Trying to understand one thing about PNP transistors

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.

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Peter Bennett, VE7CEI  
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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".

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.

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

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!

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

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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 ...

It's about as simple as you can make these things and it uses 4 NPN's like you mentioned.

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.

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Tim Wescott
Control systems and communications consulting
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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.

Basically right.

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.

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:

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RoRo

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

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.

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.

--- 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]------------------------------+

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]--- | +-----[R]-----+ | |Q2 C Q4 C 0>---[R]---B NPN NPN B---[R]----------------+-------------+

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]---

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Oops...

Should read:

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


JF

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

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.

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