Transistors (BJTs) confusion

  • #1
Hi all,

I hope you guys can help me with my confusions here:

1) I don't understand the usage of transistors as switches. Why would you use transistors as switches, and how does that work??

2) Why would you choose a PNP over an NPN transistor. I've been reading this book and I don't understand what this guy says.

http://uploader.ws/upload/200707/bjt.jpg"

any help is greatly appreciated. thanks!
 
Last edited by a moderator:

Answers and Replies

  • #2
xez
174
0
1) BJTs are used as switches because they can perform
a switching function. In a switch one generally has
a control input, and then a switched circuit path connection.
It is often desirable for a switch to take a relatively small
amount of (voltage, current, or power) on their
switch control inputs in order to control a relatively large
amount of the same quantity on their switched circuit
path. Therefore it is useful for a switch control circuit to
have high "sensitivity" or high "amplification" in order to
control a large signal by using a small signal.
Active devices have such "amplification" capability, so
such devices as BJTs, FETs, IGBTs, SCRs, TRIACs,
and various kinds of electron tubes are consequently useful
for switching functions because they permit a small control
signal to have a large effect on the controlled / switched
signal.

2) Well the point that's being made is a little esoteric and
really depends on a set of operating conditions and
concepts that are pretty unique to the circuits described.
Actually I find the way the information is presented to be
a bit inaccurate and confusing in a couple of senses.
I suppose the main point is that it's better for some
concerns to have the switched load on a BJT's collector
rather than in the emitter circuit because the voltage drop
across the load doesn't effect the base/emitter biasing
and similarly the base-emitter voltage doesn't couple
strongly to the collector-emitter voltage. Typically in
a saturated transistor the VBE voltage may well be in the
0.5 to 0.8V range, and the VCE voltage can be quite low
like 0.25, 0.35V range. If you applied only 0.35V to
the base relativeyou might not have enough base current
to saturate the transistor, so you'd end up with more
resistance (lower conductance) in the collector/emitter
circuit, and the collector/emitter voltage drop would
consequently be higher, so you'd have less 'head room'
for your signal and less current on your load. It's really
mainly an issue at quite low collector-emitter voltages.
If you're just trying to understand BJTs, work on
understanding their principles of beta hFE, alpha, IB, IC,
VBE, VCE, and then once you grasp those start analyzing
various circuits that use BJTs as switches, amplifiers,
etc. and then those analyses will make more sense for
you. Perhaps you should get a free circuit simulator
program to help you analyze the way a simple circuit like
the one presented will function across a variety of
operating load resistances, supply voltages, etc. and then
things will be more clear to you qualitatively and
quantitatively.
 
  • #3
xez
174
0
Oh and don't read too much into PNP vs. NPN from that
author's comments... PNP is really the same as NPN in
the sense of both topologies offering parts with comparable
electrical performance characteristics. The only difference
is the polarity of the required biasing and current flows.

In some cases it's preferable to switch the 'low side' of
a circuit by switching between GROUND and a LOAD
connected to some higher voltage.

In other cases it's preferable to switch the 'high side' of
a circuit by switching between the
HIGH VOLTAGE SUPPLY and the load whose other terminal
is grounded.

If your switch control logic is all running off of low voltages,
and you're switching a load connected to some much
higher positive voltage, then it's easier to make a low side
switch with a NPN transistor since its control lead (base)
need only be able to operate in the (0 to +2VDC) range
even though the collector voltage in the off state may
be positive dozens or hundreds of volts.

If you were to try to use a NPN in the same circuit as
a high side switch, the control voltage would have to
be able to go up to very near the maximum positive
supply voltage which could be positive dozens or hundreds
of volts, which is too high for common logic chips to
easily work at directly.

Similarly a PNP works well as a low-side switch for a
load whose other terminal is a negative supply voltage
since then the emitter can be grounded, the base voltage
can operate in the -0 to -2V range, and the collector can
switch negative dozens or negative hundreds of volts
efficiently.

So you see in many cases the only choice criteria
between NPN and PNP is the polarity of the available
circuit voltages for use as control and supply voltages.
 
  • #4
WOW! thanks for the very long explanation. i appreciate you taking your time. that'll take some time for me to digest. i think you're right, i dont have a good grasp on the concept of BJTs just yet. many things u said in #2 still sound like gibberish to me, like this one:

I suppose the main point is that it's better for some
concerns to have the switched load on a BJT's collector
rather than in the emitter circuit because the voltage drop
across the load doesn't effect the base/emitter biasing
and similarly the base-emitter voltage doesn't couple
strongly to the collector-emitter voltage.

why does the voltage drop across the load not affect the base/emitter biasing, and why does the base-emitter voltage not couple strongly to the collector-emitter voltage. im not even sure what biasing and coupling mean... :cry: god i suck...
 
  • #5
xez
174
0
WOW! thanks for the very long explanation. i appreciate you taking your time. that'll take some time for me to digest. i think you're right, i dont have a good grasp on the concept of BJTs just yet. many things u said in #2 still sound like gibberish to me, like this one:



why does the voltage drop across the load not affect the base/emitter biasing, and why does the base-emitter voltage not couple strongly to the collector-emitter voltage. im not even sure what biasing and coupling mean... :cry: god i suck...

You're welcome. Don't worry, it'll all make a lot more
sense after you study a little and start to intuitively
understand the way the circuit voltages and currents
relate.

In a BJT you have:

IB = Base current, which flows through the base, out
the emitter, so the IB flows both in BASE and EMITTER
since that's the path it must take.

IC = Collector current, which also flows in the EMITTER
since the path of the collector current is out through the
emitter.

IE = Emitter current
This is the sum of the base current and collector current,
and is the actual current flowing out the emitter.

VBE = Voltage between base and emitter
VCE = Voltage between collector and emitter
hFE = beta = current gain = IC/IB

The base-to-emitter junction is a diode, and so there
will not be much base current until VBE is starting to
get to around 0.5V or so in the correct polarity.
Between 0.5 and 0.8 VBE (of the appropriate polarity)
there's a very rapid (square law or exponential) increase
in base current with respect to added VBE. So that's
why that other article said it takes ~ 0.7VBE to 'turn on'
and 'saturate' a single BJT since at that level of applied
voltage there's enough base to emitter current flowing
to cause the transistor to conduct very heavily and
essentially 'saturate' where additional VBE or base current
cannot cause appreciably more collector current to flow,
and in this saturated state with high IC flowing,
VCE is at the minimum value it may typically attain.

The BJT is said to be a current operated device since
currents in one path are related to currents in another
path by relatively linear relationships. The voltages are
secondary and obey non-linear relationships and are
basically there to get the currents to flow.

When IB = 0 then IC = 0, and IE=IB+IC=0
and the transistor is 'off', and any load resistance in
the collector circuit sees no current.

A typical BJT might have hFE = Beta = IC/IB = 200,
though certainly some may be 20 and others could be
300, but let's say it's 200 for the example.

When you increase IB to perhaps 10mA, then
IC=IB*Beta = 10mA * 200 = 2000 mA so a small current
IB has switched on a large IC current 200 (Beta) times
its size!

If you put a resistor in series with the emitter the
current flow there will be IE = IC + IB, and of course
that current causes a voltage rise across the emitter
resistor of IE * R.. So if R = 100 ohms and IB=10mA
and IC=(Beta 200) * IB = 2000 mA, the voltage across
that resistor must be V=IR = 2*100 = 200 volts.
But, whoops, the BASE current must flow OUT through
the EMITTER by the BASE being about 0.7V closer to
the collector voltage than the emitter is. But now with
such a current flowing through an resistance in series
with the emitter the VE = 200V, so
VB must equal VE = VE + 0.7 (assuming NPN where
the VBE = +0.7 since in PNP VBE = -0.7 due to the
opposing polarity).
So if you had such an emitter series resistor you'd have
to supply a voltage of nearly 200.7V to the base so
that 2A could flow in your 100 ohm emitter resistor
leaving VE = 200V, and that's a bit inconvenient to have
to supply such a large base voltage in many cases.

If your resistor 100 ohm load was in series with the
collector, and your NPN BJT had the emitter at 0V,
grounded, you would apply VBE = around +0.7V to get
IB=0.01A = 10mA, so VE=0V, VB=+0.7V,
IC=IB*(Beta 200) = 2A, and VC = 0.4V,
IC=2010mA = (IB*Beta+1), and your
collector series resistor has still the 2A running through
it, and the top of the collector resistor must go to
a supply voltage of VCE+IC*R = +200.4V.
This is more convenient since now the VB only has
to be 0V = off, 0.7V = switched on, and VE never
changes from 0V, and VCE in the on state saturates
at only +0.4V, so the transistor doesn't consume much
power though it's switching on/off a 200V load at 2Amps.

The main difference between the cases is that the
base and emitter voltages don't rise due to the voltage
drop across the load resistance if you just leave the load
in the collector circuit and hence is why I said that in
such a case the base and emitter biasing voltages are
not affected by a load in series with the collector whereas
they're very much effected by a large resistive load
were it in series with the emitter since IB must flow
through IE along with IC flowing in IE (IE=IC+IB).

Biasing? Well that's just a term for whatever voltage
or currents you apply to various points on a circuit
under a given operating condition.

Switched off VB=0V, IB=0mA, so 0V is your off-state
VBE bias, and IB=0mA is your off-state base current
bias.

To switch on, bias (cause to exist from some external
circuit connection) VBE=+0.7V for NPN, to cause a
flow of IB through base and out the emitter =10mA.

Check this link, it can be handy to understand some
of the operations:
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans.html
 
  • #6
If you put a resistor in series with the emitter the
current flow there will be IE = IC + IB, and of course
that current causes a voltage rise across the emitter
resistor of IE * R.. So if R = 100 ohms and IB=10mA
and IC=(Beta 200) * IB = 2000 mA, the voltage across
that resistor must be V=IR = 2*100 = 200 volts.
But, whoops, the BASE current must flow OUT through
the EMITTER by the BASE being about 0.7V closer to
the collector voltage than the emitter is. But now with
such a current flowing through an resistance in series
with the emitter the VE = 200V, so
VB must equal VE = VE + 0.7 (assuming NPN where
the VBE = +0.7 since in PNP VBE = -0.7 due to the
opposing polarity).
So if you had such an emitter series resistor you'd have
to supply a voltage of nearly 200.7V to the base so
that 2A could flow in your 100 ohm emitter resistor
leaving VE = 200V, and that's a bit inconvenient to have
to supply such a large base voltage in many cases.

If your resistor 100 ohm load was in series with the
collector, and your NPN BJT had the emitter at 0V,
grounded, you would apply VBE = around +0.7V to get
IB=0.01A = 10mA, so VE=0V, VB=+0.7V,
IC=IB*(Beta 200) = 2A, and VC = 0.4V,
IC=2010mA = (IB*Beta+1), and your
collector series resistor has still the 2A running through
it, and the top of the collector resistor must go to
a supply voltage of VCE+IC*R = +200.4V.
This is more convenient since now the VB only has
to be 0V = off, 0.7V = switched on, and VE never
changes from 0V, and VCE in the on state saturates
at only +0.4V, so the transistor doesn't consume much
power though it's switching on/off a 200V load at 2Amps.

The main difference between the cases is that the
base and emitter voltages don't rise due to the voltage
drop across the load resistance if you just leave the load
in the collector circuit and hence is why I said that in
such a case the base and emitter biasing voltages are
not affected by a load in series with the collector whereas
they're very much effected by a large resistive load
were it in series with the emitter since IB must flow
through IE along with IC flowing in IE (IE=IC+IB).

wow ok this is much clearer now. some bulbs light up in my head.

so, in this example's PNP circuit, VE = VB + 0.7v. Since there is a Resistor in series with the emitter, assuming
R = 100 ohms,
IB = 10 mA,
IC = 10 mA * 200 = 2000 mA,
IE = 2010 mA,

VE = 100 ohms * ~2A = 200 V, and thus
VB = 199.3 V,
and so it is inconvenient to have such a large base voltage.

on the other hand, in the example's NPN circuit, VB = VE + 0.7v
since the emitter is grounded,
VE = 0v
VB = 0.7v, and obviously this is much more convenient than some 200 Volts.
I'm not sure where you got VC = 0.4v tho. Is that an arbitrary number you chose? and also IC = (IB * Beta + 1) ?

thanks very much tho. i think this is much clearer now.
 
  • #7
xez
174
0
wow ok this is much clearer now. some bulbs light up in my head.

so, in this example's PNP circuit, VE = VB + 0.7v. Since there is a Resistor in series with the emitter, assuming
R = 100 ohms,
IB = 10 mA,
IC = 10 mA * 200 = 2000 mA,
IE = 2010 mA,

VE = 100 ohms * ~2A = 200 V, and thus
VB = 199.3 V,
and so it is inconvenient to have such a large base voltage.

on the other hand, in the example's NPN circuit, VB = VE + 0.7v
since the emitter is grounded,
VE = 0v
VB = 0.7v, and obviously this is much more convenient than some 200 Volts.
I'm not sure where you got VC = 0.4v tho. Is that an arbitrary number you chose? and also IC = (IB * Beta + 1) ?

thanks very much tho. i think this is much clearer now.

Yes, that's all about right.

Maybe I made a typographic error.
IC = IB * Beta
IE = IC + IB = IB * (Beta + 1) just since
IE = (IB+Beta) [which is IC] + IB.

VCE_sat ~= 0.4V is just a typically low number
of the VCE differential voltage when a transistor is
fully saturated in a heavily conducting state.
It's not uncommon to see VCE_sat numbers in the
range of 0.2 to 0.5V.

VC on the other hand will of course just be whatever
it's going to be based on the circuit, but the differential
voltage between the collector and emitter (VCE) will
be small when saturated on. When the transistor
is off (IB=0) then IC=0 (or close to it) and the VCE
will presumably have a very large difference as if
the transistor weren't even in the circuit since the
collector is like an open non-conducting circuit in the
off state.

Another useful parameter of BJTs to know is
VBR_CEO which is the breakdown voltage of the
collector-to-emitter when the base is disconnected
(open circuit).. if you applied a voltage higher than that
the transistor's non-conduction would break down
and current would start flowing even though there's
no base current since you've exceeded the maximum
allowed voltage the particular transistor can keep at bay.

Here are a couple of links to actual data sheet
specifications for very common real transistors.
It may be instructive to look at the parameters and
a few of the graphs to get an idea of how the
voltages and currents really perform.

Small signal, fairly high gain general purpose switching
NPN:
http://www.fairchildsemi.com/ds/PN/PN2222.pdf

Small signal, fairly high gain general purpose switching
NPN:
http://www.fairchildsemi.com/ds/PN/PN2907.pdf

A medium power NPN:
http://www.fairchildsemi.com/ds/TI/TIP120.pdf

A medium power PNP:
http://www.fairchildsemi.com/ds/TI/TIP125.pdf

Hopefully those are pretty detailed with graphs
etc. though sometimes they're a bit skimpy on details
depending on the manufacturer.
 
  • #8
xez
174
0
oops another typo:
IE = (IB+Beta) [which is IC] + IB.

Should have read:
IE = (IB*Beta) [which is IC] + IB.
 
  • #9
xez
174
0
Ah too bad; I just looked at the fairchild data sheet
and it's really lacking a lot of data that it could present.
Motorola et. al. had much nicer data sheets for the same
parts once upon a time.

Anyway, I found some other data sheets for other parts
that are a lot more useful for learning how the
voltages and currents and parameters relate:
http://www.zetex.com/3.0/pdf/ZXTN25012EFH.pdf
http://www.zetex.com/3.0/pdf/ZXTP25012EFH.pdf

And others e.g.:
http://www.zetex.com/3.0/3-3-2b.asp?rid=1
http://www.zetex.com/3.0/3-3-2b.asp?rid=2
 
  • #11
xez
174
0
Where's the diagram / schematic of the inverter? I don't see any image, link, drawing, or so on.
 
  • #13
xez
174
0
do you mind explaining this BJT inverter (NOT Gate) to me?

http://www.ee.calpoly.edu/~dbraun/courses/ee307/F03/13/02_13_PhilippeGonzaga.html

When Vin = 5V, how is V3 (or V base) = 0.83V ? How do we know what the current and the voltage drop across that resistor is? and how does the output V4 become 0.019 Volts?

thanks!!

edit: LINK ADDED! sorry my bad...


a) "When Vin = 5V, how is V3 (or V base) = 0.83V ? How do we know what the current and the voltage drop across that resistor is?"
Well there is a resistor in series with the base (RB=10K),
and the base to emitter voltage is given as 0.83V in
this case, and we know that the applied voltage to
the series circuit of base_resistor + base_emitter_diode
is 5V. So since VBE=0.83 then 5-0.83=4.17V must
appear across the 10K resistor in series with the base.
So 4.17V/10K = 417uA of current in both the base
and the base resistor with 4.17 across the resistor and
0.83 across VBE to make the total of 5V which is
supplied to be the base-resistor-emitter circuit.
The reason VBE=0.83V at IB=417uA is just because
"that's what it is" for that particular model of a transistor
device. Different transistor devices (or models of them)
could have somewhat different values of VBE vs IB vs
temperature. Generally they'll all be in the range of
0.4 to 0.9V for reasonable values of base current, though,
for single BJTs. Darlington pairs or unusual semiconductors
(i.e. not Silicon) will have different properties.


b) "how does the output V4 become 0.019 Volts?"
RC = 1000 ohms in series with the collector
because that's what the page says.
VCE_sat=0.019V when IB=417uA, and the page
says that the conduction of the transistor is in the
saturated region. In saturated conduction VCE is
always a 'small number' i.e. pretty much the smallest
possible number for that particular transistor with
that amount of IC flowing.

Since we know that V_supply=5V to the circuit
containing 1000 ohm resistors in series with VCE=0.019V
we realize that 5V (V_supply) = VCE (0.019) + V_resistor
so V_resistor = 5-0.019=4.981V across a 1000 ohm
resistor, so I_resistor = V_resistor/R = 4.981/1000=
4.981mA must be flowing in that collector circuit resistor,
and since it's in series with the collector, 4.981mA
is also IC.

So IE=IC+IB,
and IC=IB*Beta = 417uA * 11.9, so in this transistor
model under those conditions of modeled temperature
and base current, collector current, we see that the
beta is around 12 in saturation which is not unreasonable
since the very definition of saturation is that the collector
circuit is already passing essentially the maximum current
that it could and that no further increase in base current
is capable of producing any significant further increase
in collector current. VCE is already approximately
as small as it can get in saturation mode.

One could keep adding more IB and not get much any
more IC, so the apparent gain goes down during
saturated conduction since you're already applying
'more than enough' base current, and any additional
base current doesn't really increase collector current much.


Let's look at a case where the BJT is said to be
in forward active mode near the page bottom:
RC=1000
RB=10000
V_base_supply=0.7550,
VBE=0.7501
V_RB=V_base_supply-VBE=0.7550-0.7501=4.9mV
I_RB=V_RB/RB=4.9mV/10k=0.49uA
IB=I_RB since RB and IB are in series.
V_collector_supply=5
VCE=1.0660
V_RC=V_collector_supply-VCE=5-1.066=3.934
I_RC=V_RC/RC=3.934/1000=3.934mA
IC=I_RC=3.934mA since RC and IC are in series.
Beta=IC/IB=3.934mA/0.49uA=8028
This seems suspicious since Beta is about 100x higher
than it ought to be. Maybe there's a typographical error somewhere.


FYI in the listed PSPICE model for a BJT 'BF' is the forward
gain for the transistor model being defined, and in
the web page you linked, BF=80, so that should
approximately equal beta during the
forward region of relatively linear relationship between
IC and IB i.e. above cut-off and below saturation.
Apparent Beta=12 or really any value LESS than the
maximum forward region Beta is believable in saturation,
but having Beta *extremely* higher than the nominal
value in the forward region isn't usual unless you're
dealing with a transistor operated WAY outside of its
normally specified operating environment e.g. a
RF transistor designed to work at frequency 3GHz but
being used only at 1MHz, so of course the Beta might
be quite different over those extremely different frequencies.


Let's look at another listed case where the BJT is in forward active mode:
RC=1000
RB=10000
V_base_supply=0.6550
VBE=0.6549
V_RB=V_base_supply-VBE=0.0001
I_RB=V_RB/RB=0.0001/10k=10nA
IB=I_RB=10nA since RB and IB are in series.
V_collector_supply=5
VCE=4.901
V_RC=V_collector_supply-VCE=5-4.901=0.099
I_RC=V_RC/RC=0.099/1000=99uA
IC=I_RC=99uA since RC and IC are in series.
Beta=IC/IB=99uA/10nA=9900.
That doesn't make much sense either since the
expected forward linear region Beta=PSPICE BF=80,
and this is more than 100x too large and yet it's listed
as being an example of forward active mode conduction.

I can understand why you'd be confused by the way these
listed voltages and parameters relate. They're not
what I would expect from a BJT model that ostensibly
has a Beta of 80; I would think IB would be much higher
than it seems to be at the given VBE values
in the various modeled scenarios I've done the calculations for here,
and I'd think that the apparent BETA (obtained by IC/IB calculations)
would be much closer to the expected 80 in the forward region, and less
than that value in the saturation region.

Though the particular numbers of
VBE vs IB and apparent BETA aren't what I'd consider
very typical of many real transistors, the overall topology of the
circuit would indeed perform a logic inversion type of function for
reasons that are intuitively straightforward even if the
particular parametric examples are in part bogus (IMHO).

If you apply 0 voltage to the base, IB=0, and
the transistor doesn't conduct and pull the voltage output node
at the collector to ground.
With no conduction in the collector, or base the transistor
may as well not be in the circuit, and the output voltage will be
basically the same as the collector supply voltage since
there's no real pull-down load absent collector current.

If you apply a logic high voltage e.g. 5V to the base
circuit, you'll get several milliamps of base current
through the series base resistor, which could produce a
heavy collector current by multiplication by Beta,
and the transistor will be saturated in conduction with
a very small VCE_sat voltage well under 1V appearing at
the voltage output collector terminal.

Thus a 0V on the base supply -> 5V at the collector,
and 5V at the base supply -> nearly 0V (VCE_sat) at the collector,
and an inversion of logic levels from base to collector
has taken place.

If you look at the data sheet graphs for the real ZETEX
transistors I linked to, you can see that at very small
collector currents below 10mA, VCE_sat does indeed have
very small values, so the listed numbers in the examples
on your web page aren't horribly out of line with what
one might expect at small collector currents.

The model's VBE_on vs IB seems rather unusually large,
though, as does the apparent Beta. That's almost
the kind of BETA one might expect for a cascaded PAIR
of BJT's in a darlington configuration, and in that case,
the 'VBE' would be close to double what one would expect
for a single normal BJT too, but that doesn't explain
why BF=80 is listed in the PSPICE model on the page...

So who knows what the model really is for that transistor
and why the modeled parameters seem a bit unexpected in
some cases relative to the model parameters shown.

It's also strange that they didn't do a MUCH better job
indicating the CURRENTS in their graphed data points
since this is about understanding BJTs, which are really
current operated devices even though it's true that
a logic inverter is usually designed to be
defined by the voltages of its inputs and outputs.
 
  • #14
Xez, thanks for your reply. I havent had the time to digest everything you wrote, yet, but i just wanted to say thank you :)
 

Related Threads on Transistors (BJTs) confusion

  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
2
Replies
26
Views
15K
Replies
8
Views
6K
  • Last Post
Replies
10
Views
4K
Replies
2
Views
981
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
20
Views
3K
  • Last Post
Replies
17
Views
2K
Top