Transistor switch

A

Andy

Guest
Will this work?


+v/\ ____ /\+v
| | | |
in|___|reg |____out |
|____| |
| |
|______/\/\_____|
com|
|
|c
__/\/\___/
\.e
|
|
_____
___
_


My theory is that if i present a voltage to the transistors input
resistor it will enable the output of the regulator and removing the
the voltage will dissable it?

Thanx Andy.
 
Andy wrote:
Will this work?


+v/\ ____ /\+v
| | | |
in|___|reg |____out |
|____| |
| |
|______/\/\_____|
com|
|
|c
__/\/\___/
\.e
|
|
_____
___
_


My theory is that if i present a voltage to the transistors input
resistor it will enable the output of the regulator and removing the
the voltage will dissable it?

Thanx Andy.
Click to expand...
Most regulators work by maintaining a fixed voltage difference between
the 'out' and 'com' terminals. At best this circuit will allow the
voltage on the 'out' pin to float up somewhere between the intended
voltage and +v (probably one or two diode drops below +v). At worst
it'll take out the regulator in the process.

There are 8-pin linear regulators with a shutoff pin, why are you
messing with this?

If you _must_ use a 3-terminal regulator, precede it with a PNP or
P-channel FET and pull the base/gate low to turn things on.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
"Andy" <maccyd10@hotmail.com> wrote in message
news:f2bcd835.0404270806.78921d6@posting.google.com...
Will this work?
Click to expand...
No, it won't. But you can look at the sample circuits datasheet for an
LM317 to see how to do it correctly.
 
On a sunny day (27 Apr 2004 09:06:07 -0700) it happened maccyd10@hotmail.com
(Andy) wrote in <f2bcd835.0404270806.78921d6@posting.google.com>:

Will this work?


+v/\ ____ /\+v
| | | |
in|___|reg |____out |
|____| |
| |
|______/\/\_____|
com|
|
|c
__/\/\___/
\.e
|
|
_____
___
_


My theory is that if i present a voltage to the transistors input
resistor it will enable the output of the regulator and removing the
the voltage will dissable it?

Thanx Andy.
It depens on what regulator that is.
Click to expand...
In a say 7805 it will partly work, (output will be low I think).
Better is to have a PNP in series before the regulator (collector
on regulator), and drive that on the base via a resistor to limit the
base current with that transistor you show.
Such a series switch drops only about 0.2 volt or so, and switches of
100%.
Maybe add a base -emittor resistor too.
JP
 
"Andy" <maccyd10@hotmail.com> wrote in message
news:f2bcd835.0404270806.78921d6@posting.google.com...
Will this work?

+v/\ ____ /\+v
| | | |
in|___|reg |____out |
|____| |
| |
|______/\/\_____|
com|
|
|c
__/\/\___/
\.e
|
|
_____
___
_


My theory is that if i present a voltage to the transistors input
resistor it will enable the output of the regulator and removing the
the voltage will dissable it?
Click to expand...
It will NOT switch the regulator off, it is more likely to make it act as if
shorted.
For a fixed regulator, as I believe that you are thinking of here, the
output is it's rated voltage above the common pin.

For example, take a 7812...
With the common pin grounded, you have 12 volts out.
Put a diode in series with that common lead, so the common pin is held at
0.7v, and the output will be 12.7 volts.
If you leave the common pin floating (as with that transistor off), the
common pin will be at the input voltage and the output will then match that.
This will be like replacing the reg with a wire link... possibly damaging
the circuit relying on that regulator.
 
On a sunny day (Fri, 30 Apr 2004 06:03:18 +0100) it happened "L. Fiar" <_@_._>
wrote in <4091f690.0@entanet>:

For example, take a 7812...
With the common pin grounded, you have 12 volts out.
Put a diode in series with that common lead, so the common pin is held at
0.7v, and the output will be 12.7 volts.
If you leave the common pin floating (as with that transistor off), the
common pin will be at the input voltage and the output will then match that.
This will be like replacing the reg with a wire link... possibly damaging
the circuit relying on that regulator.
Not my experience, I noticed disconnecting ground on a 7805 dropped the voltage.
Click to expand...
The diode thing is corect, it will raise thh output by 0.7 V or so.
Think I looked it up once in the diagram of that chip.
JP
 
"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c6u3rq$1d5$1@news.wplus.net...
Not my experience, I noticed disconnecting ground on a 7805 dropped the
voltage.
The diode thing is corect, it will raise thh output by 0.7 V or so.
Click to expand...
It is often used by manufacturers. One point is where the output is followed
by some other semiconductor like a diode or transistor, as this drops 0.7
volts. The diode in the regulator will compensate for this.
It can also be raised by more using either a Zener or LED.

Think I looked it up once in the diagram of that chip.
Click to expand...
I have just looked in a couple of TI and NS data books, and both 78 series
designs have a path from the input to output without the common being
connected.
Neither appear to show any circuits where the common would be allowed to
float, although they do show the use of a voltage on the common to increase
the output voltage and a design to use the device as a current regulator,
but the common is still connected and not floating.

It is possible that, in the suggested situation, the output voltage will be
dependant upon the load at the output.
The exact way a requlator will act in an abnormal situation will depend upon
the manufacturers design. One may act different to another. As it is not an
accepted way to connect the device, nothing can be guaranteed to work either
in the short or long term.
Even if something seems to work, without long term tests, you cannot be sure
of whether the device will hold up over time, and whether it will work with
all such devices.


LF.
 
"Andy" <maccyd10@hotmail.com> wrote in message
news:f2bcd835.0404270806.78921d6@posting.google.com...
Will this work?


+v/\ ____ /\+v
| | | |
in|___|reg |____out |
|____| |
| |
|______/\/\_____|
com|
|
|c
__/\/\___/
\.e
|
|
_____
___
_
Click to expand...

I am not used to ASCII drawing, but here goes...

Vin
|
|
|
| / c
___/\/\/\___|/ NPN, rated at same
|\ current as regulator.
| \ e
|
| _____
|_____| |______ Vout
|_____|
|
|
|
ŻŻŻŻŻ
ŻŻŻ
Ż

This totally cuts the supply to the reg, but it does not rely on some
uncertain or unreliable capabilities of a regulator.
The transistor could be placed in the regulator output, but you would then
want a diode in the regulator common lead to compensate for the drop from
the transistor.


LF.
 
On a sunny day (Sun, 2 May 2004 15:28:26 +0100) it happened "L. Fiar" <_@_._>
wrote in <4095409c.0@entanet>:

"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c6u3rq$1d5$1@news.wplus.net...

Not my experience, I noticed disconnecting ground on a 7805 dropped the
voltage.
The diode thing is corect, it will raise thh output by 0.7 V or so.

It is often used by manufacturers. One point is where the output is followed
by some other semiconductor like a diode or transistor, as this drops 0.7
volts. The diode in the regulator will compensate for this.
It can also be raised by more using either a Zener or LED.

Think I looked it up once in the diagram of that chip.

I have just looked in a couple of TI and NS data books, and both 78 series
designs have a path from the input to output without the common being
connected.
Neither appear to show any circuits where the common would be allowed to
float, although they do show the use of a voltage on the common to increase
the output voltage and a design to use the device as a current regulator,
but the common is still connected and not floating.

It is possible that, in the suggested situation, the output voltage will be
dependant upon the load at the output.
The exact way a requlator will act in an abnormal situation will depend upon
the manufacturers design. One may act different to another. As it is not an
accepted way to connect the device, nothing can be guaranteed to work either
in the short or long term.
Even if something seems to work, without long term tests, you cannot be sure
of whether the device will hold up over time, and whether it will work with
all such devices.


LF.
Well I know the voltage drops because i t was an error in a circuit, i found
Click to expand...
the 5 V was low, and it turned out to be the lose middle pen on a 7805.
i have used the diode several times to get 5.7V.
It seems this chip needs 'some' current to flow to ground to open up the series
regulator.
JP
 
"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c73nod$m41$1@news.epidc.co.kr...
Well I know the voltage drops because i t was an error in a circuit, i
found the 5 V was low, and it turned out to be the lose middle pen
on a 7805.
i have used the diode several times to get 5.7V.
It seems this chip needs 'some' current to flow to ground to open up the
series regulator.
Click to expand...
I think you missed my point. Let's put it another way...
I just asked someone if they voted for Blair at the last election - and they
said no. So, nobody voted for Blair and the election was rigged?

You may think that electronics is different, that components are designed to
be identical, but that would be wrong.
Components are designed to do a specific function within a stated spec. Some
may work beyond that spec, some far beyond it, but nothing is guaranteed
outside of the stated spec.
Two items with the same part number, even from the same manufacturer, are
not identical unless sold as a matching pair.

As an example, a manufacturer designed a circuit for a specific function and
with set specifications. They tested the circuit on prototypes, and all
worked within the spec.
On the production line, 50% were failing test... although the components
were as on the prototypes.
The cause was that the R&D tech had designed the circuit using a
semiconductor in a way which the manufacturer had not intended. Some worked
fine, just as the prototypes, others failed miserably.
As it happened, the items had to be out that night to fill an order. Company
rules meant that component types or values could not be changed from the
design without approval from higher up, such as the technical director...
and the office staff had gone home for the day.
What a bummer.
I was working until after 10pm getting as many units within spec as
possible, without changing any component outside of the range aready allowed
on the design. After that, the design was changed.


I just tested a MV 7812 and an NS LM240T12 (also marked 7812), both with the
common
pins floating.
The MV 7812 output sat at 1 volt below the input. This was checked up to
20 volts and a 600mA load. So, with 20v in, there was 19v out.
The NS LM340T12 output was also 19v with an input of 20v. However, the
output
was nearer to the input as this input got lower. Also, a load of 600mA
dropped the output voltage... but it was still above 12 volts.

I only tested one of each, which is not a good sample, so I would expect
slightly different readings from different batches of the same component,
and possibly big differences with some other manufacturers.


LF.
 
On a sunny day (Wed, 5 May 2004 07:15:39 +0100) it happened "L. Fiar" <_@_._>
wrote in <409887e3.0@entanet>:
LF.
But did you test 7805?
Click to expand...
JP
 
"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c7bn9n$1lv$1@news.epidc.co.kr...
But did you test 7805?
Click to expand...
Did the OP specify a 7805?

With the common pin floating...

7805
Unloaded:
9v in = 8.03v out.
10v in = 9.01v out.
11v in = 10.01v out.
Loaded (see note below):
9v in = 5.4v out @ 270mA load.
10v in = 6.4v out @ 330mA load.
NOTE: When under load, the regulator temperature increases. This increase in
temperature caused an increase in output voltage. Once the IC temperature
was slightly raised, the unloaded readings were also higher.

78L05
Unloaded:
9v in = 8.27v out.
10v in = 9.25v out.
11v in = 10.23v out.
Loaded (see note below):
9v in = 4.4v out @ 20mA load.
10v in = 5.4v out @ 20mA load.
11v in = 6.38v @ 20mA load.
Temperature increases can alter the output, although temperature increase
may be less of a problem at the lower currents used with this device.

7812 (motorolla)
Unloaded:
14v in = 13.43v out.
15v in = 14.43v out.
16v in = 15.42v out.
Loaded (see note below):
14v in = 12.25v out @ 300mA load.
15v in = 13.34v out @ 300mA load.
16v in = 14.31 out @ 300mA load.
The output voltage does increase with temperature, but not as much as the
tested 7805 (different manufacturer). The results were also different to the
other manufacturers 7812 that I tested.

As it can be seen from the above, the result will depend upon manufacturer,
type, input voltage and output load. They may also vary between batches from
the same manufacturer.
The results of one fault situation on one particular component is not a good
enough sample to base circuit design on. What you probably saw was not the
device being switched off, but the load being higher than that one
manufacturers device could deliver under that fault condition. Another
device, manufacturer, input voltage or output load, and the readings could
have been much different.


LF.
 
On a sunny day (Fri, 7 May 2004 07:29:08 +0100) it happened "Braěnbuster"
<braěnbuster@lineone.net> wrote in <409b2e8d.0@entanet>:

"Jan Panteltje" <pNaonStpealmtje@yahoo.com> wrote in message
news:c7bn9n$1lv$1@news.epidc.co.kr...

But did you test 7805?

Did the OP specify a 7805?

With the common pin floating...

7805
Unloaded:
9v in = 8.03v out.
10v in = 9.01v out.
11v in = 10.01v out.
Loaded (see note below):
9v in = 5.4v out @ 270mA load.
10v in = 6.4v out @ 330mA load.
NOTE: When under load, the regulator temperature increases. This increase in
temperature caused an increase in output voltage. Once the IC temperature
was slightly raised, the unloaded readings were also higher.

78L05
Unloaded:
9v in = 8.27v out.
10v in = 9.25v out.
11v in = 10.23v out.
Loaded (see note below):
9v in = 4.4v out @ 20mA load.
10v in = 5.4v out @ 20mA load.
11v in = 6.38v @ 20mA load.
Temperature increases can alter the output, although temperature increase
may be less of a problem at the lower currents used with this device.

7812 (motorolla)
Unloaded:
14v in = 13.43v out.
15v in = 14.43v out.
16v in = 15.42v out.
Loaded (see note below):
14v in = 12.25v out @ 300mA load.
15v in = 13.34v out @ 300mA load.
16v in = 14.31 out @ 300mA load.
The output voltage does increase with temperature, but not as much as the
tested 7805 (different manufacturer). The results were also different to the
other manufacturers 7812 that I tested.

As it can be seen from the above, the result will depend upon manufacturer,
type, input voltage and output load. They may also vary between batches from
the same manufacturer.
The results of one fault situation on one particular component is not a good
enough sample to base circuit design on. What you probably saw was not the
device being switched off, but the load being higher than that one
manufacturers device could deliver under that fault condition. Another
device, manufacturer, input voltage or output load, and the readings could
have been much different.


LF.
Good data! I am now very curious, and tonight I will measure some 7805,
Click to expand...
will come back on this.
JP
 
Looks like you are 100% right.
I just tested on a 7805T, what I used in the
case I thought it dropped (slightly more current capable
then a 7805), and found
input 12.49V, output 10,49V at 75 and 150 Ohm load,
when middle pin not grounded.
With middle pin grounded output is 5.00 V.
So that is important info for me, as to make sure always
double check ground on that chip.
I dunno how I measured it dropped, one would think I had
the - meter on the middle pin... but I do not think so, as
I would always use some ground PCB part, but possible.
The other thing is it had 60 TTL chips on it, that all lived,
with 9 V input, TTL would not have liked 7 V...
So I dunno.

No idea, but at least we now know a new 78xx is not safe with
no earth.
Thank you for going through the trouble of testing these, and
pointing out my error.
Looks like the OP will need a series switch before the regulator.
JP
 
Jan Panteltje wrote:
Looks like you are 100% right.
I just tested on a 7805T, what I used in the
case I thought it dropped (slightly more current capable
then a 7805), and found
input 12.49V, output 10,49V at 75 and 150 Ohm load,
when middle pin not grounded.
With middle pin grounded output is 5.00 V.
So that is important info for me, as to make sure always
double check ground on that chip.
I dunno how I measured it dropped, one would think I had
the - meter on the middle pin... but I do not think so, as
I would always use some ground PCB part, but possible.
The other thing is it had 60 TTL chips on it, that all lived,
with 9 V input, TTL would not have liked 7 V...
So I dunno.

No idea, but at least we now know a new 78xx is not safe with
no earth.
Thank you for going through the trouble of testing these, and
pointing out my error.
Looks like the OP will need a series switch before the regulator.
JP
Click to expand...
This problem with the output floating high is well documented in the
earliest National Semiconductor application notes and linear briefs on
the the 3-terminal regulators. It is the one fault condition in a live
insertion application against which there is little protection- where
the unregulated voltage makes connection before the GND-or ADJ- output
charges capacitor - then middle pin connects, discharges output through
low current regulator paths and blows it.
 
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:409CD540.3040503@nospam.com...
This problem with the output floating high is well documented in the
earliest National Semiconductor application notes and linear briefs on
the the 3-terminal regulators.
Click to expand...
I cannot find it in the app notes I have, do you have a note number for it?
What I did just find, was the NS suggested way to switch off a supply with
one of these regulators.

Application note AN-103, figure 22.
Similar to my other post, but they use a PNP transistor with an NPN
transistor controlling the base. Control signal to the NPN transistor.

I found it in their Linear Applications Handbook, but I expect it will be
available on their Web site.
http://cache.national.com/apnotes/
 
L. Fiar wrote...
Fred Bloggs wrote ...

This problem with the output floating high is well documented in the
earliest National Semiconductor application notes and linear briefs
on the the 3-terminal regulators.

I cannot find it in the app notes I have, do you have a note number
for it? What I did just find, was the NS suggested way to switch off
a supply with one of these regulators.

Application note AN-103, figure 22.
Similar to my other post, but they use a PNP transistor with an NPN
transistor controlling the base. Control signal to the NPN transistor.
Click to expand...
I'm not sure what Fred was referring to either, but the method described
in fig 22 of the app note is a standard brute-force way of high-side
switching off anything, whether a heater or a three-terminal regulator.

But these days one would prefer using a P-channel MOSFET, in order to
avoid the high base current used by a PNP pass transistor, or the high
voltage drop of a Darlington transistor. Speaking of a general-purpose
form of FET circuit, the issue of gate drive deserves special attention.
Given that a wide range of input voltages may be used, it's appealing
to drive the FET gate with a switched current source, across a resistor,
creating a fixed gate ON voltage, like this 10V gate-drive circuit:

.. IRF9Z24 _____
.. supply in IRF9Z34 | |
.. o--------+----+-A ,-----| |------
.. | _|_|_|_ |_____|
.. 2.7k ,---- |
.. | | |
.. 5V +----' gnd
.. logic |
.. enable |/
.. o------| any small
.. |\V npn BJT
.. |
.. 1.2k
.. |
.. gnd

Substitute a P-channel FET of your choice. The two parts listed are 60V,
0.28 and 0.14-ohm FETs, Digi-Key qty-10 priced at $0.79 and $1.15 resp.

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)
 
L. Fiar wrote:
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:409CD540.3040503@nospam.com...

This problem with the output floating high is well documented in the
earliest National Semiconductor application notes and linear briefs on
the the 3-terminal regulators.


I cannot find it in the app notes I have, do you have a note number for it?
What I did just find, was the NS suggested way to switch off a supply with
one of these regulators.

Application note AN-103, figure 22.
Similar to my other post, but they use a PNP transistor with an NPN
transistor controlling the base. Control signal to the NPN transistor.

I found it in their Linear Applications Handbook, but I expect it will be
available on their Web site.
http://cache.national.com/apnotes/
Click to expand...
See the last paragraph in section DIODES PROTECT AGAINST CAPACITOR
DISCHARGE, PDF page 2:
http://www.national.com/an/AN/AN-182.pdf
 
On a sunny day (11 May 2004 03:44:20 -0700) it happened Winfield Hill
<Winfield_member@newsguy.com> wrote in <c7qaq402lms@drn.newsguy.com>:

L. Fiar wrote...

Fred Bloggs wrote ...

This problem with the output floating high is well documented in the
earliest National Semiconductor application notes and linear briefs
on the the 3-terminal regulators.

I cannot find it in the app notes I have, do you have a note number
for it? What I did just find, was the NS suggested way to switch off
a supply with one of these regulators.

Application note AN-103, figure 22.
Similar to my other post, but they use a PNP transistor with an NPN
transistor controlling the base. Control signal to the NPN transistor.

I'm not sure what Fred was referring to either, but the method described
in fig 22 of the app note is a standard brute-force way of high-side
switching off anything, whether a heater or a three-terminal regulator.

But these days one would prefer using a P-channel MOSFET, in order to
avoid the high base current used by a PNP pass transistor, or the high
voltage drop of a Darlington transistor. Speaking of a general-purpose
form of FET circuit, the issue of gate drive deserves special attention.
Given that a wide range of input voltages may be used, it's appealing
to drive the FET gate with a switched current source, across a resistor,
creating a fixed gate ON voltage, like this 10V gate-drive circuit:

. IRF9Z24 _____
. supply in IRF9Z34 | |
. o--------+----+-A ,-----| |------
. | _|_|_|_ |_____|
. 2.7k ,---- |
. | | |
. 5V +----' gnd
. logic |
. enable |/
. o------| any small
. |\V npn BJT
. |
. 1.2k
. |
. gnd

Substitute a P-channel FET of your choice. The two parts listed are 60V,
0.28 and 0.14-ohm FETs, Digi-Key qty-10 priced at $0.79 and $1.15 resp.

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

Clever how you limit gate voltage with this current source :)
Click to expand...
JP
 
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:40A0C20C.90005@nospam.com...
L. Fiar wrote:
"Fred Bloggs" <nospam@nospam.com> wrote in message
news:409CD540.3040503@nospam.com...

This problem with the output floating high is well documented in the
earliest National Semiconductor application notes and linear briefs on
the the 3-terminal regulators.

I cannot find it in the app notes I have, do you have a note number for
it?

See the last paragraph in section DIODES PROTECT AGAINST CAPACITOR
DISCHARGE, PDF page 2:
http://www.national.com/an/AN/AN-182.pdf
Click to expand...
I have that application note here.
Nasty... if the ground then becomes connected, as it could with
a dry joint, then the regulator can be destroyed. Using the
switch the OP suggested could also destroy the regulator when
it is switched "on".


Thanks.
 
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