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What is the proper way to turn off a positive voltage supply with a NPN transistor?

M

Mr. J D

Jan 1, 1970
0
I want to turn off a load that requires about 200mA. I am only familiar
with the common emitter setup for a NPN transistor, which cant be used
to turn off a positive voltage supply. I am unsure of the proper
technique for the other setup that allows for the control of a positive
voltage. Thanks.
 
P

PeteS

Jan 1, 1970
0
Mr. J D said:
I want to turn off a load that requires about 200mA. I am only familiar
with the common emitter setup for a NPN transistor, which cant be used
to turn off a positive voltage supply. I am unsure of the proper
technique for the other setup that allows for the control of a positive
voltage. Thanks.

The simplest way to control the positive rail is with a PNP.

Emitter to positive rail, base to control signal via a resistor (a
typical value is [V+ - 0.5] / [Iload / 50] but it depends on the
transistor - check Hfe / DC current gain) and collector to positive
side of load.

Note it is just as easy to use a P-channel FET for this, with perhaps
better results due to lower voltage drop across the transistor (at
least at 200mA). There are some really nice P-Channel devices around
nowadays with Rds[on] of 0.03 ohm with Vgs = -4V.

Not knowing the value of your positive rail makes it difficult to
specify anything particular.

Unless you are willing to sacrifice about 1V or so of your positive
rail, you can't do this with an NPN - there would have to be extra
circuitry at that.

Cheers

PeteS
 
J

John Fields

Jan 1, 1970
0
I want to turn off a load that requires about 200mA. I am only familiar
with the common emitter setup for a NPN transistor, which cant be used
to turn off a positive voltage supply. I am unsure of the proper
technique for the other setup that allows for the control of a positive
voltage. Thanks.


---

If you have a load which is tied to the positive supply you can use
an NPN as a low-side switch like this: (View in Courier)

+V
|
[LOAD]
|
___ C
ON/OFF>---[R]---B NPN
E
|
GND



If you have a load which is tied to ground, you can use a PNP as a
high-side switch like this:


+V
|
|
__ E
ON/OFF>---[R]--B PNP
C
|
[LOAD]
|
GND


or like this:


+V +V
| |
[R] |
| E
+--[R]--B PNP
| C
___ C |
ON/OFF---[R]---B NPN [LOAD]
E |
| |
GND GND
 
J

Jonathan Kirwan

Jan 1, 1970
0
I want to turn off a load that requires about 200mA. I am only familiar
with the common emitter setup for a NPN transistor, which cant be used
to turn off a positive voltage supply. I am unsure of the proper
technique for the other setup that allows for the control of a positive
voltage. Thanks.


---

If you have a load which is tied to the positive supply you can use
an NPN as a low-side switch like this: (View in Courier)

+V
|
[LOAD]
|
___ C
ON/OFF>---[R]---B NPN
E
|
GND



If you have a load which is tied to ground, you can use a PNP as a
high-side switch like this:


+V
|
|
__ E
ON/OFF>---[R]--B PNP
C
|
[LOAD]
|
GND


or like this:


+V +V
| |
[R] |
| E
+--[R]--B PNP
| C
___ C |
ON/OFF---[R]---B NPN [LOAD]
E |
| |
GND GND

Or else:
: +V
: |
: \
: / R1 +V
: \ |
: / |
: | |<e Q2
: +---------| PNP
: | |\c
: | |
: ___ |/c Q1 |
: ON/OFF--------| NPN \
: |>e / LOAD
: | \
: | /
: \ |
: / R2 |
: \ gnd
: /
: |
: gnd

Your +V supply for the load can be higher than your ON/OFF switching
voltage supply, too.

R1 is mostly there to make sure that Q2 goes off and stays off when Q1
becomes inactive. There are some small capacitances to overcome and,
if it weren't present, the node it connects to would be fairly high
impedance and may then respond to things like touching your finger to
it. It needs to be smaller in value if you plan to run the switch
faster, but for most uses it can be something like 5k, 10k, 22k or
perhaps even 47k. The value isn't usually usually a critical one and
is often in the range I mentioned. Q1 has to pull R1 down, though,
when Q1 goes active. So making R1 smaller means that Q1 has to supply
more active current. Making R1 larger reduces this problem, but means
that switching gets slower and the node becomes more sensitive to
things like touching it. 10k is reasonable for R1.

R2 needs to be set so as to supply the necessary base current drive
for Q2, when active, _and_ any current that R1 requires at that time.
The base current drive needed by Q2 will usually be somewhere in the
range of 1/10th to 1/50th of the LOAD current. But planning on more
current is safer than planning on less, so the figure of 1/10th is a
very often-used estimate. Also, since the base-emitter voltage of Q2
will be something between .65V and perhaps .9V (with .8V being a
reasonable guess lacking better knowledge), this means that R1's
active current will be about .8V/R1. With a value for R1 in mind, you
can compute this. These two currents are added together, so that you
have I(R2) = I(LOAD)/10 + 0.8V/R1. To compute R2, though, you also
need to know the voltage across it. To get that, just estimate the
control voltage you are using (say V(on) is 3.3V or 5V or whatever it
is) and then subtract about 0.7V from it. That will be the voltage at
the emitter of Q1, which is the voltage across R2. So you then have
the equation, R2 = (V(on) - 0.7) / (I(LOAD)/10 + 0.8V/R1). Finding a
standard resistor value nearby this value is the next step. It can be
a little smaller (driving Q2 harder) or it can be a little larger
(eating into some of the conservative nature of the 1/10th estimate of
Q2's base drive beta) and things should work okay.

There is also another approach that uses the same number of parts,
too. It looks like:
: +V
: |
: \
: / R1 +V
: \ |
: / |
: | |<e Q2
: +---------| PNP
: | |\c
: | |
: |/c Q1 |
: +V----| NPN \
: |>e / LOAD
: | \
: | /
: \ |
: / R2 |
: \ gnd
: /
: __ |
: ON/OFF----------'

This inverts the sense of the control line and it will require more
current sinking capability for your control line driver. Which brings
up another point, you need to know what currents your control line
driver is capable/comfortable with. And if your LOAD requires a lot
of current, this last complementary driver circuit may require too
much from your control line driver and need yet another BJT to help
out.

Anyway, keep in mind that I'm not a designer and cannot give you a
professional quality comparison between the circuits I'm suggesting
and the one that John suggested at the bottom of his post (included
above.) One obvious difference is that my suggestion uses one less
resistor to get the job done. I can think of a few other differences
to comment on, but I'd rather let someone well informed make those
points.

Jon
 
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