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'[EE] Datasheet confusion'
2007\02\28@034703 by Rikard Bosnjakovic

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I have two datasheets describing the transistor 2N3906. Consider these
pdf-files:

www.datasheetarchive.com/datasheet.php?article=3743092
http://www.datasheetarchive.com/datasheet.php?article=3743093

The first one tells me the dissipation (Pd) is 625mW, while the second
tells me Pd is 350mW.  I'm aware of the fact that this may be because
two different companies made hem, but are components really allowed
the same brand if their characteristics differ?

I'm asking because I don't have any 2N3906 at home (and I'm not gonna
order new ones) so I'm looking for replacements. But if their values
differ - how am I supposed to know which replacements to look for?


--
- Rikard.

2007\02\28@062249 by Forrest W Christian

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Rikard Bosnjakovic wrote:

>The first one tells me the dissipation (Pd) is 625mW, while the second
>tells me Pd is 350mW.  I'm aware of the fact that this may be because
>two different companies made hem, but are components really allowed
>the same brand if their characteristics differ?
>  
>
The 2n3904/2n3906 pair are rarely used where Pd Matters, at least from
my experience.  Generally they are used as small signal amplifiers.  
More important considerations are parameters like hfe.

The Jedec part numbers (2nXXXX for example) do have a standard defined
as far as what minimum requirements a given part number has to meet.  
There isn't anything which says that a 2n3904 can't have a better Pd
rating than the standard.

Personally, I'm now using the PN100(NPN) and PN200(PNP) parts since they
are really cheap and quite available.

-forrest



2007\02\28@062937 by Rich

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The difference is small, around 4% The actual dissipation depends on the
load.  There are greater deviations in the Beta, even within lots from the
same manufacturer.  I don't see how it would make a difference unless you
are running it close to its Ic limit.  Your concern should be more with beta
if gain is important in your application.  The 2N3906 can often be replaced
by the 2N3905.  The deciding factor is what you application circuit
requirements are. That is why there are so many different parameters given
in the spec sheet.


{Original Message removed}

2007\02\28@064633 by Vasile Surducan

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w3.id.tue.nl/fileadmin/id/objects/E-Atelier/doc/Datasheets/Transistoren/2N3906.pdf

Power dissipation is also a matter of package. Take a look above.

Vasile

On 2/28/07, Rikard Bosnjakovic <spam_OUTrikard.bosnjakovicTakeThisOuTspamgmail.com> wrote:
{Quote hidden}

> -

2007\02\28@065629 by Rikard Bosnjakovic

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On 2/28/07, Forrest W Christian <.....forrestcKILLspamspam@spam@imach.com> wrote:

> The 2n3904/2n3906 pair are rarely used where Pd Matters, at least from
> my experience.  Generally they are used as small signal amplifiers.
> More important considerations are parameters like hfe.

I'm going to use the transistors as on/off-switches for 7
segment-LEDs. The reason for 2N3906 is that they used it in this
circuit: http://www.melabs.com/resources/articles/ledart.htm

According to the datasheet for my leds, each segment can eat up to
30mA. Given the worst case (all segments lit), this would be 210mA, or
1.05W at 5 volts.  The sheet also says that there's generally a 2
v-drop for each segment so I'm unsure if the dissipation is as above
or (5-2)*0.21 = 0.63W.

Feel free to correct me here where I'm thinking wrong. My main goal is
to find suitable transistors in my "goldmine" of dissorted components.
Last resort would be to buy new components.


--
- Rikard.

2007\02\28@071328 by peter green

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> I'm going to use the transistors as on/off-switches for 7
> segment-LEDs. The reason for 2N3906 is that they used it in this
> circuit: www.melabs.com/resources/articles/ledart.htm
>
> According to the datasheet for my leds, each segment can eat up to
> 30mA. Given the worst case (all segments lit), this would be 210mA, or
> 1.05W at 5 volts.  The sheet also says that there's generally a 2
> v-drop for each segment so I'm unsure if the dissipation is as above
> or (5-2)*0.21 = 0.63W.
>
> Feel free to correct me here where I'm thinking wrong
The bulk of the voltage is dropped by the resistor and the LED, assuming you drive the transistors hard into staturation there should only be a very small voltage (proablly arround 0.2V or so) accross the transistor.

The time you really have to worry about transistor power disipation is when you are running the things in an analog manner.


2007\02\28@071806 by Jinx

face picon face
> I'm going to use the transistors as on/off-switches for 7
> segment-LEDs. The reason for 2N3906 is that they used it in this
> circuit: http://www.melabs.com/resources/articles/ledart.htm

"with this kind of configuration we can build the numbers 0 - 9 and
even some letters. I should patent this idea"

Not if I get there first pal ;-)

Rikard, don't forget you're strobing the LEDs. Transistor Ic will be
the same but Pd will be less - 4 digits will reduce it to 1/4 of what
you thought it was because of the fractional on time. BC547 (100mA,
about the same as the ubiquitous 2SC945 used by the Japanese if
you fancy stripping old radios, TVs etc but watch the pin-out) or
BC337 (500mA) are handy to keep around for this sort of thing

2007\02\28@073637 by Lee Jones

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>> The 2n3904/2n3906 pair are rarely used where Pd Matters, at least from
>> my experience.  Generally they are used as small signal amplifiers.
>> More important considerations are parameters like hfe.
>
> I'm going to use the transistors as on/off-switches for 7
> segment-LEDs. The reason for 2N3906 is that they used it in this
> circuit: http://www.melabs.com/resources/articles/ledart.htm
>
> According to the datasheet for my leds, each segment can eat up to
> 30mA. Given the worst case (all segments lit), this would be 210mA, or
> 1.05W at 5 volts.  The sheet also says that there's generally a 2
> v-drop for each segment so I'm unsure if the dissipation is as above
> or (5-2)*0.21 = 0.63W.
>
> Feel free to correct me here where I'm thinking wrong. My main goal is
> to find suitable transistors in my "goldmine" of dissorted components.
> Last resort would be to buy new components.

Power dissipated in each 2N3906 is the voltage dropped within it
times the current.  In a switch application, the you are looking
at about 0.2 to 0.4 volts from collector to emitter.  Following
computations assume 0.2V drop inside the transistor.

Assume 5V supply...  Assume 2V drop per LED segment...

Each segment will only consume the amount of current _you_ allow
by your choice of dropping resistor.  In the above circuit, the 270
ohm resistors limit the current per segment to about 10mA.  If you
want to drive each segment with 30mA, then you need resistors of
(5V - 2V - 0.2V) / 0.030 = 93 ohms; so any resistor of 93 ohms or
greater will limit current to 30mA or less.

Let's choose 100 ohm resitors giving ~28mA per-segment current.

Power dissipation in each 100 ohm resistor would be 0.028A x 2.8V
(i.e. 5V - 2V - 0.2V) = 0.078 watts (so 1/8 watt or larger resistors
could handle the power).

Power dissipated in each 2N3906 (with 270 ohm resistors) would be
0.2V * (0.010A x 7 segments) = 0.014 watts.  With 100 ohm resistors,
power dissipation would be 0.2V * (0.028A x 7) = 0.039 watts.  Even
if Vce(saturation) were 0.4V, you're still well under 0.1 watts in
each transistor.

The per-segment 30mA current limit is probably the maximum power
that each segment can dissipate under continuous operation (i.e.
100% on).  Currents may be significantly higher (without burning
out a segment) if the segment is pulsed.

If you're going to have each segment driven directly by a PIC pin,
then I'd limit the per-segment drive to ~23mA (via 120 ohm resistor).
Note that the PIC itself may limit the total current because you must
assume you will drive all 7 segments at the same time.  See the PIC
datasheet for maximum.

                                               Lee Jones

2007\02\28@074900 by peter green

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> The per-segment 30mA current limit is probably the maximum power
> that each segment can dissipate under continuous operation (i.e.
> 100% on).  Currents may be significantly higher (without burning
> out a segment) if the segment is pulsed.
thats true but i wouldn't try driving segments above their continuous rated current until you are absoloutely positive the software is correct.



2007\02\28@102321 by Rikard Bosnjakovic

picon face
On 2/28/07, Lee Jones <leespamKILLspamfrumble.claremont.edu> wrote:

> Power dissipated in each 2N3906 is the voltage dropped within it
> times the current.  In a switch application, the you are looking
> at about 0.2 to 0.4 volts from collector to emitter.  Following
> computations assume 0.2V drop inside the transistor.

[...snip long good post...]

Lee, thank you for your very descriptive post. It's usually basic
things like this that tend to make things stop happening for me. I
guess the only way to really learn it (apart from having 4.0 GPA in
electrical analysis 1 and 2) is to construct more circuits on the
breadboard. I do however still have some problems understanding how
BJTs and FETs work (atleast the math-versions), but I guess it'll come
along in a nearby future.



--
- Rikard.

2007\02\28@164146 by Forrest W Christian

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Rikard Bosnjakovic wrote:

>On 2/28/07, Forrest W Christian <.....forrestcKILLspamspam.....imach.com> wrote:
>
>  
>
>According to the datasheet for my leds, each segment can eat up to
>30mA. Given the worst case (all segments lit), this would be 210mA, or
>1.05W at 5 volts.  The sheet also says that there's generally a 2
>v-drop for each segment so I'm unsure if the dissipation is as above
>or (5-2)*0.21 = 0.63W.
>  
>
Actually Pd is roughly Ic*Vce(sat).   The Collector-emitter drop at
saturation is typically less than .5V, so your total pd would be .21*.5
or around 1/10W.

I would worry more about Ic, which is rated typically around 200ma
continuous, 300ma peak.   But...  in reality you probably won't hit this
due to the duty cycle.

Remember, if you're muxing, your continuous rating can be derated since
the transistor won't be on 100% of the time.  Peak ratings may need to
be respected more.

So, all you really need is a pile of GP PNP transistors.  I have an
Atlas DCA ( http://www.peakelec.co.uk/acatalog/jz_dca55.html ) which
makes sorting the junkbox a lot easier - but again, I've recently been
re-building my collection.

-forrest

2007\02\28@171704 by Forrest W Christian

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Rikard Bosnjakovic wrote:

> I do however still have some problems understanding how
>BJTs and FETs work (atleast the math-versions), but I guess it'll come
>along in a nearby future.
>  
>
I will admit to being "FET-scared" for quite some time.  Now they're
awesome.   The 2n7000 is your friend.

On the BJT side, really the datasheet isn't all that scary.

You have only about 3-4 main parameters you need to worry about:

1) HFE or Beta.  This is the transistor's gain.    If the gain is 100,
that means that the emitter-collector current will be 100 times the
base-emitter current.  In digital electronics, everything gets driven
into saturation so you just need to make sure that the base-emitter
current is more than 1/100th of the total load.  For example, if your
load is 100ma, you need to drive 1ma into the base.  At 5v, this is a 5K
resistor.  Probably want a 1K resistor on the base.  If you use a 10K
resistor with a transistor with a beta of 100,  5 volts through 10k is
0.5ma (E=IR).  100 times that is 50ma, so it wouldn't be enough to say
turn on a 100ma load.  If you don't put enough in the base, the
transistor won't turn fully on and then you *will* be dissipating
power.  More is generally better than less, up to a certain smoke point.

2) Ice and Ibe.  These are the maximum currents you can push through the
device.   Ice is the collector-emitter junction, and the be is the
amount through the base.  If you exceed these, the device will fry.

3) Vce(sat) and Vbe.  These are the voltages across the junctions.  The
Vce(sat) figure is how many volts will be across the transistor when it
is fully on (see the HFE discussion).  The Vbe is how many volts will be
from the base to the emitter.

3) Pd.  This is the amount of power which the transistor can turn into
heat.   This *isn't* the amount of power that it can carry.  In order to
calculate this, you generally have to take the current through CE times
Vce(sat).

The reason I like FET's now are:

1) The HFE is irrelevant.  Either the voltage is high enough to turn it
on, or not.  

2) There is no current through the gate (or there isn't enough to worry
about - unless you do someting bad).

3) The resistance between the source and the drain terminals are really
really low, so you can pretty much ignore Pd.

Generally, I don't even use BJT's anymore, unless I need a totem-pole
for something.

-forrest

2007\02\28@173649 by peter green

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> 1) The HFE is irrelevant.  Either the voltage is high enough to turn it
> on, or not.  
with fets you can still get a gate voltage that is high enough to let the fet conduct some current between source and drain but not high enough to allow your loads full current.



2007\02\28@174726 by Marcel Birthelmer

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On 2/28/07, peter green <EraseMEplugwashspam_OUTspamTakeThisOuTp10link.net> wrote:
> > 1) The HFE is irrelevant.  Either the voltage is high enough to turn it
> > on, or not.
> with fets you can still get a gate voltage that is high enough to let the fet conduct some current between source and drain but not high enough to allow your loads full current.
>
>
>
> -

2007\02\28@191535 by Richard Prosser

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Marcell,
IIRC the threashold voltage is controlled by doping levels and the
like. So there is no real problem getting "logic level" FETs to
control small to reasonable currents. The tradeoff is mostly in Ron
(so there is a limitation here to some degree) and the maximum
voltages the device will withstand. Trying to find a logic level fet
for higher voltages becomes difficult and normally you'd have a
separate driver chip or circuit and drive the fet (MOSFET) at 10-15V
or so.

The other thing to watch is the switching time. This is mostly
determined by how fast you can charge or discharge the gate
capacitance. FETs (particularly MOSFETs) are very capacitive so moving
a lump of charge quickly directly via a microcontroller port limits
the available on/off times. Again, supply voltage comes into it due to
the effect the voltage gain has on "miller" capacitance so again
higher voltages are more difficult to switch quickly.

If your switching time is too slow, then the device is operating in a
linear zone for a while and can generate a lot of heat.

RP

On 01/03/07, Marcel Birthelmer <marcelb.listsspamspam_OUTgmail.com> wrote:
> On 2/28/07, peter green <@spam@plugwashKILLspamspamp10link.net> wrote:
> > > 1) The HFE is irrelevant.  Either the voltage is high enough to turn it
> > > on, or not.
> > with fets you can still get a gate voltage that is high enough to let the fet conduct some current between source and drain but not high enough to allow your loads full current.
> >
> >
> >
> > --

2007\02\28@192048 by William Chops Westfield

face picon face

>> According to the datasheet for my leds, each segment can eat up to
>> 30mA. Given the worst case (all segments lit), this would be 210mA, or
>> 1.05W at 5 volts.  The sheet also says that there's generally a 2
>> v-drop for each segment so I'm unsure if the dissipation is as above
>> or (5-2)*0.21 = 0.63W.
>>
Here, you have a 2V drop across each segment, and 210mA total,
so the power dissipation is about 420mW IN THE LED DISPLAY.

The power dissipation OF THE TRANSISTORS that are switching
the 210mA, is going to be (.21 * Vcesat), with Vcesat typically
being 0.2 to 0.5, so the dissipation is going to be about 0.1W.

Here, the transistor is being used as a switch, and most of the
parameters are pretty unimportant.  The max current Ic is probably
most important; any PNP with Ic of about 200mA or more will probably
do fine.  Popular PNP switching transistors that you can use instead
of a 3906 include the 2n2907(a) and the 2n4403...

BillW

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