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'[EE]: 5V-0V logic controlling 13V supply'
2002\11\19@050026 by tanwh

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I wish to control a 13V dc supply with a digital logic high 5V. How to
achieve this purpose by using a general purpose transistor?

Thanks in advance.

WH Tan

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2002\11\19@073001 by Roman Black

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WH Tan wrote:
>
> I wish to control a 13V dc supply with a digital logic high 5V. How to
> achieve this purpose by using a general purpose transistor?


One PNP transistor, (high side) and 3 resistors.
Connect the resistors as a 3 section voltage
divider so that when digital pin is at 5v the
voltage across B-E is only about 0.4v. When the
digital pin goes low the voltage B-E rises to
about 0.8v and PNP turns on. :o)
-Roman

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2002\11\19@074453 by Vasile Surducan

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On Tue, 19 Nov 2002, Roman Black wrote:

> WH Tan wrote:
> >
> > I wish to control a 13V dc supply with a digital logic high 5V. How to
> > achieve this purpose by using a general purpose transistor?
>
>
> One PNP transistor, (high side) and 3 resistors.
> Connect the resistors as a 3 section voltage
> divider so that when digital pin is at 5v the
> voltage across B-E is only about 0.4v. When the
> digital pin goes low the voltage B-E rises to
> about 0.8v and PNP turns on. :o)
> -Roman
>
 Nice, but not recomandable for high curents through PNP.
 Old fashion: a npn transistor or an open collector gate with 15V Vce (
unfortunately the open drain ra4 style is out of voltage spec ) which
control a power NPN ( darlinghton if necessary ). A resistor across the BC
of power transistor keep the transistor on, the open colector controlling the
base of the same transistor shut down the whole kk when is on.
Everything can be replaced with a 4 terminal controlled 12V stab with
reference adjustment floated with 1V upper the ground using a
simple resistor.

Eagle users on the list ?

by, Vasile

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2002\11\19@075740 by Olin Lathrop

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>   Old fashion: a npn transistor or an open collector gate with 15V Vce (
> unfortunately the open drain ra4 style is out of voltage spec ) which
> control a power NPN ( darlinghton if necessary ). A resistor across the
BC
> of power transistor keep the transistor on, the open colector
controlling the
> base of the same transistor shut down the whole kk when is on.
> Everything can be replaced with a 4 terminal controlled 12V stab with
> reference adjustment floated with 1V upper the ground using a
> simple resistor.

I don't like this because it results in higher voltage drop when on and
will require some current just to keep off.


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2002\11\19@080813 by Vasile Surducan
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On Tue, 19 Nov 2002, Olin Lathrop wrote:

> >   Old fashion: a npn transistor or an open collector gate with 15V Vce (
> > unfortunately the open drain ra4 style is out of voltage spec ) which
> > control a power NPN ( darlinghton if necessary ). A resistor across the
> BC
> > of power transistor keep the transistor on, the open colector
> controlling the
> > base of the same transistor shut down the whole kk when is on.
> > Everything can be replaced with a 4 terminal controlled 12V stab with
> > reference adjustment floated with 1V upper the ground using a
> > simple resistor.
>
> I don't like this because it results in higher voltage drop when on and
> will require some current just to keep off.
>
 True, nothing it's perfect.
 For 1A load, and a darlington ( let say h21e = 100 ) will be 10mA
on command. It's not very huge...

Have a nice day, mine it's over.

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2002\11\19@090054 by Olin Lathrop

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>   For 1A load, and a darlington ( let say h21e = 100 ) will be 10mA
> on command. It's not very huge...

First, that's only the on current into the base of the darlington.  The
resistor from the +V supply to this base needs to provide this 10mA with
as little voltage drop as possible, because any voltage there appears
directly accross the pass transistor.  That will therefore cause a much
larger current when the base is pulled to near 0V to shut off darlington.

Second, the darlington will have significant voltage drop even if the base
were tied directly to the supply.  1.5V drop x 1A = 1.5 watts.  That's
enough to require a heat sink with most packages.

Let's say you spend 500mV accross the base resistor to get the 10mA base
current.  500mV / 10mA = 50 ohms.  15V / 50 ohms = 300mA to pull the base
to 0V to turn off the supply.  15V * 300mA = 4.5 watts just to keep the
supply off!  At the same time the drop accross the darlington when on is
2V (1.5 for the two B-E junctions plus 500mV for the base resistor).  It
will dissipate 2W when on and drop the 15V supply to a 13V supply.

I like Roman's single PNP for the quick and dirty method.  Use a small NPN
to drive the PNP base for the next level, or a P channel FET if voltage
drop and efficiency are very important.


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2002\11\20@033324 by Vasile Surducan

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On Tue, 19 Nov 2002, Olin Lathrop wrote:

> >   For 1A load, and a darlington ( let say h21e = 100 ) will be 10mA
> > on command. It's not very huge...
>
> First, that's only the on current into the base of the darlington.  The
> resistor from the +V supply to this base needs to provide this 10mA with
> as little voltage drop as possible, because any voltage there appears
> directly accross the pass transistor.  That will therefore cause a much
> larger current when the base is pulled to near 0V to shut off darlington.
>
> Second, the darlington will have significant voltage drop even if the base
> were tied directly to the supply.  1.5V drop x 1A = 1.5 watts.  That's
> enough to require a heat sink with most packages.
>
> Let's say you spend 500mV accross the base resistor to get the 10mA base
> current.  500mV / 10mA = 50 ohms.  15V / 50 ohms = 300mA to pull the base
> to 0V to turn off the supply.  15V * 300mA = 4.5 watts just to keep the
> supply off!  At the same time the drop accross the darlington when on is
> 2V (1.5 for the two B-E junctions plus 500mV for the base resistor).  It
> will dissipate 2W when on and drop the 15V supply to a 13V supply.

 Olin dear,

 We have a romanian word : "unde dai si unde crapa !"
 more or less in english: "who you knock and who say ouch !"

 You haven't understand me ( as usual... ) The darlington is NPN, it has
*no* BE resistor, the resistor is connected from colector to base and it
has a large enough value , the colector is connected to +13V, the emiter
is connected to the load. Base of darlington is driven by an open colector
gate or just a npn to the ground. A low voltage ( zero ) on darlington base
means a zero voltage on the load.  I suggest you ( only if you have the
pleasure ) to reconsider all your computation from the next sentence :

" At the same time the drop accross the darlington when on is
 2V (1.5 for the two B-E junctions plus 500mV for the base resistor).  It
 will dissipate 2W when on and drop the 15V supply to a 13V supply. "

Vbe has nothing to do with Vce, so also with the voltage drop across the
transistor... if the electronics bases are the same in the whole world
( I'm not sure right now ... :) )
Also the Vce sat at 1A is not 1.5V only for "bulistors" ( bad-transistors
) a more resonable value is below the 0.5V

And remember, I don't want to argue, the greater fights on the list are
because of not-understandings. But I don't accept mistakes of funny
computations.

best regards,
Vasile

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2002\11\20@040520 by Alan B. Pearce

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>  You haven't understand me ( as usual... ) The darlington is NPN, it has
>*no* BE resistor, the resistor is connected from colector to base and it

Is this darlington made out of two discrete transistors, or is it a single
package darlington?

The reason I ask is that every single package darlington I have come across
has internal BE resistors, wether it is PNP or NPN.

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2002\11\20@044401 by tanwh

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> One PNP transistor, (high side) and 3 resistors.
> Connect the resistors as a 3 section voltage
> divider so that when digital pin is at 5v the
> voltage across B-E is only about 0.4v. When the
> digital pin goes low the voltage B-E rises to
> about 0.8v and PNP turns on. :o)
> -Roman


Thanks for your reply.
But I can't understand what you means '3 section voltage divider'. Could you
please point me how to connect the resistors.

WH Tan

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2002\11\20@070740 by Roman Black

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WH Tan wrote:
>
> > One PNP transistor, (high side) and 3 resistors.
> > Connect the resistors as a 3 section voltage
> > divider
>
> Thanks for your reply.
> But I can't understand what you means '3 section voltage divider'. Could you
> please point me how to connect the resistors.


OK. :o)
Here it is, the resistor divider is set so that;

digital pin=5v,  B-E voltage=0.5v,   PNP=OFF
digital pin=0v,  B-E voltage=0.83v,  PNP=ON

It has a couple of minor drawbacks, it wastes some
current through the 3R divider even when the PNP
is off, and if you need to switch significant current
you need to lower all the resistor values or maybe
use a darlington PNP.

But for a very simple low power solution (like a
way to switch programming Vpp 13v) it works fine
and saves one transistor.
:o)
-Roman


+13v -------------------------------*---------*----
                                   |         |
                            0.5v   R1        |
                         560 ohm   |         E
                                   *-------B     PNP
                                   |         C
                                   |         |
                                   R2        |
digital signal               8k2   | 7.6v    '----  switched
(or PIC ouput pin)                 |                13v out
                                   |
       *---------------------------*
  (0v or 5v signal)                |
                                   |
                             5k6   R3
                                   | 5v
                                   |
gnd  -------------------------------*--------------

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2002\11\21@010432 by tanwh

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I get it now. Thank you very much.

For those who have reply suggesting a darlington TR, I really appreciate
your time. Roman is right, darlington TR is unnecessary since I am using it
for turning on a Vpp supply to a PIC.


WH Tan


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