Driving a transistor base from a voltage divider

[Jon Danniken]

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

 

[Baron]

Jon Danniken wrote:

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.

--
Best Regards:
Baron.

 

[John Larkin]

On Sat, 11 Apr 2009 11:57:28 -0700, "Jon Danniken"
<jondanSPAMniken@yaSPAMhoo.com> wrote:

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

Assume the base is +0.7 when on. The current through R1 is
(Vin-0.7)/R1. The current through R2 is 0.7/R2. The difference current
is going into the base.

John

 

[BobW]

"Jon Danniken" <jondanSPAMniken@yaSPAMhoo.com> wrote in message
news:74c7gpF12rg7rU1@mid.individual.net...

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

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 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.

Bob
--
== All google group posts are automatically deleted due to spam ==

 

[Jon Danniken]

"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

 

[Jon Danniken]

"Baron" wrote:

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

 

[BobW]

"Jon Danniken" <jondanSPAMniken@yaSPAMhoo.com> wrote in message
news:74cjnlF11l510U1@mid.individual.net...

"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.

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).

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?

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).

Thanks again,

You're welcome.

Jon

Bob
--
== All google group posts are automatically deleted due to spam ==

 

[Greg Neill]

Jon Danniken wrote:

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

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.

 

[Phil Allison]

"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

 

[Jon Kirwan]

On Sat, 11 Apr 2009 21:32:38 -0400, "Greg Neill"
<gneillRE@MOVEsympatico.ca> wrote:

Jon Danniken wrote:
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

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)

 

[Jon Danniken]

"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

 

[Jon Danniken]

"Jon Kirwan"wrote:

"Greg Neill"> wrote:

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)

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

 

[Phil Allison]

"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

 

[Tim Wescott]

BobW wrote:

"Jon Danniken" <jondanSPAMniken@yaSPAMhoo.com> wrote in message
news:74c7gpF12rg7rU1@mid.individual.net...
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


Jon,

You can't really control the base current. It draws whatever is required to
support the emitter current that is flowing.

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.

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.

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 B-E 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).

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.

Well, at least _that_ part is correct.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Phil Allison]

"Tim Wescott"

BobW wrote:

You can't really control the base current. It draws whatever is required
to support the emitter current that is flowing.

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

 

[Jon Danniken]

"Tim Wescott" wrote:

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

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 B-E 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).

 

[Tim Wescott]

Jon Danniken wrote:

"Tim Wescott" wrote:
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?

Uhh -- no.

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.

(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
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Baron]

Jon Danniken wrote:

"Baron" wrote:

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

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.

--
Best Regards:
Baron.

 

[Baron]

Greg Neill wrote:

Jon Danniken wrote:
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

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.

--
Best Regards:
Baron.

 

[Jon Danniken]

"Tim Wescott" wrote:

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