vbe=25.865mV*(1+ln(Ib/Is+1)) @ 27C, where Is is a value found
for the simulation model of the transistor or just taken to be 10^-14
or so, blindly. For example, if Ib (base drive current from step (2))
is about 100uA, then you could compute 25.865mV*(1+ln(1+100uA/10fA)),
which works out to about 0.62V. This will be closer, probably, than
the .7V. But the .7V won't steer you much wrong, anyway.
(4) Estimate the actual control voltage. You said 5V, but unless
this is a power supply (and you've said it isn't), there is some
internal impedance and supplying the base current calculated in step
(2) will drop the voltage a little. Or, if this is a standard TTL
output, you cannot always rely upon the voltage being much better
than, say, 4V. What it is exactly will depend on your output type and
what is connected to it other than the transistor. But let's use an
estimate of 4.5V, as an example.
(5) Compute your base resistor used to limit the base current to the
value you computed in (2) on the assumption that your 5V control
voltage is low impedance enough that it can yield at least the voltage
estimated in step (4) or 4.5V. This is done by dividing the voltage
that the base resistor needs to drop by the estimated base drive
current. Something like this: (4.5V - .7V)/100uA = 38k. Use 39k, as
a standard value.
Now, suppose you are wrong about the base drive current using the
estimator of 1/10th to compute it. Suppose the actual base drive
current for the actual transistor works out to, say, 1/30th instead.
Well, the voltage drop across the base resistor will be a lot less if
that happened and that would mean that more of the 4.5V control
voltage would appear at the base of the transistor. Just for a
moment, let's assume that 1/3rd of our computed current was actually
needed -- in that case, the base resistor would drop 1/3rd of the 3.8V
we earlier used in our computation for the resistor value, or about
1.3V, instead. That would, we might guess, cause 3.2V to be at the
base of the transistor. But that won't happen. That's because, as I
mentioned before, the base voltage goes up by 60mV for each 10-fold
increase in base current. So if it really were 3.2V, that could only
happen if there were (3.2V - .7V)/60mV or about 4.2 orders of
magnitude change in the base drive current. That would be more than
10,000 times. And if you tried to send that much current through the
base resistor we computed, then the voltage drop would be impossibly
high. So what really happens is that even if the saturation beta of
the transistor _might_ be better estimated as 1/30th, than 1/10th,
having designed things with the idea of 1/10th means that it will be
very close to 1/10th, anyway. The actual transistor you use may
impact just a tiny bit the actual current flowing through the base
resistor, but it will be _very_ close to the designed value because of
this exponential behavior of base current versus base voltage. So if
you use 1/10th as your rule of thumb in your design, it will come out
pretty darned close no matter what transistor you actually use.
So no data sheet, still.
or do I go to the
datasheet for each particular transistor (which still sometimes confuses me
and not sure which callout is the designation for base volt/amp for
saturation). I've read that .7V base voltage above emitter voltage will do
it, I've also read that I need 5 volts above collector voltage will do it.
Any guidance is appreciated,
You _may_ still want to take a look at the data sheet. One thing you
need to be sure (and this is more important if you had a supply
voltage larger than 15V) of is that the transistor can "stand off"
your supply voltage when it is off. If there is no current flowing
through the load, then the load doesn't drop any voltage and the
collector will be exposed to the supply voltage. You want to be sure
that the transistor can support the Vce difference without breaking
down. For most transistors, given my hobby experience, you can
usually start worrying when the supply voltage exceeds about 25V or
so. But there are enough transistors with lower voltages that it's
worth a look just to be sure.
Jon