Trying to understand one thing about PNP transistors

[Dan]

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

-Dan

 

[Peter Bennett]

On Thu, 18 Dec 2008 15:23:08 -0500, Dan <Newcombe@mordor.clayton.edu>
wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

-Dan

One solution would be to use another NPN transistor to control the
PNP. The collector of the added NPN would connect through a resistor
to the base of the PNP, and you would need a resistor from the base of
the PNP to +12V. The base of the NPN could be driven from a logic
output via a resistor. When the logic output is low, the NPN would be
off so the resistor on the base of the PNP would pull that base
towards +12, turning the PNP off. With the logic signal high, the NPN
would be turned on, pulling the PNP's base down, and turning on the
PNP.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[DJ Delorie]

Dan <Newcombe@mordor.clayton.edu> writes:

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

Yes, the PNP is simple - you apply a voltage to the base that is .7
volts lower than the emitter (+12v in this case) and it opens the
"switch".

What I think is confusing me and is left of the diagram is what
turns those PNP's on and off.

In this case, "1" = 12v and "0" = GND. You're not using a smaller
voltage, you're using a smaller *current*. Usually, you'd use an
H-bridge pre-driver (example: freescale MC33883) that provides the
voltage conversion and MCU interface.

 

[DJ Delorie]

Dan <Newcombe@mordor.clayton.edu> writes:

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

Of course, we both meant it *closes* the switch.

 

[Dan]

On Thu, 18 Dec 2008 13:06:20 -0800, Peter Bennett
<peterbb@somewhere.invalid> wrote:

On Thu, 18 Dec 2008 15:23:08 -0500, Dan <Newcombe@mordor.clayton.edu
wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

-Dan

One solution would be to use another NPN transistor to control the
PNP. The collector of the added NPN would connect through a resistor
to the base of the PNP, and you would need a resistor from the base of
the PNP to +12V. The base of the NPN could be driven from a logic
output via a resistor. When the logic output is low, the NPN would be
off so the resistor on the base of the PNP would pull that base
towards +12, turning the PNP off. With the logic signal high, the NPN
would be turned on, pulling the PNP's base down, and turning on the
PNP.

I had thought that another NPN would take care of that somehow. I
guess then my feeble mind who has always had trouble with this thinks
why can't you just use 4 NPN's and treat them all like simple toggle
switches.

I'm guessing that it is because even though they can act as a switch,
it's not as simple as open or close, and that between the motor and
the 12V source, you would need to have 12.7V on the base to activate
the NPN, which is why you don't do that.

I'm going to have to search around for a little more detailed
schematic that doesn't try to do too much more. I usually find the
super-simple ones that leave a couple needed things out, or ones that
add half a dozen IC's - great I'm sure, but not when trying to grasp!

Thanks for the reply!

 

[BobW]

"Dan" <Newcombe@mordor.clayton.edu> wrote in message
news8clk4p4nvvhs1da6bs41pq58pgj5et2mi@4ax.com...

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

-Dan

Dan,

A transistor is a device that is used to achieve power gain. Sometimes that
involves voltage gain and current gain (e.g., in common emitter
configuration). Othertimes that involves no voltage gain and but only
current gain (e.g., in common collector (aka emitter follow) configuration).
Still othertimes, it might be only voltage gain and no current gain (e.g.,
in common base configuration).

In your H-bridge circuit, all four transistors are in common emitter
configuration. They have voltage gain because the output swings 12V with
only a small (0.7V) change at the input side (this isn't strictly true due
to the way it's usually done). They have current gain because the output
current is much higher than the input (base) currents.

Imagine, in this H-bridge, that you had two 0.7V batteries on the control
side. You could completely control the motor's 12V at (let's say) 1A needs
with only two 0.7V batteries at (let's say) 0.01A.

There's no real difference between an NPN and PNP transistor -- other than
that the direction of voltages and currents are reversed. They are both
controlled by the difference in voltage between the emitter and base that
results in a given amount of base current, or another way of looking at this
is that they're both controlled by the base current which results in a
certain amount of voltage from emitter to base.

Bob
--
== All google group posts are automatically deleted due to spam ==

 

Dan wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

-Dan

Yep. Like many so called 'tutorials' the circuit shown is pretty well
useless for understanding purposes. The author chose a good lead-in by
using relays to demonstrate the idea of reversing the motor but
totally shied away from the nitty-gritty of using real discrete
components.

Check out the excellent H bridge circuit near the bottom of this
page ...
http://www.cs.uiowa.edu/~jones/step/circuits.html
It's about as simple as you can make these things and it uses 4 NPN's
like you mentioned.

 

[Tim Wescott]

On Thu, 18 Dec 2008 15:23:08 -0500, Dan wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's at
the top, and two NPN's at the bottom - and you get direction control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the "switch".

However, where I'm getting lost is the PNP part. From what all I read,
you supply a voltage .7 volts lower than what is on the emitter (12v in
the diagram, so 11.3volts or less) to allow voltage to flow out the
collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do to
put a PNP in the 0/Off state. Sure - you could put 12V on the base, but
isn't the idea that you are using a smaller voltage to control a larger
voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am I
missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

-Dan

At the top level a PNP transistor is just an NPN transistor in reverse.
You turn an NPN transistor on by putting current into its base (by
raising its base voltage relative to its emitter), you turn a PNP
transistor on by pulling current out of its base (by lowering its base
voltage relative to its emitter).

Where your voltage gain confusion comes in is looking at that 11.3V and
thinking "my, that is a big fraction of 12V". What you need to do is
look at the 0.7V from base to emitter and think "my, that is a small
fraction of 12V". Then you'll see that the voltage gain is just as large
as with the NPN.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

 

[Robert Roland]

On Thu, 18 Dec 2008 15:23:08 -0500, Dan <Newcombe@mordor.clayton.edu>
wrote:

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

Well, technically, bipolar transistors (normal NPN and PNP) are
current controlled, not voltage controlled. At first, the difference
may seem insignificant, but the difference is important. Since you
need a voltage difference for a current to flow, voltage is, of
course, also important.

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

Basically right.

Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

Not in this case. When using a transistor as a switch like this, it
makes no sense to talk about voltage gain. When the transistor is in
full saturation, the voltage across B-E will be .6-.7 volts, while the
voltage across C-E will be .1-.2 volts.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

You are perfectly right. A driver that outputs only 5V will not be
able to turn off the PNP power transistor where E is at 12V.

The normal solution to this problem is to use an "open collector"
driver. This is essentially simply another NPN transistor that drives
the base of the power PNP transistor: When the driver NPN is open,
current will flow from E to B towards ground in the power PNP, so that
the power PNP opens. When the driver NPN is closed, there is no E-B
current in the power PNP, and therefore no E-C current either.

In practice, you use a pull-up resistor from the base of the power PNP
to +12V, to make sure the base does not get pulled low accidentally by
leakage currents or noise.

Here's a short article on open collector:
http://en.wikipedia.org/wiki/Open_collector
--
RoRo

 

[Jasen Betts]

On 2008-12-18, Dan <Newcombe@mordor.clayton.edu> wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

you are uasing a smaller voltage - the difference between 12V and 11.3V is
small.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

what I would do in that case is use some NPN transistors to control
the PNPs.

another option would be to connect the power to the the chips between
7V and 12V instead of between 0V and 5V, but that would make it hard
to control the NPNs


the NPNs that control the PNPs can also be used to control the
opposite NPNs reducing the current needed to control the bridge.

http://homepages.xnet.co.nz/~jasen/H-bridge.html

 

[Nobody]

On Thu, 18 Dec 2008 18:54:29 -0800, john wrote:

Check out the excellent H bridge circuit near the bottom of this
page ...
http://www.cs.uiowa.edu/~jones/step/circuits.html
It's about as simple as you can make these things and it uses 4 NPN's
like you mentioned.

The main problem with that circuit is that the high side transistor won't
saturate, as the base can't exceed the collector voltage, and the emitter
will be at least ~0.6V below the base, and hence ~0.6V below the
collector. That could be a significant issue if you're switching tens of
amps. Using a PNP transistor for the high side doesn't have that issue.

 

[John Fields]

On Thu, 18 Dec 2008 15:23:08 -0500, Dan <Newcombe@mordor.clayton.edu>
wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown here http://www.dprg.org/tutorials/1998-04a/ Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part. From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off. They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state. Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well? So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

---
It might help if you think of what it takes to turn a transistor on.

In either case (NPN or PNP) what does it is causing the base voltage to
move away from the emitter voltage in the direction of the collector
voltage.

In the case of an NPN, the collector is usually run more positive than
the emitter, so to turn it on you could do this: (View in Courier.)


+V>----+---------+
| |
| [R]
| |
| C
+--[R]--B
E
|
GND>-------------+


In the case of a PNP, the collector is usually run less positive than
the emitter, so this would turn it on:


+12>---+---------+
| |
| E
+--[R]--B
| C
| |
[R] [R]
| |
GND>---+---------+


Connecting both circuits in a half bridge with both transistors off
(Note that the bases are each connected to their respective emitters
through resistors) would look like this:


+V>----+---------+
| |
| E
+-[R]---B PNP
C
|
[R]
|
C
+-[R]---B NPN
| E
| |
GND>---+---------+


And with both transistors on, (note that the bases are now connected to
voltages moving away from the emitter voltage and toward the collector
voltage) like this:


+V>--------------+
|
E
GND>-----[R]---B PNP
C
|
[R]
|
C
+V>------[R]---B NPN
E
|
GND>-------------+


Now, if we call the voltages connected to the resistors '1' for +V and
'0' for GND, we'll have, for the half bridge turned off:

+V>--------------+
|
E
1>---[R]---B PNP
C
|
[R]
|
C
0>---[R]---B NPN
E
|
GND>-------------+


and, turned on:


+V>--------------+
|
E
0>---[R]---B PNP
C
|
[R]
|
C
1>---[R]---B NPN
E
|
GND>-------------+


If we rearrange things a little, our turned-on half bridge will look
like this, with I --> indicating the direction of conventional current
flow:


+V>--------------+
|
E
0>---[R]---B PNP
C
| I -->
+-----[R]-----+
|
C
B---[R]---<1
E
|
GND>---------------------------+


If we want to, we can now add another half bridge to the circuit and if
we make sure it's turned off, it'll be like it's not even there. So,
doing that, we'll now have an 'H' bridge:


+V>--------------+-------------+
| |
E Q1 E Q3
0>---[R]---B PNP PNP B---[R]---<1
C C
| I --> |
+-----[R]-----+
| |Q2
C Q4 C
0>---[R]---B NPN NPN B---[R]---<1
E E
| |
GND>-------------+-------------+


In order to reverse the direction of current through the load all we
have to do is turn off Q1 and Q2 and turn on Q3 and Q4:


+V>--------------+-------------+
| |
E Q1 E Q3
1>---[R]---B PNP PNP B---[R]---<0
C C
| <-- I |
+-----[R]-----+
| |Q2
C Q4 C
1>---[R]---B NPN NPN B---[R]---<0
E E
| |
GND>-------------+-------------+

Some caveats:

Note that even though the same signal polarity is applied to two
transistors at once, it's _not_ permissible to connect those inputs
together and drive them from a single source.

Also, there should be some dead time allowed when switching the bridge
from one state to the other.

I've not shown them but if you're going to be driving an inductive load
you need to put diodes across the transistors, as shown in the circuits
you linked to.

Finally, be careful "plugging" the motor.

Finally, finally, if you don't already have it,

LTspice is a wonderful _free_ resource you can download from:

http://www.linear.com/designtools/software/

and here's a little "H" bridge you can play around with:

Version 4
SHEET 1 2244 800
WIRE 1120 -368 16 -368
WIRE 512 -304 224 -304
WIRE 544 -272 352 -272
WIRE 352 -224 352 -272
WIRE 224 -176 224 -304
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WIRE 1712 -176 624 -176
WIRE 272 -128 144 -128
WIRE 512 -128 512 -304
WIRE 512 -128 448 -128
WIRE 1600 -128 512 -128
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WIRE 544 -48 352 -48
WIRE 144 0 144 -128
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WIRE 1664 0 576 0
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WIRE 1264 32 1120 32
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WIRE 1264 96 1264 32
WIRE 624 112 624 -176
WIRE 688 112 624 112
WIRE 1600 112 1600 -128
WIRE 1600 112 1536 112
WIRE 576 144 576 0
WIRE 688 144 576 144
WIRE 784 144 752 144
WIRE 912 144 864 144
WIRE 1360 144 1328 144
WIRE 1472 144 1440 144
WIRE 1664 144 1664 0
WIRE 1664 144 1536 144
WIRE 976 240 976 192
WIRE 1088 240 976 240
WIRE 1264 240 1264 192
WIRE 1264 240 1168 240
WIRE 976 288 976 240
WIRE 1264 288 1264 240
WIRE 576 336 576 144
WIRE 688 336 576 336
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WIRE 1664 336 1536 336
WIRE 16 352 16 -368
WIRE 144 352 144 0
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WIRE 1120 432 976 432
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WIRE 1264 432 1120 432
WIRE 16 528 16 432
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WIRE 144 528 16 528
WIRE 352 528 352 -48
WIRE 352 528 144 528
WIRE 1120 528 1120 432
WIRE 1120 528 352 528
WIRE 16 560 16 528
FLAG 16 560 0
SYMBOL npn 912 288 R0
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL npn 1328 288 M0
SYMATTR InstName Q2
SYMATTR Value 2N3904
SYMBOL pnp 912 192 M180
SYMATTR InstName Q3
SYMATTR Value 2N3906
SYMBOL pnp 1328 192 R180
SYMATTR InstName Q4
SYMATTR Value 2N3906
SYMBOL res 880 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R1
SYMATTR Value 1000
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SYMATTR Value 1000
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SYMATTR Value 1000
SYMBOL voltage 16 336 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
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SYMATTR Value 1
SYMBOL voltage 144 336 R0
WINDOW 3 24 104 Invisible 0
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SYMATTR Value PULSE(0 1 0 1e-6 1e-6 .5 1)
SYMATTR InstName V2
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SYMATTR InstName A1
SYMBOL Digital\\and 720 288 R0
SYMATTR InstName A2
SYMBOL Digital\\and 1504 64 M0
SYMATTR InstName A3
SYMBOL Digital\\and 1504 288 M0
SYMATTR InstName A4
TEXT -16 648 Left 0 !.tran 5
TEXT -16 672 Left 0 !.option noopiter


JF

 

[John Fields]

On Fri, 19 Dec 2008 08:48:20 -0600, John Fields
<jfields@austininstruments.com> wrote:

In the case of a PNP, the collector is usually run less positive than
the emitter, so this would turn it on:


+12>---+---------+
| |
| E
+--[R]--B
| C
| |
[R] [R]
| |
GND>---+---------+

---
Oops...

Should read:

+12>-------------+
|
E
+--[R]--B
| C
| |
[R] [R]
| |
GND>---+---------+


JF

 

On Dec 18, 3:23 pm, Dan <Newco...@mordor.clayton.edu> wrote:

I'm trying to wrap my head around a simple H-bridge circuit, such as
shown herehttp://www.dprg.org/tutorials/1998-04a/  Simply two PNP's
at the top, and two NPN's at the bottom - and you get direction
control.

The NPN is simple - you apply a voltage to the base that is .7 volts
higher than the emitter (ground in this case) and it opens the
"switch".

However, where I'm getting lost is the PNP part.  From what all I
read, you supply a voltage .7 volts lower than what is on the emitter
(12v in the diagram, so 11.3volts or less) to allow voltage to flow
out the collector.

What I think is confusing me and is left of the diagram is what turns
those PNP's on and off.  They show a logic table of 0 and 1 for the
state of each swich, but I am trying to figure out what you would do
to put a PNP in the 0/Off state.  Sure - you could put 12V on the
base, but isn't the idea that you are using a smaller voltage to
control a larger voltage.    

If you had an IC that was putting out 0V for off, and 5V for on,
wouldn't the 0V be less than 11.3V and turn it on as well?  So what am
I missing in the middle?

Thanks - this has been confusing me for a while trying to get into all
this.

        -Dan

The pnp operation is just like the npn but thepolarity is opposite.
With an npn, in order to get collector current, electrons must be
injected from the emitter region towards the base region by forward
biasing the b-e junction, which is basically just a p-n diode
junction. When fwd biased, the base emits holes towards the emitter,
and the emitter emits electrons towards the base. But, the b-c
junction is reverse biased, and an electric field exists in the b-c
depletion region.. The polarity of said E field is such that holes
move from collector to base, and electrons move from base to
collector. The electrons just emitted from the emitter to the base
enter the base region, and the strong E field attracts them into the
collector. This is for npn polarity.

With pnp, the b-e junction is fwd biased, and the base emits electrons
towards the emitter, and the emitter emits holes towards the base.
The b-c junction is reverse biased with a strong E field. Holes
emitted from the emitter reach the base and are drawn into the
collector by the b-c depletion zone E field. The pnp is a little
trickier to envision because the charge carrier being transported from
emitter to base then onward to the collector is a **hole**. A good
semiconductor physics text will explain the "hole" concept to those
who aren't familiar with it. A good scholarly online source can also
explain what holes are all about. Make sure that the info is peer-
reviewed and reliable. Trustworthy sources include university sites,
and semiconductor maker sites. If a source on the web takes a view
contrary to what Fairchild. Tex Instr, Natl Semi, etc. teach, that
source should be considered suspect. No one knows more about npn and
pnp devices, than the makers of integrated and discrete semiconductor
devices. Their info is then passed on to the universities.

Have I helped you at all? BR.

Claude

 

Nobody wrote:

On Thu, 18 Dec 2008 18:54:29 -0800, john wrote:

Check out the excellent H bridge circuit near the bottom of this
page ...
http://www.cs.uiowa.edu/~jones/step/circuits.html
It's about as simple as you can make these things and it uses 4 NPN's
like you mentioned.

The main problem with that circuit is that the high side transistor won't
saturate, as the base can't exceed the collector voltage, and the emitter
will be at least ~0.6V below the base, and hence ~0.6V below the
collector. That could be a significant issue if you're switching tens of
amps. Using a PNP transistor for the high side doesn't have that issue.

Yes.
Even worse, I believe the Tr's are actually Darlingtons, so we''ll
lose ~2V across each come what may!. For functional simplicity though
in driving small (1~2 amp) loads, I think the arrangement takes some
beating.
Nfets with bootstrapping is normal for these kind of jobs but basic
'operational' clarity is lost.

 

[Rich Grise]

On Thu, 18 Dec 2008 16:18:11 -0500, Dan wrote:

On Thu, 18 Dec 2008 13:06:20 -0800, Peter Bennett

One solution would be to use another NPN transistor to control the PNP.
The collector of the added NPN would connect through a resistor to the
base of the PNP, and you would need a resistor from the base of the PNP
to +12V. The base of the NPN could be driven from a logic output via a
resistor. When the logic output is low, the NPN would be off so the
resistor on the base of the PNP would pull that base towards +12, turning
the PNP off. With the logic signal high, the NPN would be turned on,
pulling the PNP's base down, and turning on the PNP.

I had thought that another NPN would take care of that somehow. I guess
then my feeble mind who has always had trouble with this thinks why can't
you just use 4 NPN's and treat them all like simple toggle switches.

Get yourself a pencil and piece of paper, and draw out Peter's circuit,
and it should be obvious what's going on.

Good Luck!
Rich