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'[PIC] +12V high-side transistor switch?'
2008\06\11@125337 by PicDude

flavicon
face

I need to switch on/off a 12V load, high-side, from a PIC running on 5V, and
tried this circuit, but the output is always on.  With the PIC I/O floating
(set as input), the load should be switched off, and with the I/O low, the
load should be switched on.  I'm using 1k for R1 and 10K for R2.  The load
will be up to ~250mA, btw.

Any idea what I'm doing wrong here?  The circuit...
http://www.narwani.net/neil/electronics/PIC-HS12VSwitch-01.gif

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2008\06\11@130834 by Dario Greggio

face picon face
PicDude wrote:

> I need to switch on/off a 12V load, high-side, from a PIC running on 5V, and
> tried this circuit, but the output is always on.  With the PIC I/O floating
> (set as input), the load should be switched off, and with the I/O low, the
> load should be switched on.  I'm using 1k for R1 and 10K for R2.  The load
> will be up to ~250mA, btw.

Internal protection diodes of PIC's pins will make your transistor
conduct all the time - since they shunt to 5.3Vcc (VDD+0.3 or so).

You should use a true Open Collector pin. Or place an intermediate
transistor in between.

--
Ciao, Dario

2008\06\11@131705 by Bob Blick

face
flavicon
face
part 1 788 bytes content-type:text/plain; charset="ISO-8859-1" (decoded 7bit)


On Wed, 11 Jun 2008 09:53:12 -0700 (PDT), "PicDude"
<spam_OUTpicdude2TakeThisOuTspamavn-tech.com> said:
{Quote hidden}

Hi Neil,

On old 16C parts you do it with RA4 because the old process could
tolerate the voltage. But you'll need to use an extra transistor to
level shift. Attached is an example.

Cheerful regards,

Bob

--
http://www.fastmail.fm - The way an email service should be



part 2 4144 bytes content-type:image/gif; name="hiside.gif" (decode)


part 3 35 bytes content-type:text/plain; charset="us-ascii"
(decoded 7bit)

2008\06\11@131948 by Marcos Fabricio Bezerra

flavicon
face
Hi, Picdude.
The internal protection diodes are keeping the transistor on. Even with the
I/O floating, the voltage at the pin will be 5.7V, what keeps the PNP
transistor ON. You can move R2 to the base of the transistor, and calculate
the resistor divider to switch on the transistor only with less than 5V on
the PIC I/O.
Fabricio.


{Original Message removed}

2008\06\11@132939 by Marcos Fabricio Bezerra

flavicon
face
Hi, Neil.
Beside changing the R2 position, you can use a series zener diode connected
to R1 and I/O pin. I think that a 9V1 would do the trick.
Cheers,
Fabricio.



{Original Message removed}

2008\06\11@143235 by Spehro Pefhany

picon face
Quoting PicDude <.....picdude2KILLspamspam@spam@avn-tech.com>:

>
> I need to switch on/off a 12V load, high-side, from a PIC running on 5V, and
> tried this circuit, but the output is always on.  With the PIC I/O floating
> (set as input), the load should be switched off, and with the I/O low, the
> load should be switched on.  I'm using 1k for R1 and 10K for R2.  The load
> will be up to ~250mA, btw.

Use Bob's method only. Eg.

R1 1K 1/4W
R2 10K
R3 6K8
NPN: 2N/MMBT4401
PNP: 2N/MMBT4403

In most cases you'll also want to add a diode or something similar  
across the load to tame inductive spikes, even for something like a  
nominally resistive load with long leads.

Best regards,
Spehro Pefhany
--
"it's the network..."                          "The Journey is the reward"
s...spamKILLspaminterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

2008\06\11@173549 by olin piclist

face picon face
Bob Blick wrote:
> On old 16C parts you do it with RA4 because the old process could
> tolerate the voltage. But you'll need to use an extra transistor to
> level shift. Attached is an example.

You can simplify this by moving R1 to the emitter of Q1 and losing R3
altogether.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2008\06\11@174623 by olin piclist

face picon face
Spehro Pefhany wrote:
>> The load will be up to ~250mA, btw.
>
> Use Bob's method only. Eg.
>
> R1 1K 1/4W
> R2 10K
> R3 6K8
> NPN: 2N/MMBT4401
> PNP: 2N/MMBT4403

You had better make sure the load will use all its available voltage else
the PNP transistor will fry.  Note that the MMTB4403 can only handle 500mV
accross it at 250mA.  This circuit will also be quite slow turning off, but
that may not matter.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2008\06\11@232430 by Mike Hagen

flavicon
face
IR has high side switches, over current protected, over temperature
protected.  I have had good results with them!



PicDude wrote:
> I need to switch on/off a 12V load, high-side, from a PIC running on 5V, and
> tried this circuit, but the output is always on.  With the PIC I/O floating
> (set as input), the load should be switched off, and with the I/O low, the
> load should be switched on.  I'm using 1k for R1 and 10K for R2.  The load
> will be up to ~250mA, btw.
>
> Any idea what I'm doing wrong here?  The circuit...
> http://www.narwani.net/neil/electronics/PIC-HS12VSwitch-01.gif
>
>  

2008\06\12@003606 by PicDude

flavicon
face

Hadn't factored in the internal protection diodes, thanks.
I need to do this with 8 outputs, and only one OC output.  I started with a
2 transistor circuit (1 NPN + 1 PNP), but thought I could simplify it this
way.

Cheers,
-Neil.



Dario Greggio (in giro) wrote:
>
>
> Internal protection diodes of PIC's pins will make your transistor
> conduct all the time - since they shunt to 5.3Vcc (VDD+0.3 or so).
>
> You should use a true Open Collector pin. Or place an intermediate
> transistor in between.
>
> --
> Ciao, Dario
> --

2008\06\12@003953 by PicDude

flavicon
face

Hmmm... zener you say, huh.... I'll have to try this. Thanks.



Marcos Fabricio Bezerra wrote:
>
> Hi, Neil.
> Beside changing the R2 position, you can use a series zener diode
> connected
> to R1 and I/O pin. I think that a 9V1 would do the trick.
> Cheers,
> Fabricio.
>
>
>
> {Original Message removed}

2008\06\12@004306 by PicDude

flavicon
face

That's pretty much where I started (except I put R2 on the NPN collector,
instead of the PNP base), but thought I could simplify it with one
transistor.  I hadn't realized the protection diodes would get in the way.
I need to do this on 8 outputs simultaneously, so RA4 is not the full answer
here.

Cheers,
-Neil.



Bob Blick-4 wrote:
{Quote hidden}

> --

2008\06\12@043439 by Dario Greggio

face picon face
PicDude wrote:

> Hadn't factored in the internal protection diodes, thanks.
> I need to do this with 8 outputs, and only one OC output.  I started with a
> 2 transistor circuit (1 NPN + 1 PNP), but thought I could simplify it this
> way.

Yeah, or add a ULN2003/2803 ?

--
Ciao, Dario

2008\06\12@051513 by Apptech

face
flavicon
face
>> Hadn't factored in the internal protection diodes,
>> thanks.
>> I need to do this with 8 outputs, and only one OC output.
>> I started with a
>> 2 transistor circuit (1 NPN + 1 PNP), but thought I could
>> simplify it this
>> way.
>
> Yeah, or add a ULN2003/2803 ?

The ULN280x/ULN200x products are well priced and extremely
useful. They come in SOIC and DIP (at least).

   www.st.com/stonline/products/literature/ds/1536/uln2805a.pdf
   http://focus.ti.com/lit/ds/symlink/ulq2004a.pdf

A few points to note:

- They are "darlington" output so the output can never
saturate - you get at least 1 x Vbe drop to ground.
Typically this is 0.6V at about zero output current and will
rise with Iout.

- They do not have amazingly high capabilitty outputs. The
data sheet shown the ULN200x as having Vout=1V typical at
200 mA and a rather undesirable 1.5V at 500 mA. Actual Vout
varies with input drive but the above is at a forced beta of
10 ie Iin = 500 uA, and it's worse at lower drives. YMMV but
may not.

- The ULN280x are no better.

- While the ULN2003 / ULN2803 are the most common the others
in the series can be very useful. The ULN2x04 may work well
with no input resistor and 5V drive. Examine the spec sheets
carefully to see if a particular one meets your need.

- At under $1 in 1's and under $US0.30  1000 up it is hard
to match the price using discretes, although discrete
drivers will allow you achieve a better spec for a similar
price with care.

- The suggested 2 transistor driver allows grounded load and
no polarity inversion. The ULN2x0x give high side connected
load and arguably polarity inversion, depending which supply
your brain/frame of reference is wired to.





       Russell McMahon

2008\06\12@052755 by Jinx

face picon face
> Yeah, or add a ULN2003/2803 ?

http://www.ceid.upatras.gr/courses/microlab/ask4/uln2003.pdf

In the equipment I strip or service the 200x are very common
logic-to-power i/f chips. It's not usual to see discrete transistor
arrarys, for whatever reason. The pdf includes an internal
schematic if you want to make extra Darlington drivers over the
7 in the chip. If you're short of I/O you could drive the drivers
with a shift register, but let's not get carried away just yet

2008\06\12@053243 by Jinx

face picon face
> Yeah, or add a ULN2003/2803 ?

Another option might be the 7406

http://focus.ti.com/docs/prod/folders/print/sn7406.html

It doesn't have 250mA sink capability but it'll do the voltage level
conversion for a high-side switch

2008\06\12@055853 by Dario Greggio

face picon face
Jinx wrote:

>>Yeah, or add a ULN2003/2803 ?
>
> Another option might be the 7406

yep, the oldie TTL :)
you're right.

--
Ciao, Dario

2008\06\12@075029 by Spehro Pefhany

picon face
At 05:48 PM 6/11/2008, you wrote:
>Spehro Pefhany wrote:
> >> The load will be up to ~250mA, btw.
> >
> > Use Bob's method only. Eg.
> >
> > R1 1K 1/4W
> > R2 10K
> > R3 6K8
> > NPN: 2N/MMBT4401
> > PNP: 2N/MMBT4403
>
>You had better make sure the load will use all its available voltage else
>the PNP transistor will fry.

What is that supposed to mean? None of the high-side drivers presented
are short-circuit proof.

>  Note that the MMTB4403 can only handle 500mV
>accross it at 250mA.  This circuit will also be quite slow turning off, but
>that may not matter.

?? 'Handle 500mV accross' it?   You talking about SOA? Static Pd(max) at Ta?

'Quite slow' turn-off? tf will be << 1usec.

Where is Olin, and what have you done with him?

>Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
EraseMEspeffspam_OUTspamTakeThisOuTinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



2008\06\12@083010 by Harold Hallikainen

face
flavicon
face

> Hi, Neil.
> Beside changing the R2 position, you can use a series zener diode
> connected
> to R1 and I/O pin. I think that a 9V1 would do the trick.
> Cheers,
> Fabricio.

I've used this "zener trick" before. It has an advantage over pure
resistive level shifting in that the top of the zener still has a 5Vpp
change while the top of a resistor would not. I did a similar circuit to
PWM 5V fans with a 3.3V PIC. There, instead of a zener, I used an LED.
When the fan should be running, the LED glows.

Harold


--
FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
opportunities available!

2008\06\12@084227 by Spehro Pefhany

picon face
At 05:37 PM 6/11/2008, you wrote:
>Bob Blick wrote:
> > On old 16C parts you do it with RA4 because the old process could
> > tolerate the voltage. But you'll need to use an extra transistor to
> > level shift. Attached is an example.
>
>You can simplify this by moving R1 to the emitter of Q1 and losing R3
>altogether.

Sure, at the expense of greatly increasing the Pd of Q1 and exposing the
PIC to damage from transients on the +12V power rail.

Also, although I wouldn't too much worry about it for a hobby design, and
probably Rds(on) of the p-channel in the PIC will suffice, but this seminar
explains how a simple emitter-follower can turn into a Colpitts oscillator
if the gods of poles and zeros don't smile upon you (and, as they point out,
there's never a guaranteed *maximum* ft)

http://ee.eng.usf.edu/gradcourses/wireless-seminar/sp05/abrahams-1-26-05-pres.pdf


>Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speffspamspam_OUTinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



2008\06\12@084952 by Spehro Pefhany

picon face
At 04:15 AM 6/12/2008, you wrote:
>PicDude wrote:
>
> > Hadn't factored in the internal protection diodes, thanks.
> > I need to do this with 8 outputs, and only one OC output.  I started with a
> > 2 transistor circuit (1 NPN + 1 PNP), but thought I could simplify it this
> > way.
>
>Yeah, or add a ULN2003/2803 ?

Sure.

For a single channel, there are integrated transistor/resistor combos that
will do the NPN thing. Some duals might even do the whole thing, maybe
with an external resistor to control Ib, but the choices are getting a bit
thinner for reliable operation at 250mA. If you don't mind non-jellybean
parts, there's even an integrated SMT N-channel MOSFET (with pull-down)
and PNP BJT that comes close (with an external resistor for Ib).

No good integrated bipolar high-side drivers AFIR, since it's difficult
to make a good PNP transistor in an IC.

>Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
@spam@speffKILLspamspaminterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



2008\06\12@094443 by Apptech

face
flavicon
face
>> >> The load will be up to ~250mA, btw.
>> >
>> > Use Bob's method only. Eg.
>> >
>> > R1 1K 1/4W
>> > R2 10K
>> > R3 6K8
>> > NPN: 2N/MMBT4401
>> > PNP: 2N/MMBT4403
>>
>>You had better make sure the load will use all its
>>available voltage else
>>the PNP transistor will fry.
>
> What is that supposed to mean? None of the high-side
> drivers presented
> are short-circuit proof.

I took it to mean (possibly wrongly) that the PNP needed to
be in saturation or close to it.

Without looking back I don't know if the 250 mA was
specified when those resistor values were given. As will be
seen below, the circuit is in trouble as shown at 250 mA
with that transistor and resistors. It can almost be
redeemed but ideally either a higher current gain transistor
for Q2 or even better a MOSFET would make it work well.

Because:

Drive current will be about (12v-Vbe_Q2)/R1 = 11.4/10k = 1
mA+
For 250 mA load the beta of Q2 wants to be 300 or better.
Beta of an xx4403 is under 100 guaranteed at 250 mA so you
need say 3x as much base drive for Q2 if you want to support
250 mA.
R2 = 3k3 or 2k7.
E&OE check the sums.
Dissipation in R2 is still minimal.

Vce_sat max can be well over 1 volt for this device at 250
mA so Pd at 250 mA = > 250 mW.
That would get the TO92 part hot (2N4403) and the SMD part
very hot (MMBT..)

If I was doing this I'd start with my favourite jellybean
BC327-40.
500mA to 700 mA rated depending on manufacturer.
The -40 means mean beta in batch of 400.
(~= Geometric mean of 250 to 600 range).
This is good for most such tasks.
You'd need to work through the above example to be sure.

An easier way to run cool is to use a FET.
Either a high side P Channel or a single N channel if you
can tolerate the low side drive to the load.

Type depends on drive voltage.
The beautiful but not too common in the US CES2310 that I
mentioned a while ago would do it with ease.
Failing that the FDN337 with Rdson = 0.082 typ at 2.5V
drive.
Headline Rdsons are always a lie compared to DC steady state
but at 3V gate you'd get well under 0.1V on voltage at 250
mA for 25 mW dissipation.
With Tja of 250 C/W it would run touch cold.
Probably :-)

$US0.44/1 Digikey
$US0.12/1000.
A bargain, if you can't get CES2310's :-)

       http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=FDN337NCT-ND

       http://www.fairchildsemi.com/ds/FD/FDN337N.pdf

It's FDN338 P Channel stablemate is also your friend at
about the same price.
As expected, not quite as good an Rdson as N channel part,
but still very good in this task.

       http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=FDN338PCT-ND

       http://www.fairchildsemi.com/ds/FD/FDN338P.pdf

Use this in place of Q2 in the original circuit and Bob's
your uncle.
R1 10k or whatever.
R2 100k or  >>> R2.
R3 almost anything.
Speed gets slower with higher R.
A speedup cap and some R tailoring could make this very fast
if required (for some values of very).

>>  Note that the MMTB4403 can only handle 500mV
>>accross it at 250mA.

A valid concern - as above.
Beta won't support required current with provided drive



           Russell McMahon


2008\06\12@094719 by olin piclist

face picon face
PicDude wrote:
> Hadn't factored in the internal protection diodes, thanks.
> I need to do this with 8 outputs, and only one OC output.  I started
> with a 2 transistor circuit (1 NPN + 1 PNP), but thought I could
> simplify it this way.

If your voltages are all well controlled, then you can set up a resistor
divider so that the base of the PNP transistor is on when the PIC output is
0V and off when 5V.  I haven't done the math to see what base current drive
level you end up with even if you max out the PIC pin current to 25mA when
low.

I don't really like this idea and would be very unlikely to put it into
production.  You only need one extra transistor to do it robustly.  This is
a jellybean NPN that costs about $.02 in volume.  The total robust solution
is only 2 transistors and 2 resistors.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2008\06\12@094948 by Daniel Serpell

picon face
Hi!

On Thu, Jun 12, 2008 at 8:53 AM, Spehro Pefhany <KILLspamspeffKILLspamspaminterlog.com> wrote:
>
> For a single channel, there are integrated transistor/resistor combos that
> will do the NPN thing. Some duals might even do the whole thing, maybe
> with an external resistor to control Ib, but the choices are getting a bit
> thinner for reliable operation at 250mA. If you don't mind non-jellybean
> parts, there's even an integrated SMT N-channel MOSFET (with pull-down)
> and PNP BJT that comes close (with an external resistor for Ib).
>
> No good integrated bipolar high-side drivers AFIR, since it's difficult
> to make a good PNP transistor in an IC.

You have the L293: 4 chanels, push-pull, 1A max.
  http://focus.ti.com/docs/prod/folders/print/l293.html

     Daniel.

2008\06\12@101704 by olin piclist

face picon face
Spehro Pefhany wrote:
> What is that supposed to mean? None of the high-side drivers presented
> are short-circuit proof.
>
>>  Note that the MMTB4403 can only handle 500mV
>> accross it at 250mA.  This circuit will also be quite slow turning
>> off, but that may not matter.
>
> ?? 'Handle 500mV accross' it?   You talking about SOA? Static Pd(max)
> at Ta?

You specified MMTB4403, which comes in a SOT-23 package.  At 25C ambient the
junction will reach 150C at 350mW.  That is the absolute maximum, and I was
using the more "normal" dissipation guideline of 1/8 W or 125mW for a SOT-23
for general practise.  The current spec is 250mA.  125mW / 250mA = 500mV,
which is the most voltage drop the transistor should therefore be subjected
to when on.  Even at the 350mW guaranteed to fry threshold, you only get
1.4V.

This is so far from short-circuit proof that I thought it was worth pointing
out.  For example, if the load limited its voltage to 10V at 250mA, the
transistor would dissipate 500mW, which is in the guaranteed to fry range
even though it's far from what most people would consider a short circuit.

> 'Quite slow' turn-off? tf will be << 1usec.

I was trying to point out that the turnoff of the circuit would be much much
slower than what you might otherwise expect from these transistors.  Both
transistors will be saturated when on, and the relatively high resistances
of 10Kohms and 6.8Kohms will not quickly force the transistors to turn off.

Neil never said what kind of load he was driving or how fast or how often he
needs to switch it, so we don't have enough information to know whether any
of this matters, which is why I pointed it out.

In any case, it sill makes sense to get rid of the R3 and put R1 between the
emitter of Q1 and ground.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2008\06\12@102924 by Spehro Pefhany

picon face
Quoting Apptech <RemoveMEapptechTakeThisOuTspamparadise.net.nz>:

{Quote hidden}

Well, there's no reason for it to be out of saturation unless the
load is too low resistance. Not with Ic/Ib ~- 20. Well, I suppose you
could drop it into a dewar filled with liquid N2.

{Quote hidden}

I gave a value of 1K 0.25W, which yields 11.3mA, so the
remainder of the calculation won't be valid. The criteria
used is Ic/Ib ~= 20.  Actual power dissipation is 128mW.

> For 250 mA load the beta of Q2 wants to be 300 or better.
> Beta of an xx4403 is under 100 guaranteed at 250 mA so you
> need say 3x as much base drive for Q2 if you want to support
> 250 mA.

You should expect to operate *well* into saturation in this sort of
application. Ic/Ib = 10 to Ic/Ib = 20, NOT Ic/Ib = 100!

> An easier way to run cool is to use a FET.
> Either a high side P Channel or a single N channel if you
> can tolerate the low side drive to the load.

Yeah, or a dual N&P channel, but care is required as to
maximum gate voltage on the P-channel (might require a protection
zener diode), it tends to be slower and much more expensive. The
P-channel you mention below has a Vgs(max) of only 8VDC.
 It's also a lot easier to get it
into a linear mode where it will burn up due to a low supply rail,
so you might have to add UVLO circuitry. BJTs, being current
driven devices (although they model better as voltage driven , but
that's another discussion) don't generally have that issue since
base current scales with a low supply rail unless you go nuts with
low B-E resistance.

> Type depends on drive voltage.
> The beautiful but not too common in the US CES2310 that I
> mentioned a while ago would do it with ease.

Ideal, aside from the detail that they're unavailable. ;-)

{Quote hidden}

Might be a feature if you want it to switch very slowly, provided it
doesn't kill the output transistor in the process. SOA, & Pd.

> A speedup cap and some R tailoring could make this very fast
> if required (for some values of very).

And some values of "could". ;-) Think about it-- this is not
particularly easy to do reliably without making or buying a full-fledged
gate driver. The level shifting stage has asymmetric output impedance.

A couple of mA gate drive will make it pretty fast though (microsecond
range) for a *small* p-channel MOSFET (a couple of nC gate charge).

--

Best regards,
Spehro Pefhany
--
"it's the network..."                          "The Journey is the reward"
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2008\06\12@102957 by olin piclist

face picon face
Spehro Pefhany wrote:
> Sure, at the expense of greatly increasing the Pd of Q1

But Q1 is only seeing the base current of Q2 plus the 750mV or so drop
accross R2, so this would be well within its capabilities.  Let's say you
want to make sure Q2 is solidly on (and perhaps use a differnt transistor)
and give it 1/20 the base current of the maximum collector current.  250mA /
20 = 12.5mA.  Let's say R2 is 1Kohms, so that's another 750uA for a total of
13.3mA.  When on, Q1 would have about 7V accross it, for a dissipation of
93mW.  That's well within the safe range of a MMBT4401.

> and exposing
> the PIC to damage from transients on the +12V power rail.

I don't recall Neil saying anything about transients or automotive use.  But
in any case Q1 is operating as a controlled current sink when on, so can
take momentary collector voltages up to its max collector breakdown spec,
which is around 40V if I remember right.  None of this is going to get back
to the PIC.


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2008\06\12@104104 by olin piclist

face picon face
Apptech wrote:
>>>> R1 1K 1/4W
>>>> R2 10K
>>>> R3 6K8
>>>> NPN: 2N/MMBT4401
>>>> PNP: 2N/MMBT4403
>
> I took it to mean (possibly wrongly) that the PNP needed to
> be in saturation or close to it.

Yes, that's exactly what I meant.

> Without looking back I don't know if the 250 mA was
> specified when those resistor values were given.

It was specified by the OP.

> As will be
> seen below, the circuit is in trouble as shown at 250 mA
> with that transistor and resistors.

No, you flipped R2 and R1 apparently.  The base current is 12V accross R1
(1Kohm) minus a B-E drop and saturated C-E drop.  Figuring 750mV for B-E
drop of Q2 and 200mV for the saturated on voltage of Q1, that gives a base
current of (12V - 750mV - 200mV) / 1Kohm = 11V / 1Kohm = 11mA.  At 250mA max
load current, Q2 must have a gain of at least 250mZ / 11mA = 23, which is
quite safe for a MMBT4403 at 250mA.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2008\06\12@113450 by Apptech

face
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face
> No, you flipped R2 and R1 apparently.

OK. That makes more sense.

{Quote hidden}

You stil have to watch the saturation, alas.
It is quite high.
You could get 250 mW or so Pd (graph reading).
Tja is high for the SMD pkg.
Maybe OK with lots of PCB copper.

A highside FET like the FDN338 would be happier but the
bipolar will work.



       Russell

2008\06\18@120253 by Martin

face
flavicon
face
Hey Neil,
I've been trying to send you an email privately but I don't have your
non-piclist email address anymore. Could you send me an email?
Thanks,
Martin


PicDude wrote:
> That's pretty much where I started (except I put R2 on the NPN collector,
> instead of the PNP base), but thought I could simplify it with one
> transistor.  I hadn't realized the protection diodes would get in the way.
> I need to do this on 8 outputs simultaneously, so RA4 is not the full answer
> here.
>
> Cheers,
> -Neil.
>

2008\06\24@232033 by PicDude

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face

FWIW, I went back to the original 2-transistor design (pretty much Bob's
circuit).  Guess I couldn't get away with the simplification  :( but I did
manage to re-layout the board tighter and fit it in the same enclosure I
originally intended.

The load btw is strings of LED's for automobile taillights that I built for
someone.  The taillights will be sequenced in 3 steps for the turn
indicators.  Current on this application is ~80mA per output, but I may use
this for someone else in the future, so the 250mA capability is an
conservative overhead.

Cheers,
-Neil.





PicDude wrote:
{Quote hidden}

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2008\06\25@031018 by Vasile Surducan

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On 6/12/08, Apptech <TakeThisOuTapptechEraseMEspamspam_OUTparadise.net.nz> wrote:
> >> >> The load will be up to ~250mA, btw.
> >> >
> >> > Use Bob's method only. Eg.
> >> >
> >> > R1 1K 1/4W
> >> > R2 10K
> >> > R3 6K8
> >> > NPN: 2N/MMBT4401
> >> > PNP: 2N/MMBT4403

Only for the experts in analogic computation which has answered to
this subject.
There is a methode which is able to drive a +140V directly from a PIC
pin , without any NPN. Any Nixie guys here ?
:)

greetings,
Vasile

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