Question about using PNP as 12V switch, but driven by CMOS

J

James W

Guest
I need to switch a 12Vdc load. I'd like to use a PNP transistor, because
this would allow me to simply ground the base(though a resistor) to
turns things on. However, this circuit will generally be driven by a
microprocessor.

My understanding is that one would typically use a PNP as a top-side
switch, i.e. connect the emitter to 12V, and the collector to the load,
and finally, the load to ground.

A pullup on the base would switch the transistor OFF, and pulling the
base down through a resistor, would turn the transistor ON.

However, this means that there will be 12V on the base, and the
microprocessor is at 5V, so things are not good.

Clearly, one doesn't have this issue using a bottom-side NPN transistor,
but then one has a tougher time turning the transistor ON when the
microprocessor is not present.

(If this 'switch stuff' is confusing.. just realize that I'd like to be
albe to 'override' the microprocessor with a physical switch from time
to time)


So, is there any simple design that allows me to have a simple "ground
this pin with a switch' override of a microprocessor controlled
transistor switch for 12V?

p.s. Load current is ~150mA inductive (a coil relay to be specific)
 
James W wrote:

I need to switch a 12Vdc load. I'd like to use a PNP transistor, because
this would allow me to simply ground the base(though a resistor) to
turns things on. However, this circuit will generally be driven by a
microprocessor.

My understanding is that one would typically use a PNP as a top-side
switch, i.e. connect the emitter to 12V, and the collector to the load,
and finally, the load to ground.

A pullup on the base would switch the transistor OFF, and pulling the
base down through a resistor, would turn the transistor ON.

However, this means that there will be 12V on the base, and the
microprocessor is at 5V, so things are not good.

Clearly, one doesn't have this issue using a bottom-side NPN transistor,
but then one has a tougher time turning the transistor ON when the
microprocessor is not present.
Click to expand...
I don't see why it is more difficult, you just need to pull the base
high instead of low. Why can't you connect the base to +12V (or +5v)
through a resistor? Maybe you don't have a positive voltage where you
want the switch or something like that?

So, is there any simple design that allows me to have a simple "ground
this pin with a switch' override of a microprocessor controlled
transistor switch for 12V?
Click to expand...
Is there any reason why you can't just bypass the transistor with the
switch so that the switch directly activates the relay? I mean use an
NPN transistor between the relay and ground, and connect the switch
between the relay and ground.

If, for some reason, you can't do this you can use an NPN transistor
controlled by the microcontroller to pull the base of the PNP transistor
low.

You could also do it the other way round and have an NPN transistor
activate the relay, and pull the base high using a PNP transistor. The
PNP transistor being turned on by connecting a resistor to ground.

Gareth.

--
-----------------------------------------------------------------------
To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris
 
James W wrote:


p.s. Load current is ~150mA inductive (a coil relay to be specific)
Click to expand...
Don't forget to add a diode when switching an inductive load.

--
-----------------------------------------------------------------------
To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris
 
"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:104a242nd6g7pbe@corp.supernews.com...
I need to switch a 12Vdc load. I'd like to use a PNP transistor, because
this would allow me to simply ground the base(though a resistor) to
turns things on. However, this circuit will generally be driven by a
microprocessor.

My understanding is that one would typically use a PNP as a top-side
switch, i.e. connect the emitter to 12V, and the collector to the load,
and finally, the load to ground.

A pullup on the base would switch the transistor OFF, and pulling the
base down through a resistor, would turn the transistor ON.

However, this means that there will be 12V on the base, and the
microprocessor is at 5V, so things are not good.

Clearly, one doesn't have this issue using a bottom-side NPN transistor,
but then one has a tougher time turning the transistor ON when the
microprocessor is not present.

(If this 'switch stuff' is confusing.. just realize that I'd like to be
albe to 'override' the microprocessor with a physical switch from time
to time)


So, is there any simple design that allows me to have a simple "ground
this pin with a switch' override of a microprocessor controlled
transistor switch for 12V?

p.s. Load current is ~150mA inductive (a coil relay to be specific)
Click to expand...
Use an NPN transistor/PNP transistor combination as follows:


+-----+ VCC
.-. |
10k| | |
| | |
'-' |
| |e
+--b|
1k | |c
___ |c |
uC -|___|-b| |
|e .--|--.
| | |
| |Load |
| | |
| '--|--'
| |
+-----+ GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


uC = 5V means load is powered
uC = 0V means load isn't powered

Another way is to use a zener diode:

+---VCC----+
| |
.-. |
| |1k |
| | |
'-' |
| ___ |e
+-|___|-b|
| 1k |c
- |
^8.2 Znr |
| Load
uC +---+ |
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


Regards,
Bob Monsen
 
hmmm, this doesn't look right... There's nothing to limit the base
current through the PNP, is there?

- jim

Robert C Monsen wrote:

Use an NPN transistor/PNP transistor combination as follows:


+-----+ VCC
.-. |
10k| | |
| | |
'-' |
| |e
+--b|
1k | |c
___ |c |
uC -|___|-b| |
|e .--|--.
| | |
| |Load |
| | |
| '--|--'
| |
+-----+ GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


uC = 5V means load is powered
uC = 0V means load isn't powered
Click to expand...
In this one, seems like we might run into trouble if VCC were to rise
above 13.2V. I know I said 12V, but in fact, this is an automotive
installation, so VCC could rise to ~14 or so volts.


Another way is to use a zener diode:

+---VCC----+
| |
.-. |
| |1k |
| | |
'-' |
| ___ |e
+-|___|-b|
| 1k |c
- |
^8.2 Znr |
| Load
uC +---+ |
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


Regards,
Bob Monsen
Click to expand...
 
If I connect the base to +5, then I'll have quite a bit of base
current.. No?

Your idea of using the switch to bypass the circuit is interesting. I'll
think about that.

thanks
- jim

Gareth wrote:
James W wrote:

I need to switch a 12Vdc load. I'd like to use a PNP transistor,
because this would allow me to simply ground the base(though a
resistor) to turns things on. However, this circuit will generally be
driven by a microprocessor.

My understanding is that one would typically use a PNP as a top-side
switch, i.e. connect the emitter to 12V, and the collector to the
load, and finally, the load to ground.

A pullup on the base would switch the transistor OFF, and pulling the
base down through a resistor, would turn the transistor ON.

However, this means that there will be 12V on the base, and the
microprocessor is at 5V, so things are not good.

Clearly, one doesn't have this issue using a bottom-side NPN
transistor, but then one has a tougher time turning the transistor ON
when the microprocessor is not present.


I don't see why it is more difficult, you just need to pull the base
high instead of low. Why can't you connect the base to +12V (or +5v)
through a resistor? Maybe you don't have a positive voltage where you
want the switch or something like that?

So, is there any simple design that allows me to have a simple "ground
this pin with a switch' override of a microprocessor controlled
transistor switch for 12V?


Is there any reason why you can't just bypass the transistor with the
switch so that the switch directly activates the relay? I mean use an
NPN transistor between the relay and ground, and connect the switch
between the relay and ground.

If, for some reason, you can't do this you can use an NPN transistor
controlled by the microcontroller to pull the base of the PNP transistor
low.

You could also do it the other way round and have an NPN transistor
activate the relay, and pull the base high using a PNP transistor. The
PNP transistor being turned on by connecting a resistor to ground.

Gareth.
Click to expand...
 
"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:104amhdml3tkk76@corp.supernews.com...
hmmm, this doesn't look right... There's nothing to limit the base
current through the PNP, is there?
Click to expand...
Right, increase the 1k from the uC to the base to 47k. 1k might allow too
much current to flow through the upper PN junction. Or, you could use an
emitter resistor.

- jim

Robert C Monsen wrote:


Use an NPN transistor/PNP transistor combination as follows:


+-----+ VCC
.-. |
10k| | |
| | |
'-' |
| |e
+--b|
1k | |c
___ |c |
uC -|___|-b| |
|e .--|--.
| | |
| |Load |
| | |
| '--|--'
| |
+-----+ GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


uC = 5V means load is powered
uC = 0V means load isn't powered


In this one, seems like we might run into trouble if VCC were to rise
above 13.2V. I know I said 12V, but in fact, this is an automotive
installation, so VCC could rise to ~14 or so volts.
Click to expand...
OK. This won't work with VCC above about 13.8V. A zener that puts you in the
range of VCC to VCC - 0.7 when uC goes from 5V to 0V would be required, but
if you can't predict VCC with any accuracy, this method isn't very useful.

Another way is to use a zener diode:

+---VCC----+
| |
.-. |
| |1k |
| | |
'-' |
| ___ |e
+-|___|-b|
| 1k |c
- |
^8.2 Znr |
| Load
uC +---+ |
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


Regards,
Bob Monsen
Click to expand...
 
James W wrote:
I need to switch a 12Vdc load. I'd like to use a PNP transistor, because
this would allow me to simply ground the base(though a resistor) to
turns things on. However, this circuit will generally be driven by a
microprocessor.

My understanding is that one would typically use a PNP as a top-side
switch, i.e. connect the emitter to 12V, and the collector to the load,
and finally, the load to ground.

A pullup on the base would switch the transistor OFF, and pulling the
base down through a resistor, would turn the transistor ON.

However, this means that there will be 12V on the base, and the
microprocessor is at 5V, so things are not good.

Clearly, one doesn't have this issue using a bottom-side NPN transistor,
but then one has a tougher time turning the transistor ON when the
microprocessor is not present.

(If this 'switch stuff' is confusing.. just realize that I'd like to be
albe to 'override' the microprocessor with a physical switch from time
to time)

So, is there any simple design that allows me to have a simple "ground
this pin with a switch' override of a microprocessor controlled
transistor switch for 12V?

p.s. Load current is ~150mA inductive (a coil relay to be specific)
Click to expand...
You probably don't need any additional current gain from the level
shifting circuit to change the 0 to 5 volt swing from the micro to the
12 to 11.4 volt base swing on the PNP. So I suggest you use a second
transistor, an NPN as a level shifter on the output of the micro. Tie
the base directly to the +5 volt supply and connect the micro output
to the emitter through a current limiting resistor. Micros are
usually better at pulling down than up, so this should transfer the
micro pull down directly to the PNP base (which gets connected
directly to the NPN collector. You should still keep the PNP base to
emitter resistor, selected to drain about 10% of the current through
the NPN to make sure the PNP turns off completely. Don't forget a
diode across the inductive load to let its current run down with only
a diode drop across it when the PNP switches off. A 1N4148 or
something else smallish is fine.

A 560 ohm emitter resistor on the NPN will allow about 4v/560=7.1mA to
pass through the NPN to the base of the PNP, while a 1k base to
emitter resistor will detour about .6v/1000=.6mA leaving about 7.1mA
-.6mA= 6.5mA to drive the 150 mA load with a gain required of
150/6.5=23. Use a PNP rated for several times the 150 mA to have
plenty of gain at the expected load to get a low saturation drop.
Something rated for 600 ma or more should do fine.

The thing I like about this arrangement is that it puts no load on the
5 volt supply and the PNP base current is regulated independently of
small variations in the 12 volt supply.
--
John Popelish
 
"electricked" <no_emails_please> wrote in message
news:memdnaKlgsa9-tjdRVn-hw@comcast.com...
I'm new but I think that's what the 10K and the 1K resistors are there
for.

--Viktor
Click to expand...
No, he is right. I wasn't paying attention initially. Using a 1k resistor,
when the uC brings up the voltage to 5V, then the base of the NPN will sink
about (5 - .7)/1000 = 4.3mA. If the NPN has a beta of 100, that means it'll
sink 430mA. The PNP transistor acts like a diode from VCC to the collector
of the NPN transistor, so all 430mA (minus the 1.2mA that goes through the
10k resistor) goes through the base. Thats too much, obviously.

By increasing the size of the base resistor, the current allowed through the
NPN transistor is decreased. You can say that the current at the collector
of the PNP transistor is

Ic = beta^2 * 4.3/R

assuming the same beta for both of the transistors. Assume 100. Then we have

Ic = 43k/R

So, by putting in a 47k resistor, the current at the collector of the PNP
transistor is limited to a bit less than an amp, which is probably OK. More
resistance will limit it even further.

Note that Vcc doesn't enter into the equation, thus one can use this to
level shift with impunity.

Regards
Bob Monsen

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:104amhdml3tkk76@corp.supernews.com...
hmmm, this doesn't look right... There's nothing to limit the base
current through the PNP, is there?

- jim

Robert C Monsen wrote:


Use an NPN transistor/PNP transistor combination as follows:


+-----+ VCC
.-. |
10k| | |
| | |
'-' |
| |e
+--b|
1k | |c
___ |c |
uC -|___|-b| |
|e .--|--.
| | |
| |Load |
| | |
| '--|--'
| |
+-----+ GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


uC = 5V means load is powered
uC = 0V means load isn't powered


In this one, seems like we might run into trouble if VCC were to rise
above 13.2V. I know I said 12V, but in fact, this is an automotive
installation, so VCC could rise to ~14 or so volts.


Another way is to use a zener diode:

+---VCC----+
| |
.-. |
| |1k |
| | |
'-' |
| ___ |e
+-|___|-b|
| 1k |c
- |
^8.2 Znr |
| Load
uC +---+ |
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


Regards,
Bob Monsen
Click to expand...
 
I'm new but I think that's what the 10K and the 1K resistors are there for.

--Viktor

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> wrote in message
news:104amhdml3tkk76@corp.supernews.com...
hmmm, this doesn't look right... There's nothing to limit the base
current through the PNP, is there?

- jim

Robert C Monsen wrote:


Use an NPN transistor/PNP transistor combination as follows:


+-----+ VCC
.-. |
10k| | |
| | |
'-' |
| |e
+--b|
1k | |c
___ |c |
uC -|___|-b| |
|e .--|--.
| | |
| |Load |
| | |
| '--|--'
| |
+-----+ GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


uC = 5V means load is powered
uC = 0V means load isn't powered


In this one, seems like we might run into trouble if VCC were to rise
above 13.2V. I know I said 12V, but in fact, this is an automotive
installation, so VCC could rise to ~14 or so volts.


Another way is to use a zener diode:

+---VCC----+
| |
.-. |
| |1k |
| | |
'-' |
| ___ |e
+-|___|-b|
| 1k |c
- |
^8.2 Znr |
| Load
uC +---+ |
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


Regards,
Bob Monsen
Click to expand...
 
James W wrote:
If I connect the base to +5, then I'll have quite a bit of base
current.. No?
Click to expand...
You need a resistor between the +V and the base of the NPN transistor,
just as you would need a resistor between the base of the PNP transistor
and ground.

Your idea of using the switch to bypass the circuit is interesting. I'll
think about that.
Click to expand...
You could also bypass the relay with a mechanical switch.

--
-----------------------------------------------------------------------
To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris
 
"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:3df1c.169382$uV3.726953@attbi_s51...
No, he is right. I wasn't paying attention initially. Using a 1k resistor,
when the uC brings up the voltage to 5V, then the base of the NPN will
sink
about (5 - .7)/1000 = 4.3mA. If the NPN has a beta of 100, that means
it'll
sink 430mA. The PNP transistor acts like a diode from VCC to the collector
of the NPN transistor, so all 430mA (minus the 1.2mA that goes through the
10k resistor) goes through the base. Thats too much, obviously.

By increasing the size of the base resistor, the current allowed through
the
NPN transistor is decreased. You can say that the current at the collector
of the PNP transistor is

Ic = beta^2 * 4.3/R

assuming the same beta for both of the transistors. Assume 100. Then we
have

Ic = 43k/R

So, by putting in a 47k resistor, the current at the collector of the PNP
transistor is limited to a bit less than an amp, which is probably OK.
More
resistance will limit it even further.
Click to expand...
Limiting the current is a necessity for sure, but doing so with a high
resistance in a base circuit of the previous stage seems the worst possible
thing to me. It depends on a Beta that can vary over a wide range especially
with a rising temperature, so the transistor may well go into thermal
runaway under some conditions. Since the whole thing is gonna be run off
spiked voltage with poor regulation and inductive loads facilitating
oscillation, this method looks like relying on the accuracy of a weather
forecast for 12.09.2525 :).

Won't the usual solution "limit the current where it flows" be more
appropriate? Something like placing a 470R-1K resistor in the NPN's
collector circuit maybe.

Dimitrij
 
"Dimitrij Klingbeil" <dimitrijklingbeil@arcor.de> wrote in message
news:40465419_1@news.arcor-ip.de...
"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:3df1c.169382$uV3.726953@attbi_s51...
No, he is right. I wasn't paying attention initially. Using a 1k
resistor,
when the uC brings up the voltage to 5V, then the base of the NPN will
sink
about (5 - .7)/1000 = 4.3mA. If the NPN has a beta of 100, that means
it'll
sink 430mA. The PNP transistor acts like a diode from VCC to the
collector
of the NPN transistor, so all 430mA (minus the 1.2mA that goes through
the
10k resistor) goes through the base. Thats too much, obviously.

By increasing the size of the base resistor, the current allowed through
the
NPN transistor is decreased. You can say that the current at the
collector
of the PNP transistor is

Ic = beta^2 * 4.3/R

assuming the same beta for both of the transistors. Assume 100. Then we
have

Ic = 43k/R

So, by putting in a 47k resistor, the current at the collector of the
PNP
transistor is limited to a bit less than an amp, which is probably OK.
More
resistance will limit it even further.

Limiting the current is a necessity for sure, but doing so with a high
resistance in a base circuit of the previous stage seems the worst
possible
thing to me. It depends on a Beta that can vary over a wide range
especially
with a rising temperature, so the transistor may well go into thermal
runaway under some conditions. Since the whole thing is gonna be run off
spiked voltage with poor regulation and inductive loads facilitating
oscillation, this method looks like relying on the accuracy of a weather
forecast for 12.09.2525 :).

Won't the usual solution "limit the current where it flows" be more
appropriate? Something like placing a 470R-1K resistor in the NPN's
collector circuit maybe.
Click to expand...
If you do that, the final current would then depend on VCC, and thus be
dependent on those spikes you were worrying about. You could put a resistor
between the emitter of the NPN and ground, if you are worried about the laws
of physics suddenly failing.

Using a 150k resistor, for example, means that for his application, assuming
a beta of 200 (which is high) the current through the collector of the NPN
would be limited to something less than 6mA. I'm guessing that with this
current, thermal runaway isn't an issue. With a more likely beta, the
current is even less.

Dimitrij
Click to expand...
- Bob Monsen
 
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