# Transistors - making me ill

Discussion in 'Electronic Basics' started by Jon, Jan 28, 2004.

1. ### JonGuest

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

2. ### Tim DicusGuest

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?
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.
You're welcome!

Tim

3. ### John PopelishGuest

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

4. ### Robert C MonsenGuest

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.
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.
Many LEDs will light with 0.5mA

I then adjusted the current to the Base
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

5. ### JonGuest

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

6. ### JonGuest

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

current gain". High gain at low current. Low gain at high
gain of about 30. What was the current through the LED again?
usually they do have a small amount of resistance for
the emitter and ground, that would change the bias on the
difference may have caused the reduction in current.
simultaneously. One to measure the collector curent, and the

7. ### JonGuest

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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

8. ### John LarkinGuest

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.

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

9. ### Tim DicusGuest

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.
No problem!

Tim

10. ### Keith R. WilliamsGuest

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. ;-)
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.

11. ### John PopelishGuest

Under that condition, the transistor is acting less like a switch and
more like a current regulator, wasting whatever voltage is necessary
to limit the flow of collector current. It is just a different
application of a transistor.
That means that the transistor is dangerously close to changing from a
switch (providing whatever current is required at a nearly zero
voltage drop) to a current regulator. It is fine for an experiment,
but bad form for a design, where the big variation in current gain
between one device and another and even the small changes in gain
caused by temperature and resistor aging might seriously alter the
voltage drop collector to emitter. It is just not a stable switch.

12. ### John LarkinGuest

5 volts on the base in the forward direction (positive for an NPN)
would surely fry the transistor. Vebo is the max reverse voltage.

Vebo means Voltage, emitter-to-base, with collector open. "Emitter to
base" is the reverse of base-to-emitter, so positive Vebo means the
base is negative relative to the emitter! That's just a secret
shorthand notation to keep competition down.
That's right. It's unlikely that many transistors would ever actually
need 1.8 volts Vbe, except maybe a monster power transistor conducting
lots of amps.

John

13. ### Tim DicusGuest

I am sure you meant 1ma to the base and attempt 200ma through the collector-emitter. That would be 200 times the current. My
previous answer was based on the percentage.

I will also assume you meant connect a resistor from V+ to the collector that would pass 200ma if connected directly from V+ to the
emitter/ground.

Since you are putting 1ma through the base-emitter, the collector-emitter current should be a little over 100ma according to the
chart. We'll say 100ma as a nice round number.

In this case, the voltage across the resistor and the collector-emitter voltage should be the same (V+/2).

Tim

14. ### Tim DicusGuest

You are correct. Base voltage below the emitter.

What I was refering to was the phrase "base-emitter saturation voltage". I should have been more specific. Are you familiar with
that rating?

Tim

15. ### Robert C MonsenGuest

You are right; the voltage at the base is influenced by the base current
into the transistor. Thats why we chose voltage divider resistances that
were fairly small.

Since the current into the base is small compared to the current going down
the bottom leg of the divider, this effect is very small. You can compute
it, but most of the time its unimportant and can be neglected.
view.

In order to predict exactly where the base voltage will sit, you should use
a better model of the transistor. Do a google search for ebers-moll, there
should be some good references. Typically, though, you don't need to predict
things this precisely.
Well, whats going on is that the thing you positively know is that the
emitter will be 0.7v below the base. However, you can't predict where the
collector voltage will be unless you know the collector resistance and
current. Thus, if you set the base voltage at a known value (by using that
voltage divider) you know where the emitter voltage will be. At 0.7V below
that value! By ohms law, Ve/Re = the current that can flow through the
emitter. The current through the base can be neglected, as we said before.
Thus, the collector current is essentially equal to the emitter current, and
is Ve/Re = (Vb - 0.7)/Re. Thus, knowing both Vb and Re, you can predict the
collector current.

Using a collector resistor, you can limit the current through the
transistor; however, you can't predict what it will be, you can simply set
the upper bound.
No problem. Its fun to help out a fellow explorer...

Regards,
Bob Monsen

16. ### JonGuest

I've been trying to analyze and understand a circuit for a voltage regulator
built from discrete components (www.rason.org/Projects/discreg/discreg.htm).
It is built without any ICs, using various transistors, resistors, and
capacitors.

Does anyone have some suggestions for how to approach this?

Jon

17. ### JeffMGuest

trying to analyze...a circuit for a voltage regulator
Give it a try. Break it down. What do you recognize?
Voltage divider? Darlington? Current sensing resistor?
Feedback path?

18. ### Keith R. WilliamsGuest

Nope, I'm not familliar with that rating. Can you point me to a spec
sheet or some context where this is used?

19. ### Tim DicusGuest

I would be happy to do that for ya !^)

http://www.fairchildsemi.com/ds/TI/TIP31.pdf

Check under "Electrical Characteristics" near the bottom.

Tim

20. ### JonGuest

I'm going to try again tonight. I'm thinking that I'll start with the
voltage divider and assume that nothing else is in the circuit. Then, I'll
start adding one component at a time and think through how it affects the
whole circuit.

Good plan?

Jon