Transistors - making me ill

I've spent the past few evenings, and part of the days, too, learning about
transistors. I've surfed the web for information, have read and re-read my
"Basic Electronics" book, and have purchased a TIP31 NPN transistor from
Radio Shack to play around with. I'm getting there, but am a bit lost.

I've been experimenting with using the transistor as an on/off switch. I've
hooked up an LED, set my voltage to 5.3V, and added enough resistance to get
about 15 mA through it. I then connected the negative lead of the LED to
the Collector pin of the transistor.

My first question was how much current/voltage to send to the Base pin. I
found a tutorial that suggested I divide my LED current by the minimum hFE,
and add about 30%. My transistor is labeled 10-50 for the hFE, so I divided
my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
realized that I could use way less current because I should be able to push
50X the Base current through the Collector. In fact, it seems to work fine
with way less current. What is the proper formula for determining current
to the Base pin?

My second question has to do with saturation. To test when the transistor
would shut off, I added a pot between the Base pin and ground, and varied
the resistance to bleed current away from the Base pin. This is where
things started to get strange. First, even with 0.01 mA flowing to the Base
pin, (the lower limit of what my digital MM can measure), the LED lit,
though barely. I guess that is OK. I then adjusted the current to the Base
pin so that the LED just barely fully illuminated. I measured the current
through the LED, and found it to be just over 14mA. I then measured the
current from the Emitter pin to ground, and it was just over 8 mA. Where
did the 6 mA difference go? The current to the Base pin was less than 1 mA.
When I reset the current to the base pin higher, to a point where I get the
full 15 mA through the LED, the current from the Emitter to ground is always
higher than the current through the LED, as I think it should be. I most
certainly did not have enough current through the base pin to saturate the
transistor. But where did that 6mA of current go?

Thanks!

Jon

"Jon" <(E-Mail Removed)> wrote in message news:2IGRb.124308$(E-Mail Removed)...
> I've spent the past few evenings, and part of the days, too, learning about
> transistors. I've surfed the web for information, have read and re-read my
> "Basic Electronics" book, and have purchased a TIP31 NPN transistor from
> Radio Shack to play around with. I'm getting there, but am a bit lost.
>
> I've been experimenting with using the transistor as an on/off switch. I've
> hooked up an LED, set my voltage to 5.3V, and added enough resistance to get
> about 15 mA through it. I then connected the negative lead of the LED to
> the Collector pin of the transistor.
>
> My first question was how much current/voltage to send to the Base pin. I
> found a tutorial that suggested I divide my LED current by the minimum hFE,
> and add about 30%. My transistor is labeled 10-50 for the hFE, so I divided
> my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
> realized that I could use way less current because I should be able to push
> 50X the Base current through the Collector. In fact, it seems to work fine
> with way less current. What is the proper formula for determining current
> to the Base pin?

It isn't a formula that you need. It is a datasheet:
http://www.fairchildsemi.com/ds/TI/TIP31.pdf

Under "Typical Characteristics", take a look at the chart labeled "DC current gain". High gain at low current. Low gain at high
current. Up to about an amp, it has a gain of 100. At 5 amps, it has a gain of about 30. What was the current through the LED again?

>
> My second question has to do with saturation. To test when the transistor
> would shut off, I added a pot between the Base pin and ground, and varied
> the resistance to bleed current away from the Base pin. This is where
> things started to get strange. First, even with 0.01 mA flowing to the Base
> pin, (the lower limit of what my digital MM can measure), the LED lit,
> though barely. I guess that is OK. I then adjusted the current to the Base
> pin so that the LED just barely fully illuminated. I measured the current
> through the LED, and found it to be just over 14mA. I then measured the
> current from the Emitter pin to ground, and it was just over 8 mA. Where
> did the 6 mA difference go? The current to the Base pin was less than 1 mA.
> When I reset the current to the base pin higher, to a point where I get the
> full 15 mA through the LED, the current from the Emitter to ground is always
> higher than the current through the LED, as I think it should be. I most
> certainly did not have enough current through the base pin to saturate the
> transistor. But where did that 6mA of current go?
>

Maybe it didn't go anywhere. I am not familiar with your voltmeter, but usually they do have a small amount of resistance for
current measurements. If it caused even a small voltage difference between the emitter and ground, that would change the bias on the
base by raising the voltage level there the same amount. That new voltage difference may have caused the reduction in current.

If you know somebody with a voltmeter, try connecting both voltmeters simultaneously. One to measure the collector curent, and the
other to measure the emitter current.

> Thanks!
>
> Jon
>

You're welcome!

Tim


>
>
>

Jon wrote:
>
> I've spent the past few evenings, and part of the days, too, learning about
> transistors. I've surfed the web for information, have read and re-read my
> "Basic Electronics" book, and have purchased a TIP31 NPN transistor from
> Radio Shack to play around with. I'm getting there, but am a bit lost.
>
> I've been experimenting with using the transistor as an on/off switch. I've
> hooked up an LED, set my voltage to 5.3V, and added enough resistance to get
> about 15 mA through it. I then connected the negative lead of the LED to
> the Collector pin of the transistor.
>
> My first question was how much current/voltage to send to the Base pin. I
> found a tutorial that suggested I divide my LED current by the minimum hFE,
> and add about 30%. My transistor is labeled 10-50 for the hFE, so I divided
> my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
> realized that I could use way less current because I should be able to push
> 50X the Base current through the Collector. In fact, it seems to work fine
> with way less current. What is the proper formula for determining current
> to the Base pin?

It depends on whether you are trying to get a particular transistor to
work as desired, or to design a circuit that will work with any
transistor that meets the full range of specifications on the data
sheet. They are not all alike.

> My second question has to do with saturation. To test when the transistor
> would shut off, I added a pot between the Base pin and ground, and varied
> the resistance to bleed current away from the Base pin. This is where
> things started to get strange. First, even with 0.01 mA flowing to the Base
> pin, (the lower limit of what my digital MM can measure), the LED lit,
> though barely. I guess that is OK.

Especially if your particular example of this transistor has a current
gain of 50. .01 ma times 50 is .5 ma, definitely enough current to
visibly light almost any LED.

> I then adjusted the current to the Base
> pin so that the LED just barely fully illuminated. I measured the current
> through the LED, and found it to be just over 14mA. I then measured the
> current from the Emitter pin to ground, and it was just over 8 mA. Where
> did the 6 mA difference go? The current to the Base pin was less than 1 mA.

The milliamp meter has enough resistance that when you inserted it
into the emitter path, it reduced the total current passing through
the transistor, mostly by raising the emitter voltage, making the base
look less positive in comparison. To do this right, you need to have
several meters all in the circuit at the same time, so conditions are
all the same while each measurement is taken.

> When I reset the current to the base pin higher, to a point where I get the
> full 15 mA through the LED, the current from the Emitter to ground is always
> higher than the current through the LED, as I think it should be.

Right. The emitter must pass both the collector current (that passes
through the LED) and also the base current.

> I most
> certainly did not have enough current through the base pin to saturate the
> transistor. But where did that 6mA of current go?

A transistor is generally assumed to be in saturation when the
collector voltage is less than the base voltage. Next time, measure
the base current that is needed to drop the collector voltage to just
about equal to the base voltage. Again, it is handy to have two
meters, so you don't have to deal with how relocating one meter
changes the situation.

Some variations of your circuit are a lot less sensitive to having the
amp meter put in series. For instance, using a very high resistance
variable resistor between the positive supply and the base with no
resistor base to minus supply rail acts a lot more like a current
source (though it is not a very stable way to bias a transistor).
This approach will not change the base current significantly if the
emitter voltage changes a few millivolts or you insert the milliamp
meter in series with this resistor to measure the base current. Just
be careful you never run the resistance all the way to zero, or you
will burn out the transistor with excess base current. A good way to
do this is to put a minimum fixed resistance in series with the
variable one. This way you can measure a current and then switch to a
voltage range and check the base to emitter voltage and the collector
to emitter voltage or vice versa.

--
John Popelish

"Jon" <(E-Mail Removed)> wrote in message
news:2IGRb.124308$(E-Mail Removed)...
> I've spent the past few evenings, and part of the days, too, learning
about
> transistors. I've surfed the web for information, have read and re-read
my

Ah, less than a year ago I was totally baffled by transistors! They don't
behave like you initially expect them to behave. If you work with them a
bit, however, they become much easier to understand.

> "Basic Electronics" book, and have purchased a TIP31 NPN transistor from
> Radio Shack to play around with. I'm getting there, but am a bit lost.
>
> I've been experimenting with using the transistor as an on/off switch.
I've
> hooked up an LED, set my voltage to 5.3V, and added enough resistance to
get
> about 15 mA through it. I then connected the negative lead of the LED to
> the Collector pin of the transistor.
>
> My first question was how much current/voltage to send to the Base pin. I
> found a tutorial that suggested I divide my LED current by the minimum
hFE,
> and add about 30%. My transistor is labeled 10-50 for the hFE, so I
divided
> my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
> realized that I could use way less current because I should be able to
push
> 50X the Base current through the Collector. In fact, it seems to work
fine
> with way less current. What is the proper formula for determining current
> to the Base pin?

Don't think about it as current into the base pin. Think about it as setting
the voltage on the base pin relative to the emitter pin. Thats called
'biasing' the transistor. Since beta varies so widely, its difficult to
predict. There is a better way.

Assuming you want a constant 10mA through the LED, here are the steps.

First, figure out a resistance that will give you 1V with 10mA. This is

R = 1/0.010 = 100

Call this your emitter resistor Re

2) Set the voltage at the base to 1.7V, using a resistive voltage divider.
The emitter is always around 0.7V below the base, so that will force the
emitter to be at 1V. You do this by 'biasing' the base using a voltage
divider:

Vcc ---/\/\/\---+---/\/\/\--- Gnd
R1 | R2
Vb

The formula for using a voltage divider is

R2
Vb = Vcc --------
R1 + R2

Make the smaller resistor (the bottom one) at most 5x the emitter resistor
Re above, so the current flowing into the base doesn't cause the base
voltage to change much. Thus:

R2 = 470 (a standard value), and assuming Vcc = 5V

1.7 470
--- = ---
5 470+ R1

So R1 = 912. Pick 820 (another standard value). Then, our transistor current
limit looks like this (view with font=courier):

5V
+----+----+
| |
.-. |
R1 | |820 V LED
| | -
'-' |
| b |/ collector
|-------|
| |> emitter
| |
| |
.-. .-.
R2 | |470 | |Re=100
| | | |
'-' '-'
| |
| |
+----+----+
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

I substituted standard resistor values. This circuit will pass ~10mA through
the LED very reliably. In fact, it doesn't matter if the LED is there. If
you put the positive probe of your multimeter on the 5V supply, and the
negative on the collector, and put it into current measurement mode, it'll
measure 10mA. Your multimeter is effectively a short circuit from collector
to 5V. This circuit is called a current source (actually a sink), because it
sinks a predictable current into the collector no matter what the load
resistance (until the resistance of the load gets large enough to drop the
voltage to near 1.2V. Thats called 'saturation', and the current will
decrease.)

If you want to vary the current through the LED, and thus the brightness,
then its easy to see that the most accurate way is to vary the resistance of
the 100 ohm resistor. Since we've set the voltage at the emitter to be 1V,
then varying Re between 500 and 50 ohms gives us a range of 2mA to 20mA
through the transistor, and thus the LED. A 1k trimmer wheel would work
nicely; however, make sure that there is at least 50 ohms, or you'll burn
out your LED (since with R < 50 ohms, I > 20mA). I'd put a 47 ohm resistor
in series with the trimmer to prevent this, since most LEDs will fry with
much more than 20mA.

This construction is done using a single rule: that the voltage drop between
the base and the emitter is about 0.7V. This is nearly true over a wide
range of currents, because current through a BJT varies with e^V. Thus,
voltage varies much more slowly than current, and consequently current can
vary over a wide range when voltage is 'near' its nominal value of 0.7V.

>
> My second question has to do with saturation. To test when the transistor
> would shut off, I added a pot between the Base pin and ground, and varied
> the resistance to bleed current away from the Base pin. This is where
> things started to get strange. First, even with 0.01 mA flowing to the
Base
> pin, (the lower limit of what my digital MM can measure), the LED lit,
> though barely. I guess that is OK.

Many LEDs will light with 0.5mA

I then adjusted the current to the Base
> pin so that the LED just barely fully illuminated. I measured the current
> through the LED, and found it to be just over 14mA.

I think you measured wrong. You measure current by putting your multimeter
in series with the current you are measuring, and selecting current.

Current + >>>>>> + + >>>>>>>>> -
| |
+- MM -+

I then measured the
> current from the Emitter pin to ground, and it was just over 8 mA. Where
> did the 6 mA difference go? The current to the Base pin was less than 1
mA.
> When I reset the current to the base pin higher, to a point where I get
the
> full 15 mA through the LED, the current from the Emitter to ground is
always
> higher than the current through the LED, as I think it should be. I most
> certainly did not have enough current through the base pin to saturate the
> transistor. But where did that 6mA of current go?
>
> Thanks!
>
> Jon
>
>
>
>

Wow! Thanks for the lesson, Robert.

I set the circuit up as you described, only my smallest resistor was a 220
ohm. I used that as my emitter resistor (Re). In order to get 10mA, I
upped the voltage through Re to 2.0V, and the Vb to 2.7V. I set R1 and R2
with potentiometers.

Everything seems to work, though my measured numbers are slightly off from
my predicted numbers. I measure a voltage drop of 0.576V between Vb and Ve.
My Vb is 2.687V,and my Ve is around 2.11V. No big deal, but I'm not sure
whether that is measurement error, or something else.

Specifically, I'm having a problem analyzing transistor effects in circuits.
When I tap off the 2.7V from the voltage divider, does that not change the
effective resistance, and therefore the voltage drop across R1? How would I
calculate the predicted current through R2, and through the transistor base?
I know to do this if I have a resistance between the voltage divider and
ground, but I cannot seem to translate the voltage to the transistor into an
effective resistance.

I know that the effect is quite small in this case, but I'm really trying to
understand transistors from both a theoretical and practical point of view.

I also have a question about putting the 220 ohm current limiting resistor
on the emitter side rather than the collector side. I've racked my brain,
and just cannot seem to understand the effect this has on the circuit. Does
it also serve to control the current that is pulled through the base?

Thanks a million for your help on this!

-Jonathan



"Robert C Monsen" <(E-Mail Removed)> wrote in message
news:HKIRb.132185$sv6.704027@attbi_s52...
> "Jon" <(E-Mail Removed)> wrote in message
> news:2IGRb.124308$(E-Mail Removed)...
> > I've spent the past few evenings, and part of the days, too, learning
> about
> > transistors. I've surfed the web for information, have read and re-read
> my
>
> Ah, less than a year ago I was totally baffled by transistors! They don't
> behave like you initially expect them to behave. If you work with them a
> bit, however, they become much easier to understand.
>
> > "Basic Electronics" book, and have purchased a TIP31 NPN transistor from
> > Radio Shack to play around with. I'm getting there, but am a bit lost.
> >
> > I've been experimenting with using the transistor as an on/off switch.
> I've
> > hooked up an LED, set my voltage to 5.3V, and added enough resistance to
> get
> > about 15 mA through it. I then connected the negative lead of the LED
to
> > the Collector pin of the transistor.
> >
> > My first question was how much current/voltage to send to the Base pin.
I
> > found a tutorial that suggested I divide my LED current by the minimum
> hFE,
> > and add about 30%. My transistor is labeled 10-50 for the hFE, so I
> divided
> > my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
> > realized that I could use way less current because I should be able to
> push
> > 50X the Base current through the Collector. In fact, it seems to work
> fine
> > with way less current. What is the proper formula for determining
current
> > to the Base pin?
>
> Don't think about it as current into the base pin. Think about it as
setting
> the voltage on the base pin relative to the emitter pin. Thats called
> 'biasing' the transistor. Since beta varies so widely, its difficult to
> predict. There is a better way.
>
> Assuming you want a constant 10mA through the LED, here are the steps.
>
> First, figure out a resistance that will give you 1V with 10mA. This is
>
> R = 1/0.010 = 100
>
> Call this your emitter resistor Re
>
> 2) Set the voltage at the base to 1.7V, using a resistive voltage divider.
> The emitter is always around 0.7V below the base, so that will force the
> emitter to be at 1V. You do this by 'biasing' the base using a voltage
> divider:
>
> Vcc ---/\/\/\---+---/\/\/\--- Gnd
> R1 | R2
> Vb
>
> The formula for using a voltage divider is
>
> R2
> Vb = Vcc --------
> R1 + R2
>
> Make the smaller resistor (the bottom one) at most 5x the emitter resistor
> Re above, so the current flowing into the base doesn't cause the base
> voltage to change much. Thus:
>
> R2 = 470 (a standard value), and assuming Vcc = 5V
>
> 1.7 470
> --- = ---
> 5 470+ R1
>
> So R1 = 912. Pick 820 (another standard value). Then, our transistor
current
> limit looks like this (view with font=courier):
>
> 5V
> +----+----+
> | |
> .-. |
> R1 | |820 V LED
> | | -
> '-' |
> | b |/ collector
> |-------|
> | |> emitter
> | |
> | |
> .-. .-.
> R2 | |470 | |Re=100
> | | | |
> '-' '-'
> | |
> | |
> +----+----+
> GND
> created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
>
> I substituted standard resistor values. This circuit will pass ~10mA
through
> the LED very reliably. In fact, it doesn't matter if the LED is there. If
> you put the positive probe of your multimeter on the 5V supply, and the
> negative on the collector, and put it into current measurement mode, it'll
> measure 10mA. Your multimeter is effectively a short circuit from
collector
> to 5V. This circuit is called a current source (actually a sink), because
it
> sinks a predictable current into the collector no matter what the load
> resistance (until the resistance of the load gets large enough to drop the
> voltage to near 1.2V. Thats called 'saturation', and the current will
> decrease.)
>
> If you want to vary the current through the LED, and thus the brightness,
> then its easy to see that the most accurate way is to vary the resistance
of
> the 100 ohm resistor. Since we've set the voltage at the emitter to be 1V,
> then varying Re between 500 and 50 ohms gives us a range of 2mA to 20mA
> through the transistor, and thus the LED. A 1k trimmer wheel would work
> nicely; however, make sure that there is at least 50 ohms, or you'll burn
> out your LED (since with R < 50 ohms, I > 20mA). I'd put a 47 ohm resistor
> in series with the trimmer to prevent this, since most LEDs will fry with
> much more than 20mA.
>
> This construction is done using a single rule: that the voltage drop
between
> the base and the emitter is about 0.7V. This is nearly true over a wide
> range of currents, because current through a BJT varies with e^V. Thus,
> voltage varies much more slowly than current, and consequently current can
> vary over a wide range when voltage is 'near' its nominal value of 0.7V.
>
> >
> > My second question has to do with saturation. To test when the
transistor
> > would shut off, I added a pot between the Base pin and ground, and
varied
> > the resistance to bleed current away from the Base pin. This is where
> > things started to get strange. First, even with 0.01 mA flowing to the
> Base
> > pin, (the lower limit of what my digital MM can measure), the LED lit,
> > though barely. I guess that is OK.
>
> Many LEDs will light with 0.5mA
>
> I then adjusted the current to the Base
> > pin so that the LED just barely fully illuminated. I measured the
current
> > through the LED, and found it to be just over 14mA.
>
> I think you measured wrong. You measure current by putting your multimeter
> in series with the current you are measuring, and selecting current.
>
> Current + >>>>>> + + >>>>>>>>> -
> | |
> +- MM -+
>
> I then measured the
> > current from the Emitter pin to ground, and it was just over 8 mA.
Where
> > did the 6 mA difference go? The current to the Base pin was less than 1
> mA.
> > When I reset the current to the base pin higher, to a point where I get
> the
> > full 15 mA through the LED, the current from the Emitter to ground is
> always
> > higher than the current through the LED, as I think it should be. I
most
> > certainly did not have enough current through the base pin to saturate
the
> > transistor. But where did that 6mA of current go?
> >
> > Thanks!
> >
> > Jon
> >
> >
> >
> >
>
>

Thanks for the link, Tim. I had pulled down a similar datasheet, but I
guess that I'm just not yet too comfortable reading these things. I cannot
believe that the current gain is over 100. That is pretty remarkable.
These are pretty sensitive little switches!

If I have a maximum current gain of 100, what would happen if I tried to
pass a current that is 200 times greater than the base current? Would the
transistor simply limit it to 100 times? For example, let's say I'm
supplying 10mA to the base, but I try to pass 200ma through the
collector-emitter. Would the transistor simply limit the current to the
value on the DC current gain chart?

The datasheet shows a maximum Vebo of 5V for this transistor. I read that
that is the break-down voltage between the base and the emitter, but I'm not
too clear on what that means. I also see that the base-emitter saturation
voltage is 1.8V max, and I'm not too clear on that, either.

Thanks for you help!

Jon



"Tim Dicus" <(E-Mail Removed)> wrote in message
news:64HRb.2291$CJ1.488@lakeread01...
> "Jon" <(E-Mail Removed)> wrote in message
news:2IGRb.124308$(E-Mail Removed)...
> > I've spent the past few evenings, and part of the days, too, learning
about
> > transistors. I've surfed the web for information, have read and re-read
my
> > "Basic Electronics" book, and have purchased a TIP31 NPN transistor from
> > Radio Shack to play around with. I'm getting there, but am a bit lost.
> >
> > I've been experimenting with using the transistor as an on/off switch.
I've
> > hooked up an LED, set my voltage to 5.3V, and added enough resistance to
get
> > about 15 mA through it. I then connected the negative lead of the LED
to
> > the Collector pin of the transistor.
> >
> > My first question was how much current/voltage to send to the Base pin.
I
> > found a tutorial that suggested I divide my LED current by the minimum
hFE,
> > and add about 30%. My transistor is labeled 10-50 for the hFE, so I
divided
> > my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
> > realized that I could use way less current because I should be able to
push
> > 50X the Base current through the Collector. In fact, it seems to work
fine
> > with way less current. What is the proper formula for determining
current
> > to the Base pin?
>
> It isn't a formula that you need. It is a datasheet:
> http://www.fairchildsemi.com/ds/TI/TIP31.pdf
>
> Under "Typical Characteristics", take a look at the chart labeled "DC
current gain". High gain at low current. Low gain at high
> current. Up to about an amp, it has a gain of 100. At 5 amps, it has a
gain of about 30. What was the current through the LED again?
>
> >
> > My second question has to do with saturation. To test when the
transistor
> > would shut off, I added a pot between the Base pin and ground, and
varied
> > the resistance to bleed current away from the Base pin. This is where
> > things started to get strange. First, even with 0.01 mA flowing to the
Base
> > pin, (the lower limit of what my digital MM can measure), the LED lit,
> > though barely. I guess that is OK. I then adjusted the current to the
Base
> > pin so that the LED just barely fully illuminated. I measured the
current
> > through the LED, and found it to be just over 14mA. I then measured the
> > current from the Emitter pin to ground, and it was just over 8 mA.
Where
> > did the 6 mA difference go? The current to the Base pin was less than 1
mA.
> > When I reset the current to the base pin higher, to a point where I get
the
> > full 15 mA through the LED, the current from the Emitter to ground is
always
> > higher than the current through the LED, as I think it should be. I
most
> > certainly did not have enough current through the base pin to saturate
the
> > transistor. But where did that 6mA of current go?
> >
>
> Maybe it didn't go anywhere. I am not familiar with your voltmeter, but
usually they do have a small amount of resistance for
> current measurements. If it caused even a small voltage difference between
the emitter and ground, that would change the bias on the
> base by raising the voltage level there the same amount. That new voltage
difference may have caused the reduction in current.
>
> If you know somebody with a voltmeter, try connecting both voltmeters
simultaneously. One to measure the collector curent, and the
> other to measure the emitter current.
>
> > Thanks!
> >
> > Jon
> >
>
> You're welcome!
>
> Tim
>
>
> >
> >
> >
>
>

"John Popelish" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
..
..
..
<deleted>
..
..
..

>
> A transistor is generally assumed to be in saturation when the
> collector voltage is less than the base voltage. Next time, measure
> the base current that is needed to drop the collector voltage to just
> about equal to the base voltage. Again, it is handy to have two
> meters, so you don't have to deal with how relocating one meter
> changes the situation.


Thanks for all your responses, John.

If I run the transistor with a collector voltage that is higher than the
base voltage, am I going to run into problems? What will happen? In the
circuit I put together (as per my description in my response to Robert), my
collector voltage is about 0.35V higher than my base voltage.


Jon

On Wed, 28 Jan 2004 18:16:50 GMT, "Jon" <(E-Mail Removed)> wrote:

>Thanks for the link, Tim. I had pulled down a similar datasheet, but I
>guess that I'm just not yet too comfortable reading these things. I cannot
>believe that the current gain is over 100. That is pretty remarkable.
>These are pretty sensitive little switches!
>
>If I have a maximum current gain of 100, what would happen if I tried to
>pass a current that is 200 times greater than the base current? Would the
>transistor simply limit it to 100 times? For example, let's say I'm
>supplying 10mA to the base, but I try to pass 200ma through the
>collector-emitter. Would the transistor simply limit the current to the
>value on the DC current gain chart?
>

Right. The transistor may then get pretty hot, depending on the C-E
voltage. This Ic will vary a lot between transistors, as beta is
usually spec'd over a pretty wide range, part to part.


>The datasheet shows a maximum Vebo of 5V for this transistor. I read that
>that is the break-down voltage between the base and the emitter, but I'm not
>too clear on what that means. I also see that the base-emitter saturation
>voltage is 1.8V max, and I'm not too clear on that, either.

Vebo is the maximum *reverse* (negative for an NPN) voltage you should
apply to the base; that turns the transistor off. If you go more
negative on the base, you will zener the b-e junction, which can
damage the transistor.


John

"Jon" <(E-Mail Removed)> wrote in message news:m0TRb.28982$(E-Mail Removed)...
> Thanks for the link, Tim. I had pulled down a similar datasheet, but I
> guess that I'm just not yet too comfortable reading these things. I cannot
> believe that the current gain is over 100. That is pretty remarkable.
> These are pretty sensitive little switches!
>
> If I have a maximum current gain of 100, what would happen if I tried to
> pass a current that is 200 times greater than the base current? Would the
> transistor simply limit it to 100 times? For example, let's say I'm
> supplying 10mA to the base, but I try to pass 200ma through the
> collector-emitter. Would the transistor simply limit the current to the
> value on the DC current gain chart?
>

Yes.

> The datasheet shows a maximum Vebo of 5V for this transistor. I read that
> that is the break-down voltage between the base and the emitter, but I'm not
> too clear on what that means. I also see that the base-emitter saturation
> voltage is 1.8V max, and I'm not too clear on that, either.
>

That means you can put 5 volts more than the emitter into the base. It would be in serious saturation at that point.

The saturation voltage, as I recall, is the base-emitter voltage at which the transistor junction is saturated at its max
collector-emitter current rating. Please correct me if I am wrong.

> Thanks for you help!
>
> Jon
>

No problem!

Tim


>
>
> "Tim Dicus" <(E-Mail Removed)> wrote in message
> news:64HRb.2291$CJ1.488@lakeread01...
> > "Jon" <(E-Mail Removed)> wrote in message
> news:2IGRb.124308$(E-Mail Removed)...
> > > I've spent the past few evenings, and part of the days, too, learning
> about
> > > transistors. I've surfed the web for information, have read and re-read
> my
> > > "Basic Electronics" book, and have purchased a TIP31 NPN transistor from
> > > Radio Shack to play around with. I'm getting there, but am a bit lost.
> > >
> > > I've been experimenting with using the transistor as an on/off switch.
> I've
> > > hooked up an LED, set my voltage to 5.3V, and added enough resistance to
> get
> > > about 15 mA through it. I then connected the negative lead of the LED
> to
> > > the Collector pin of the transistor.
> > >
> > > My first question was how much current/voltage to send to the Base pin.
> I
> > > found a tutorial that suggested I divide my LED current by the minimum
> hFE,
> > > and add about 30%. My transistor is labeled 10-50 for the hFE, so I
> divided
> > > my 15 mA by 10, added 30%, and came up with 1.95mA. That works, but I
> > > realized that I could use way less current because I should be able to
> push
> > > 50X the Base current through the Collector. In fact, it seems to work
> fine
> > > with way less current. What is the proper formula for determining
> current
> > > to the Base pin?
> >
> > It isn't a formula that you need. It is a datasheet:
> > http://www.fairchildsemi.com/ds/TI/TIP31.pdf
> >
> > Under "Typical Characteristics", take a look at the chart labeled "DC
> current gain". High gain at low current. Low gain at high
> > current. Up to about an amp, it has a gain of 100. At 5 amps, it has a
> gain of about 30. What was the current through the LED again?
> >
> > >
> > > My second question has to do with saturation. To test when the
> transistor
> > > would shut off, I added a pot between the Base pin and ground, and
> varied
> > > the resistance to bleed current away from the Base pin. This is where
> > > things started to get strange. First, even with 0.01 mA flowing to the
> Base
> > > pin, (the lower limit of what my digital MM can measure), the LED lit,
> > > though barely. I guess that is OK. I then adjusted the current to the
> Base
> > > pin so that the LED just barely fully illuminated. I measured the
> current
> > > through the LED, and found it to be just over 14mA. I then measured the
> > > current from the Emitter pin to ground, and it was just over 8 mA.
> Where
> > > did the 6 mA difference go? The current to the Base pin was less than 1
> mA.
> > > When I reset the current to the base pin higher, to a point where I get
> the
> > > full 15 mA through the LED, the current from the Emitter to ground is
> always
> > > higher than the current through the LED, as I think it should be. I
> most
> > > certainly did not have enough current through the base pin to saturate
> the
> > > transistor. But where did that 6mA of current go?
> > >
> >
> > Maybe it didn't go anywhere. I am not familiar with your voltmeter, but
> usually they do have a small amount of resistance for
> > current measurements. If it caused even a small voltage difference between
> the emitter and ground, that would change the bias on the
> > base by raising the voltage level there the same amount. That new voltage
> difference may have caused the reduction in current.
> >
> > If you know somebody with a voltmeter, try connecting both voltmeters
> simultaneously. One to measure the collector curent, and the
> > other to measure the emitter current.
> >
> > > Thanks!
> > >
> > > Jon
> > >
> >
> > You're welcome!
> >
> > Tim
> >
> >
> > >
> > >
> > >
> >
> >
>
>

In article <kTURb.2382$CJ1.347@lakeread01>, (E-Mail Removed) says...
> "Jon" <(E-Mail Removed)> wrote in message news:m0TRb.28982$(E-Mail Removed)...
> > Thanks for the link, Tim. I had pulled down a similar datasheet, but I
> > guess that I'm just not yet too comfortable reading these things. I cannot
> > believe that the current gain is over 100. That is pretty remarkable.
> > These are pretty sensitive little switches!
> >
> > If I have a maximum current gain of 100, what would happen if I tried to
> > pass a current that is 200 times greater than the base current? Would the
> > transistor simply limit it to 100 times? For example, let's say I'm
> > supplying 10mA to the base, but I try to pass 200ma through the
> > collector-emitter. Would the transistor simply limit the current to the
> > value on the DC current gain chart?
> >
>
> Yes.
>
> > The datasheet shows a maximum Vebo of 5V for this transistor. I read that
> > that is the break-down voltage between the base and the emitter, but I'm not
> > too clear on what that means. I also see that the base-emitter saturation
> > voltage is 1.8V max, and I'm not too clear on that, either.
> >
>
> That means you can put 5 volts more than the emitter into the base. It would be in serious saturation at that point.

No, BVebo is the breakdown voltage from the emitter to the base (with
the collector open) with the junction *reverse* biased. In an NPN this
would mean the maximum voltage base can be driven *below* the emitter.

If you drive the base voltage 5V *above* the emitter, you'll let out
the magic smoke. ;-)

> The saturation voltage, as I recall, is the base-emitter voltage at which the transistor junction is saturated at its max
> collector-emitter current rating. Please correct me if I am wrong.

No, a transistor is in saturation when its base-collector junction is
forward biased. The saturation voltage is then the voltage from
collector to emitter, which will be lower than the base-emitter
(I.e. < ~.7V, typically <.4V), since both junctions are forward biased.

--
Keith