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Driving a transistor base from a voltage divider

J

Jon Danniken

Jan 1, 1970
0
Hi, I am trying to determine the operating specs for operating a transistor
base (as a switch) from a voltage divider.

What I am trying to figure out is the appropriate values of R1 and R2, given
a desired Base current and source voltage/source current available (I know
the source voltage, and how much current I can pull from it).

If I eliminate the "bottom" (R2) resistor in the voltage divider and only
use one resistor, I can easily determine the current that flows through that
resistor (E/R), and therefore the current that flows through the base
((E-0.7)/R), which is just computing the value of a Base resistor.

Right now I am getting hung up on what effect the "bottom" resistor has on
the base current. I know that without the transistor, I can figure out the
voltage at the center point as a ratio of the resistances, and the current
through both resistors, but what current is available from the center point
to drive a transistor base?

The closest I have come is by calculating a "resistance" for the transistor
Base/Emitter junction by dividing 0.7V by the desired base current, and
treating the transistor as a resistance in parallel with R1, but I'm not
very confident with that solution.

BTW, here is an ascii of what I have:

---+
|
|
\
R1 / C/
\ | /
| |/
|-----|
| B|\
\ | \
R2 / E\
\
|
|
gnd

Thanks for any help,

Jon
 

neon

Oct 21, 2006
1,325
Joined
Oct 21, 2006
Messages
1,325
voltage diveder or not you will need a substantial ,current ot turn it on. my rule is take the Ic diide by ten [ force beta 10] and that is the current that your divider must supply for a switch. since the base current also increase Vsat consider the ramification of beta 10
 
B

Baron

Jan 1, 1970
0
Jon said:
Hi, I am trying to determine the operating specs for operating a
transistor base (as a switch) from a voltage divider.

What I am trying to figure out is the appropriate values of R1 and R2,
given a desired Base current and source voltage/source current
available (I know the source voltage, and how much current I can pull
from it).

If I eliminate the "bottom" (R2) resistor in the voltage divider and
only use one resistor, I can easily determine the current that flows
through that resistor (E/R), and therefore the current that flows
through the base ((E-0.7)/R), which is just computing the value of a
Base resistor.

Right now I am getting hung up on what effect the "bottom" resistor
has on
the base current. I know that without the transistor, I can figure
out the voltage at the center point as a ratio of the resistances, and
the current through both resistors, but what current is available from
the center point to drive a transistor base?

The closest I have come is by calculating a "resistance" for the
transistor Base/Emitter junction by dividing 0.7V by the desired base
current, and treating the transistor as a resistance in parallel with
R1, but I'm not very confident with that solution.

BTW, here is an ascii of what I have:

---+
|
|
\
R1 / C/
\ | /
| |/
|-----|
| B|\
\ | \
R2 / E\
\
|
|
gnd

Thanks for any help,

Jon

If the current through R1 + R2 is much larger than the required base
current then the junction will set the base voltage. That will allow
you to set the collector current anywhere between zero and maximum.

However for a switch the transistor would be hard on or hard off. In
which case a single resistor will limit base current to a safe level,
at the point where its switched hard on.
 
J

Jon Danniken

Jan 1, 1970
0
"BobW"
You can't really control the base current. It draws whatever is required
to support the emitter current that is flowing.

In your resistor divider setup, R1 will provide current to the base of the
transistor AND to R2. So, if you measure the current through R1 (volts
across R1 divided by the value of R1) and subtract the current through R2
(volts across R2 divided by the value of R2) then you'll have the base
current. The only problem with this technique is that you need to know the
value of the voltages and resistances very accurately. An easier approach
would be to add a small resistor in series with the base of the transistor
and measure the current by reading the volts across this resistor and
dividing it by the value of the resistor. Or, you could simply put a
current meter in series with the base of the transistor.

The current that is available at the connection of R1 and R2 depends on
how many volts (from gnd) you desire. At zero volts (from gnd), the
current you'll get is simply V+/R1. To better visualize what you really
have, first remove the transistor from your thoughts. Now, just look at
the voltage divider. It can be reduced to (thought of as) a voltage source
with a series resistor coming from it. The value of the voltage source is
the open circuit voltage you would get at the R1-R2 connection. That is,
you would have a voltage source equal to (V+) * (R2/(R1+R2)). The
effective series resistor would be equal to the parallel combination of R1
and R2, and would be equal to 1/( (1/R1)+(1/R2) ).

For example, if V+ is 12V, and R1 is 1K ohm and R2 is 400 ohms, the
"equivalent" circuit would "look like" a 3.43V battery with a single
series resistor of 286 ohms. So, its open circuit voltage is 3.43V. Its
short circuit current is 12mA (note that this 12mA value is the same as
you get from the "real" circuit -- that is 12V/1K). Now, when you hook up
your transistor, it's easier to see what the base voltage would be for a
given amount of base current.

Thanks, Bob, much appreciated. I think I was thinking of the voltage source
as I do with a linear power supply, ie so many volts *while* delivering a
certain current level.

Perhaps instead I should be thinking of voltage and current as opposites
given open or closed circuit behavior, as in when plotting a load line for a
tube? If that is the case, I'm guessing a minimum voltage being 0.7V and
the short-circuit current as whatever I need to get sufficient base current
to switch the transistor on?

Thanks again,

Jon
 
J

Jon Danniken

Jan 1, 1970
0
Baron said:
If the current through R1 + R2 is much larger than the required base
current then the junction will set the base voltage. That will allow
you to set the collector current anywhere between zero and maximum.

However for a switch the transistor would be hard on or hard off. In
which case a single resistor will limit base current to a safe level,
at the point where its switched hard on.

Thanks Baron, that is appreciated.

Speaking of switching the transistor "hard on", how (in the absence of load
charts for a particular transistor) do I determine a sufficient base current
to do this?

Thanks again,

Jon
 
P

Phil Allison

Jan 1, 1970
0
"Jon Danniken"
Speaking of switching the transistor "hard on", how (in the absence of
load charts for a particular transistor) do I determine a sufficient base
current to do this?


** The base current needs to be at least 10% of the collector current to
ensure the transistor is biased hard on - make sure that current level is
within the capabilities of the device first ( see the data sheet).

The base-emitter resistor ( R2) plays no part in biasing the transistor
on - its role is helping it to switch off fast, but it will also take some
drive current away from the base of the transistor.

So, the design begins with deciding on the value of R2 and the known
collector current, Ic .

For non-critical applications, R2 can be between 100 ohms and 1kohms.



...... Phil
 
J

Jon Kirwan

Jan 1, 1970
0
Find the Thevenin equivalent of the base bias network
for starters. Then you'll be working with a single
equivalent resistance for setting base current. Once
that's done you can look at the formula for the Thevenin
resistance (how its composed of R1 and R2) to decide on
what values can be chosen for R1 and R2 to set the total
current draw as desired.

I think that's the answer the OP seemed to want.

Jon

P.S.
Vth = V*R2/(R1+R2)
Rth = R1*R2/(R1+R2)
Assuming grounded emitter and Vth > 0.7V:
Ib = (Vth-0.7V)/Rth
Ic <= beta_sat*Ib (compliance)
 
J

Jon Danniken

Jan 1, 1970
0
"BobW"
If the voltage at your emitter is fixed (e.g. at gnd) then it makes the
analysis easier because the base will be (about) 0.7V away from the
emitter (for any decent level of base current). So, if this is the case
for your circuit, then the base current will be easy to calculate using
the previous ideas that I presented.

Determining how much base current is sufficient to switch the transistor
on is a little more complicated. You need to know:

1 - The maximum collector current when the transistor is on.
2 - The minimum Beta (Ic/Ib) of your transistor when the collector is
close to being in saturation (e.g. when Vce = 1V)

If you know these two quantities then you can easily calculate the base
current required. When I do this, I usually double or triple the base
drive current to insure that the device is fully on (i.e. in full
saturation).

Ah, thanks Bob, I hadn't made that leap of using the beta and Ic max to
figure out the point of base current required for saturation.

Thanks again,

Jon
 
J

Jon Danniken

Jan 1, 1970
0
"Jon Kirwan"wrote:
I think that's the answer the OP seemed to want.

Jon

P.S.
Vth = V*R2/(R1+R2)
Rth = R1*R2/(R1+R2)
Assuming grounded emitter and Vth > 0.7V:
Ib = (Vth-0.7V)/Rth
Ic <= beta_sat*Ib (compliance)

Thanks, Greg, and Jon. And Jon you are correct, that is what I needed to
know, I just didn't know the name of it then. I think that is going to help
make a lot of sense out of this.

Jon
 
P

Phil Allison

Jan 1, 1970
0
"Jon Danniken"
Thanks, Greg, and Jon. And Jon you are correct, that is what I needed to
know, I just didn't know the name of it then. I think that is going to
help make a lot of sense out of this.


** Until YOU supply a few actual details of your design - NOBODY can
supply any real help.

What is the transistor type ?

What Ic are you trying to switch ?

What is the switching frequency ?

You post is nothing but a STUPID DUMB TROLL without these essential
facts supplied.



..... Phil
 
P

Phil Allison

Jan 1, 1970
0
"Tim Wescott"
Absolutely, positively, wrong.


** Nope - it is correct.

Cos BobW is commenting on the OP's grossly ambiguous schem ( with C and E
uncommitted) which looked like an emitter follower ( rather than a
saturated switch ) to him.





...... Phil
 
J

Jon Danniken

Jan 1, 1970
0
Tim Wescott said:
The base voltage has a mild dependency on current, a moderate dependency
on temperature (so pay attention if you want it to work outside of room
temperature!) and a moderate to strong dependency on the transistor
construction.

For a silicon transistor go with John Larkin's 0.7V.

Thanks, Tim, I appreciate it. I hadn't realized about the temperature
dependency, and will keep it in mind if I'll be in a temperature extreme.

And yes, this is a switch, so I'll be looking to run it with the base
saturated.

BTW, am I correct in understanding that the current needed to saturate the
base is independant upon the collector current?

Jon
 
B

Baron

Jan 1, 1970
0
Jon said:
Thanks Baron, that is appreciated.

Speaking of switching the transistor "hard on", how (in the absence of
load charts for a particular transistor) do I determine a sufficient
base current to do this?

Thanks again,

Jon

Others have answered far better than I ! The data sheet for the
particular device you want to use should be the first point of
reference.
 
B

Baron

Jan 1, 1970
0
Greg said:
Find the Thevenin equivalent of the base bias network
for starters. Then you'll be working with a single
equivalent resistance for setting base current. Once
that's done you can look at the formula for the Thevenin
resistance (how its composed of R1 and R2) to decide on
what values can be chosen for R1 and R2 to set the total
current draw as desired.

I remember many years ago books of "Transistor Bias Tables" ! One for
Germanium and one for Silicon transistors. I vaguely remember that
they contained data for gain(hfe) and temperature along with the
resistor ratios to obtain set currents.
 
J

Jon Danniken

Jan 1, 1970
0
Tim Wescott said:
Saturation happens when the B-C junction starts forward biasing, and that
depends on the collector current, indeed it does. Hold the C-E potential
to 5V and the transistor won't saturate, no matter how much current you
pour into the base. There will be some Really Bad Things happening to the
transistor for a very short while, but it won't saturate.

To do this really right you figure out the current gain when the
transistor is in saturation (it'll be on the data sheet, and it's usually
way less than the best value of current gain), then put in more than
enough base current to provide your desired collector current with that
current gain.

Okay, gotcha, thanks Tim. So if I am reading you correctly, I should using
the minimum Beta for when the transistor is at maximum rated Ic (indeed
given by the specs, and definitely less than the best Beta), and using that
Beta, figure out the Base current for my desired Ic to be switched? I hope
I got that right.

Thanks again for your help,

Jon
 
W

whit3rd

Jan 1, 1970
0
Hi, I am trying to determine the operating specs for operating a transistor
base (as a switch) from a voltage divider.

Your voltage divider and the input signal define a 'load line', and
you
can graph the I/V characteristic of this combination.

The transistor base (with emitter grounded) also has an I/V
characteristic,
which has some perhaps non-negligible temperature dependence.
Graph that.

Overlay the graphs. Where they intersect, is the base current so high
that the transistor will burn up? Is it so low that the transistor
will not
switch on?

Since the input is a 'signal', it has both high and low excursions;
look
at the low excursion and graph the load line for that case. Overlay
the graphs, and determine if the base is reverse biased beyond the
spec sheet limits (usually about -5V). Is the base current low enough
to really switch the transistor OFF in this condition?

The answers to these questions, if satisfactory, will tell you that
the
divider is going to switch the transistor. The exact values of the
components can influence ANY of these multiple tests, there is
no single right combination.

You can eliminate R1 values that, ignoring R2, don't deliver enough
current.
You can eliminate R2 values that would dissipate excessive heat
at transistor turn-on voltages.
 
J

Jon Kirwan

Jan 1, 1970
0
Okay, gotcha, thanks Tim. So if I am reading you correctly, I should using
the minimum Beta for when the transistor is at maximum rated Ic (indeed
given by the specs, and definitely less than the best Beta), and using that
Beta, figure out the Base current for my desired Ic to be switched? I hope
I got that right.
<snip>

Various transistors have various saturation betas, but many datasheets
will use a beta factor of 10 as their assumed "saturation case" partly
because it is a commonly used value (well recognized) and the factor
of 10 works most places. You can go to the datasheet itself to check,
though.

For example, if you look for a spec on the collector-emitter voltage
(labeled as Vcesat, often), you will see that they often specify it
with an Ic that is 10 times the Ib. That's a clue that you may want
to use that value. If you see it specified with an IC that is 5 times
the Ib, you should take that as a clue to use 5, not 10. For most
small signal BJTs I've looked at (not so many, really), 10 is the
common factor used. But if you examine, for example, the datasheet
from OnSemi for the 2N3055, you will see two entries -- one for a beta
of 10 (Ic=4A, Ib=0.4A) and a beta of 3 (Ic=10A, Ib=3.3A). This should
be a clue to think a little more about it.

Beta is a slippery thing and none of this means you must use 10 (or 5)
for a saturation beta. Actually, depending on what you can tolerate
for Vce in your application, you might actually want to use a larger
number for the beta because the Vce you actually get may be quite
fine. So now go over to the "Collector Saturation Region" curves, if
you can find them included in the datasheet. For example, in the
2N2222 sheet from OnSemi I'm looking at right now, there are four
curves for a fixed Ic -- 1ma, 10mA, 150mA, and 500mA. The y-axis will
be Vce and the x-axis will be a log-scale of Ib. I find that at a
Vce=0.2V, which might be acceptable to me, that at an Ic=10ma, Ib=55uA
-- that's a beta of about 180!! So using a beta of 10 might be
overkill. Of course, the curve is for the "typical case" so you need
to keep that in mind, too. So here, I might choose a beta of 30
instead of 10 or 180. Just to be safe, yet not require quite so much
base current.

This can an important thing to think about when a microcontroller is
driving things. The pins may be only able to sink or source some
particular level of milliamps with any guarantee of the output voltage
being in a well-defined range. And you may be forced into using two
BJTs to get the desired Vce when switched on if you assumed a beta of
10 as an absolute rule, when in fact you would find out you can get
away with one BJT if you only had looked at an actual curve.

By the way, on that curve of Ic=10mA, a beta of 10 suggests Vce=30mV.
So you can see that using a smaller beta assumption means achieving a
really low (well, low for most of us normal humans) Vce. But
typically only a difference of 170mW between a beta of 180 and 10. You
may not need that kind of extreme difference in Vce.

The datasheet has a lot of stuff and you should spend a little time
familiarizing yourself. Gradually, it will make increasing sense what
to look for and where to look for it.

Also, different manufacturers make different things even when they use
the same number. The 2N2222 comes in Vceo of 40V and 60V, but from
different manufacturers. Different processes and topologies, I guess.
(I'm a hobbyist, not a professional, so I don't need to know exactly
why.) So it helps, in getting a "feel" for the 2N2222 to examine a
variety of datasheets to see where they seem alike and where they seem
different enough to notice.

Jon
 
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