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 Posted on
 Jon Danniken
April 11, 2009, 6:57 pm
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
((E0.7)/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
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
((E0.7)/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
Re: Driving a transistor base from a voltage divider
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.

Best Regards:
Baron.
Best Regards:
Baron.
Re: Driving a transistor base from a voltage divider
"Jon Danniken"
** 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 baseemitter 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 noncritical applications, R2 can be between 100 ohms and 1kohms.
..... Phil
Re: Driving a transistor base from a voltage divider
Jon,
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 R1R2 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.
Bob

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Re: Driving a transistor base from a voltage divider
"BobW"
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 shortcircuit current as whatever I need to get sufficient base current
to switch the transistor on?
Thanks again,
Jon
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 shortcircuit current as whatever I need to get sufficient base current
to switch the transistor on?
Thanks again,
Jon
Re: Driving a transistor base from a voltage divider
In a linear power supply (or any type of power supply), its output voltage
and current are independant quantities. For a constant voltage power supply,
the output current is only determined by the load that's connected to that
power supply. If the load is a resistor then the output voltage and current
will be related (I=V/R), but it's not that simple for other types of loads
(like the base of a transistor).
I'm not sure it's helpful to think of voltage and current as being
opposites, in this case. Generally speaking, however, the output voltage
delivered will decrease as the output current increases. However, it's best
to try and understand each circuit on its own.
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).
You're welcome.
Bob

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Re: Driving a transistor base from a voltage divider
Absolutely, positively, wrong.
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.
Absolutely, positively, unnecessary (but at least it's not wrong).
Yes, the base voltage will vary somewhat, and you'll have to take this
into account if you want the thing to work over temperature and
manufacturing variations. But you just can't trust a transistor to
maintain it's BE voltage well enough that measuring the base current
would do you any good at all. Instead, you need to design your circuit
so that you always have sufficient base current (I'm assuming the OP
wants it to be saturated, otherwise he needs to post again and ask for
clarification).
Well, at least _that_ part is correct.

Tim Wescott
Wescott Design Services
Tim Wescott
Wescott Design Services
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Re: Driving a transistor base from a voltage divider
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
Re: Driving a transistor base from a voltage divider
Uhh  no.
Saturation happens when the BC junction starts forward biasing, and
that depends on the collector current, indeed it does. Hold the CE
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.
(By the way  you'll sometimes see circuits for saturated switches that
leave off the resistor to ground. This is bad both because it may not
provide a positive "off" to the transistor, and because a saturated
transistor turns off faster if there's something pulling charge out of
the base. If the resistor is hooked to something that'll sink current,
like a CMOS logic output, it's far less bad, but still not as fast as
your circuit).

Tim Wescott
Wescott Design Services
Tim Wescott
Wescott Design Services
We've slightly trimmed the long signature. Click to see the full one.
Re: Driving a transistor base from a voltage divider
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
Re: Driving a transistor base from a voltage divider
On Sun, 12 Apr 2009 08:26:40 0700, "Jon Danniken"
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 collectoremitter 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 (Ic4A%, Ib=0.4A) and a beta of 3 (Ic10%A, 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 yaxis will
be Vce and the xaxis will be a logscale of Ib. I find that at a
Vce=0.2V, which might be acceptable to me, that at an Ic10%ma, Ib55%uA
 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 welldefined 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 Ic10%mA, a beta of 10 suggests Vce30%mV.
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
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 collectoremitter 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 (Ic4A%, Ib=0.4A) and a beta of 3 (Ic10%A, 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 yaxis will
be Vce and the xaxis will be a logscale of Ib. I find that at a
Vce=0.2V, which might be acceptable to me, that at an Ic10%ma, Ib55%uA
 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 welldefined 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 Ic10%mA, a beta of 10 suggests Vce30%mV.
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
Re: Driving a transistor base from a voltage divider
Wrong? Unless the transistor is in saturation then it is absolutely,
positively, correct.
Now, if the transistor is in its active region and if you drive the base
with a current source then you've got control of its base current. However,
this is not a common situation.
Bob

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Re: Driving a transistor base from a voltage divider
transistor
given
that
the
point
transistor
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.
Re: Driving a transistor base from a voltage divider
On Sat, 11 Apr 2009 21:32:38 0400, "Greg Neill"
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 = (Vth0.7V)/Rth
Ic <= beta_sat*Ib (compliance)
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 = (Vth0.7V)/Rth
Ic <= beta_sat*Ib (compliance)
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