PNP as switch with different emitter voltage

[Jon Kirwan]

Good student work, Jon ;-)

Sadly, I've never taken or audited a single course in
elecronics. But let it never be said I cease being a
student.

Now, what about a bit of a snubber resistor on the base of
Q1, too?

Jon

 

[Jamie]

The problem is that an NPN on this output will give me the opposite from
what I am after. I am trying to switch on the load from a 0V signal, and
turn it off with a 5V signal.

NPN would work if I was trying to switch the load with a 5V signal, but that
is not what I am doing.

Jon

http://www.fairchildsemi.com/ds/FO/FOD3180.pdf

That is what you need..
Its a 2A Gate driven output via an optical input, which you can drive
with your circuit.

This optical has either low or high side, or use both of you wish
1 R to drive it from your logic circuit. If you want high side, connect
the Vcc to your 18 volts. If low side is what you want? connect the Vee
to the common and cycle the 18 volts through your buzzer to the output (6)

P.S.
I wouldn't push that 2A's continuos but I think you can drive that
buzzer of yours which should be far under that load. And you may want to
put a snubber on it.

Jamie

 

[Jon Danniken]

: +18V
: |
: |
: \
: / R2
: \ 22k +18V
: / |
: | |
: | |<e Q2
: +----------|
: | |\c
: nailed | |
: to |/c Q1 |
: +5V ---| |
: |>e |
: | \
: | / Piezo
: | \ at 15mA
: | / (~1.2k)
: \ |
: / R1 |
: \ 12k |
: / gnd
: |
: |
: 0/5 on/off >-'

Assume Q2 nearing saturation sufficiently well with beta=50,
then the Q2 base current should be about 15mA/50 or 300uA.
The on/off control needs to be able to sink 300uA, plus a
little more for R2 (say 10% of 200uA) and Q1 base (likely
only 2-3uA.) R1 needs to supply that when the on/off control
is at or near 0V. Assume about 1V for compliances needed by
Q1's Vbe and the sinking output near 0V... so:

R1 = (5V-1V)/(300uA+30uA+3uA) = 12k
R2 = 0.7V / 30uA = near 22k

Close enough for student work. Could add a speed-up cap
across R1.

Interesting, thanks Jon, I hadn't seen a switched emitter before.

I'd still like to do this with only one transistor, though. I'm wondering if
something like this would work:

: +18V
: |
: |
: \
: / R2
: \
: /
: |
: |
: +------------
: | |
: | |
: Rb |/c Q1 |
: IC--\/\/\---| |
: |>e |
: | \
: | / Piezo
: | \ at 15mA
: | / (~1.2k)
: | |
: | |
: | |
: gnd gnd

When the IC output was 0, the piezo would draw current through R2. When the
IC output was high, the transistor would conduct, pulling down the voltage
at the collector.

Dunno if that might work, though. It'd depend on if R2 would be small
enough to still allow enough current through the piezo, while not burning
out the transistor, but still big enough to pull down the collector enough
to shut down the piezo (~3 volts).

Jon

 

[Jamie]

Something wrong with using a transistor to drive the LED ? THe idea
was to keep what he had and put that PNP on the logic supply line to
drive the LED via a R instead of the buzzer. The output of the coupler
then could be connected to what ever he pleases with in it's range..

you did notice I said "with your circuit"


Jamie

 

[Sylvia Else]

On 16/01/2011 11:43 PM, John Fields wrote:
On Sun, 16 Jan 2011 23:13:14 +1100, Sylvia Else

On 16/01/2011 4:24 PM, Jon Danniken wrote:
Hi, I am trying to take advantage of an extra flip/flop output in an
existing circuit. The driving signal is either 0V or +5V, but the
emitter
voltage (to drive the switched load) is +18V.

This is the circuit I am planning to use:

+18V
|
Rb |
___ |/
On=0V ---|___|-----| PNP
Off=5V |<
\
|
.-.
| | Load (15mA)
| |
'-'
|
|
===
GND

I am going to try getting the base resistor to be such at the
transistor is
off when the IC is at 5 VDC, and on when the IC is at 0 volts.

Besides fiddling with the base resistor, is there a different
topology, or
modification I can make to the existing one to make this work better?

I do realize that putting an NPN transistor off of the IC's output
would
solve this nicely, but space is at a premium, and I would like to do
this
with just the one transistor.

Any ideas?

Thanks,

Jon



I assume you've put the transistor the wrong way up by mistake.

If you attach the emitter to +18V and the collector to the load, and
then stick a resistor between the base and +18V, then by a suitable
choice of resistor values, you can make the transistor switch.

---
I don't think so. View with a fixed-pitch font:

+18-----------+------+
| |
[R2] |
| E
VIN>----[R1]--+----B
C
|
[LOAD]
|
GND>-----------------+

Assuming that what's on the left hand side of VIN can sink _and_
source current, in order to turn the switch ON with Vin = 0V and turn
it OFF with Vin = 5V, what would be the ratio of R2:R1?

---
JF

What's on the left of Vin is never asked to source current, but
presumably there's no doubt it can sink current when it's driven low.
The doubt would be as to its behaviour when driven high - usually it
would expect to source current in this state, so will it still sink it?
But you're assuming that it will, so...

Consider Veb when the transistor is on to be 0.6V (Veb because it's PNP).

Suppose we want to switch when Vin traverses 2.5V.

Then:

0.6 = (18 - 2.5) * R2/(R1 + R2).
0.6R1 + 0.6R2 = 15.5R2

0.6R1 = 14.9R2

R1 = approx 24 * R2

Vb would only rise to 17.5 (Veb = 0.5) volts when Vin is 5v, but that
should be enough to turn the transistor off.

The main issue I see with this arrangement is whether it is possible to
sink enough base current to saturate the transistor given the value of
load, which depends on the thing IC driving Vin, and the transitor beta,
but I said that.

If it's an ideal IC, and can sink an infinite amount of current in both
states, then it's just a matter of choosing the resistor values with
knowledge of the worst case beta.

It would be more difficult if the OP wanted the opposite switching
behaviour.

Sylvia.

I would add that I wouldn't want to use this approach in a circuit
intended for mass production. A combination of adverse IC output
characteristic, resistor value and transistor variation would mean it
was asking for trouble.

But as a one off...?

He could get more leeway by putting a diode between the emitter and 18v,
and the cost of reducing the voltage across his load, which is now
revealed to be a piezo beeper.

However, that pushes up the component count again, and space was stated
to be an issue.

Sylvia.

 

[Sylvia Else]

Indeed I did, I went with what the ascii conversion software gave me instead
of flipping and mirroring it. My apologies for not noticing that.


It's coming off of a 4013 flip flop, and I did actually measure the outputs
to be +5 or 0. The other output from the IC is inverted, and in use
(driving an NPN), hence my desire to use the PNP on the other output.

Are you sure you can't drive both transistors off that same output?
You're guaranteed 0.44 mA, which would be easily be enough to drive two
typical low power NPNs into saturation with 15mA loads.

Sylvia.

 

[Jon Danniken]

http://www.fairchildsemi.com/ds/FO/FOD3180.pdf

That is what you need..
Its a 2A Gate driven output via an optical input, which you can drive
with your circuit.

This optical has either low or high side, or use both of you wish
1 R to drive it from your logic circuit. If you want high side,
connect the Vcc to your 18 volts. If low side is what you want?
connect the Vee to the common and cycle the 18 volts through your
buzzer to the output (6)
P.S.
I wouldn't push that 2A's continuos but I think you can drive that
buzzer of yours which should be far under that load. And you may want
to put a snubber on it.

Thanks Jamie, that would work, but it seems a bit overkill (to me) for
driving a little piezo beeper at 15mA. I'd also have to order it, as there
is no NTE sub for that part, and I don't have that in my parts bin.

Jon

 

[Jamie]

Do you need to ground one side of the load? If you can run the piezo
between V+ and a switch to ground, all you need is an NPN and a
resistor. Even better, just an N channel mosfet, 2N7002 maybe.

John

That would throw out the idea of using logic 0 to activate it.
I still like my idea of using his existing circuit to drive that nice
opto. Either way, 2 active components are going to be needed here,
unless he elects to use a zener of ~ 15 volts in line with Rb.

Jamie

 

[Jamie]

Are you sure you can't drive both transistors off that same output?
You're guaranteed 0.44 mA, which would be easily be enough to drive two
typical low power NPNs into saturation with 15mA loads.

Sylvia.

You know, that's a good point...

 

[Jon Kirwan]

Interesting, thanks Jon, I hadn't seen a switched emitter before.

I'd still like to do this with only one transistor, though. I'm wondering if
something like this would work:



When the IC output was 0, the piezo would draw current through R2. When the
IC output was high, the transistor would conduct, pulling down the voltage
at the collector.

Dunno if that might work, though. It'd depend on if R2 would be small
enough to still allow enough current through the piezo, while not burning
out the transistor, but still big enough to pull down the collector enough
to shut down the piezo (~3 volts).

I don't think so.

Assume the piezo requires Ip=15mA and Vp=? volts, for now.
When the piezo is sounding at 15mA, your R2 must be <=
(18-Vp)/Ip or [1200-(200/3)*Vp]. Of course, if Vp is 18 then
R2=0 and you are obviously in trouble. So you need to get by
on less than Vp=18V with the buzzer. I suspect you can. But
let's continue on a different path.

Now assume the piezo is off and Q1 is turned on. At this
point you've got to cause Q1's collector to drop below some
voltage required for the buzzer to work. The arrangement you
have above doesn't really allow you to do anything but treat
the BJT as a switch, so you need to figure that there will be
18V across R2 in order to turn off the piezo, so Ic*R2=18.

I haven't looked up your output capability when high, but I
recall reading Sylvia saying you are guaranteed 440uA for HI.
Assume it is for now. You can plan on say beta=30 for the
BJT.

Then Ic can be as much as 30*440uA or 13.2mA. But also
recall that Ic*R2=18. So from this we compute R2=18/13.2mA
or 1364 ohms.

Which is a problem because 1364 ohms * the piezo's 15mA is
more than 18V drop across R2. Which means the piezo will be
off either way. It just won't work.

Even if you assumed that beta=100, R2 would be about 410 ohms
and 410 * 15mA = 6.15V. Which means your piezo would have to
work okay with less than 12V. Assuming it still drew 15mA,
then. Likely, it draws maybe a little less meaning the
voltage across it is a little higher and it finds the balance
point at something less than 15mA and something more than 12V
to work with.

You could try it, I suppose. Maybe try R2=390 or 450?

Jon

 

[Sylvia Else]

Yes, I could drive both transistors from the same CMOS output.
Yes, I could also piggyback on the LED.
Yes, I could also use two transistors (which I stated in my initial post).

I had already considered those solutions, but I decided against them for a
number of reasons before I made my initial post. While I did mention the
last one, I apologize for not listing the first two in my initial post.

What I am attempting to do is to make use of the inverted output coming from
the CMOS IC, using just one transistor (and associated circuitry to run that
additional transistor). So far, Pimpom's suggestion of using a zener on the
base of a PNP seems to be the only solution proffered.

I can't find any statement that the IC will sink current when it's
output is high, much less how much current it will sink. Maybe it's
obvious from the FETs in the output stage that it will sink current (I'm
not at all familiar with FETs) but there's still the question of how
much, and in particular, how much it can sink and remain operational
over the long term.

Sylvia.

 

[Pimpom]

I can't find any statement that the IC will sink current when
it's
output is high, much less how much current it will sink. Maybe
it's
obvious from the FETs in the output stage that it will sink
current
(I'm not at all familiar with FETs) but there's still the
question of
how much, and in particular, how much it can sink and remain
operational over the long term.

Sylvia.

About that last statement, just for information: The output FETs
in 4xxx ICs can drive several mAs reliably. The output current is
partly limited by the FETs' output resistance. The original
datasheets by RCA- also adopted by TI, etc - show output curves
for typical and worst-case devices.

For example, those curves show that the short-circuit output
current for a typical device with a 5V supply is about 4.2mA.
That's 21mW dissipation per logic unit. With a 10V supply, this
goes up to 20mA and 200mW.

 

[Pimpom]

You're right. I missed that post. Sorry, no plagiarism intended.

 

[Jon Danniken]

About that last statement, just for information: The output FETs
in 4xxx ICs can drive several mAs reliably. The output current is
partly limited by the FETs' output resistance. The original
datasheets by RCA- also adopted by TI, etc - show output curves
for typical and worst-case devices.

For example, those curves show that the short-circuit output
current for a typical device with a 5V supply is about 4.2mA.
That's 21mW dissipation per logic unit. With a 10V supply, this
goes up to 20mA and 200mW.

The specific chip is a TC4013BP, datasheet here:
http://www.toshiba.com/taec/components2/Datasheet_Sync//152/35.pdf

My chip is running off of a 5V supply.

I tested the transistor (KN2907) I will be using, and I was able to
successfully run the 15mA beeper with a base current of 0.8mA, using a base
resistance of 22k. I could probably go lower, given that the Hfe of the
transistor is measured at 170.

Which brings up another question: what percentage of Hfe is usually
considered adequate to drive the transistor into saturation (how low of a
base current could I practically go to reliably drive it into saturation)?

Jon

 

[Jon Danniken]

For example, a BC547B or a 2N3906 can switch two 15 mA loads with
~1 mA base current. A 2.2k-3.3k resistor plus the 4013's internal
output resistance will supply more than enough base current to
saturate the transistor with a 30mA load. If it's not convenient
to parallel the two loads and drive them both with the existing
NPN transistor, you can add the extra NPN and drive it with the
same FF output via a 3.3-4.7k base resistor. There's no need to
go to the trouble of using a PNP on the other FF output.

I discarded that route initially, as it would require too much reworking of
the original board (more reworking than I wish to do). There already exists
an empty solder pad coming off of the inverted output, hence my desire to
make use of it.

Jon

 

[Spehro Pefhany]

He could get more leeway by putting a diode between the emitter and 18v,
and the cost of reducing the voltage across his load, which is now
revealed to be a piezo beeper.

However, that pushes up the component count again, and space was stated
to be an issue.

---
Indeed; that's why I'll stick with my original solution of an N MOSFET
(with, conceivably, no gate resistor) being driven by the 4013's Q
output or an NPN with a gate resistor being driven from the same
point.

That is, if Q isn't being loaded so heavily by what's already there
that my scheme won't work.
_
If it won't, and he's forced to use Q, then I'd bite the real-estate
bullet and use two NPN's, like this:

+18
+18 |
| [BUZZER]
[10k] |
| C
+--[10k]--B 2N3904
| E
_ C |
Q>--[10k]--B 2N3904 GND
E
|
GND

No Zeners, no tricks, no marginal operation, just a rock-solid switch.

You don't need the 10K between C and B of the 2N3904s.

He could also use one of these, plus one resistor:

http://www.onsemi.com/pub_link/Collateral/MUN5211DW1T1-D.PDF

 

[Spehro Pefhany]

The specific chip is a TC4013BP, datasheet here:
http://www.toshiba.com/taec/components2/Datasheet_Sync//152/35.pdf

My chip is running off of a 5V supply.

I tested the transistor (KN2907) I will be using, and I was able to
successfully run the 15mA beeper with a base current of 0.8mA, using a base
resistance of 22k. I could probably go lower, given that the Hfe of the
transistor is measured at 170.

Which brings up another question: what percentage of Hfe is usually
considered adequate to drive the transistor into saturation (how low of a
base current could I practically go to reliably drive it into saturation)?

Where it is specified is more than adequate, usually Ic/Ib = 10. I
often use 20, unless operation at temperature extremes is required
(and even then, self heating may help the cause, but it requires
analysis).

 

[Pimpom]

No. The OP wants to turn the beeper on when the FF output is low.

 

[Jon Kirwan]

No. The OP wants to turn the beeper on when the FF output is low.

---
That, however, doesn't diminish its excellence.

Here's Larkins's suggestion:

. +18
. |
. [PIEZO]
. |
. +----+ D
. | Q|----G 2N7002
. | _| E
. | Q| |
. +----+ GND


And here's Kirwan's, arguably the best 2 transistor submission:


. +5 +18 +18
. +----+ | | |
. | Q| | [22k] |
. | _| B | E
. | Q|O---[12k]--E C----+----B PNP
. +----+ NPN C
. |
. [PIEZO]
. |
. GND

In both examples the piezo works at the proper time, and in view of
the OP's statement that he had PCB real estate issues and wanted to do
the least possible amount of modification to the PCB, which would you
recommend: two transistors and three resistors or a single transistor?

Well. Two resistors (unless the +5 isn't readily available
and a resistor divider is to be used to develop that base
voltage -- then 4.) Regardless, the OP also said the reason
for the /Q choice is that there is a pad just sitting there
open and unused on /Q. Since the OP said "no" to two BJTs,
too, neither of the above appears to be "a solution."

So, no choice is possible given the above two choices and the
OP's boundary criteria.

By the way, Jim suggested the idea although I've used that
exact method many times, myself. So I immediately recognized
it. I actually like it and have discussed it more than a few
times in sci.electronics.*, though sometimes pulling current
via the emitter isn't workable if the heavy load itself is on
the collector and the driver has to sink all of it.

I also took a look at the OP's own idea about using an NPN to
bypass the piezo. It _might_ work, but cannot be guaranteed
to do so without more information. Testing would be required
if he's dead-set on one BJT and the /Q pad. Hope it's not
powered by a battery, though -- it always draws current. (I
don't remember if the OP mentioned how it was powered, but
18V might be a pair of 9V batteries.)

Jon

 

[Jamie]

I can't find any statement that the IC will sink current when it's
output is high, much less how much current it will sink. Maybe it's
obvious from the FETs in the output stage that it will sink current (I'm
not at all familiar with FETs) but there's still the question of how
much, and in particular, how much it can sink and remain operational
over the long term.

Sylvia.

not much!

Jamie