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Topic: Newbie puzzled by H-Bridge (Read 19363 times) previous topic - next topic

Ischia

Hi,

I've read this tutorial as well as several posts in this forum about H-Bridges, but for the life of me I can't understand why the "top" 2 transistors in the classic circuit need to be PNP while the bottom two should be NPN. The way I see it (for what it's worth), if I apply a voltage to the base, current should flow between the emitter and collector. If I do that to diagonally-opposed NPN transistors, what would be the problem? In other words, why can't I use 4 NPN transistors?

Thank you.

Chagrin

If you tie the signals for R1/R2 and R3/R4 together then you avoid the possibility of creating short circuits.

terryking228

Hi,
Quote
why can't I use 4 NPN transistors?


You CAN use all NPN transistors, but it makes the design and implementation of the driving circuitry more difficult and expensive.

It's a reasonable question...
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HOW-TO: http://ArduinoInfo.Info

Docedison

Terry the real problem is the "Top npn is an 'Emitter follower' and the bottom one a true switch thus there will never be more than Vcc- .6 - .7V applied to the load. Where when using a complimentary pair (Ptop and Nbot) both are switches, the N switches ground and the P switches + supply. You 'pull' current from the P base to turn it on and 'push' current to the N to switch the bridge to conduct. It is hard sometimes to visualize the reversal of polarity in the PNP. I do it by remembering that N is always base positive to the emitter to conduct and the P is always base neg to the emitter to conduct so one pulls down the P and pulls up the N to turn on the bridge... hope it helps...

Doc
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Ischia

Still not getting it, I'm afraid. Based on the tutorial link in my original post, I drew up something similar (ignoring the diodes) using only NPNs. Could someone please explains in simple terms the reasons why this fails/sizzles/explodes or otherwise (assuming I only apply positive voltage to diagonally opposed pairs at a time)?

Thank you again.

pwillard

#5
Jun 08, 2012, 07:25 pm Last Edit: Jun 08, 2012, 07:29 pm by pwillard Reason: 1
As doc said above...  only the BOTTOM pair of transistors are able to act like "switches" in your drawing... and you really want all 4 to behave that way (that's why you want PNP on the positive voltage side).  If the transistors on top won't act like a switch...   well, put another way... if you got it to work for you... you would then come back  to us and ask... "why is my motor only going s l o w ?"

zoomkat

One concept for the N issues being on top might be in the realm of the N on top emitter flow has to flow thru the load itself which might result in lower conduction issues across the transistor. A lower base voltage than collector voltage may also play into lower performance.  There are MOSFET H-bridges that use N MOSFETS on the high side. These H-bridges have boot strap charge pump type circuits that ensure the MOSFET gate voltage is higher than the voltage being switched. 
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terryking228

Quote
These H-bridges have boot strap charge pump type circuits that ensure the MOSFET gate voltage is higher than the voltage being switched. 


Exactly. Same deal for NPN high side switches.

As somebody, hopefully me, said:
Quote
it makes the design and implementation of the driving circuitry more difficult and expensive.


Transistors don't know where their next meal is coming from, or who feeds them base current.

For a high-side NPN to be fully turned on, just like a low-side NPN, there needs to be a voltage source higher than the collector supply.  An NPN in saturation (The Emitter-Collector voltage is .2 to .4 volts, say) has a base voltage higher than the collector voltage.  It's gotta come from somewhere.

Some power supply circuits, like my venerable HP bench supplies, have a separate higher-voltage supply just to be able to turn the NPN output transistors full on.

Ischia, you asked a perfectly good question, and the fact that you are not immediately satisfied with the answers tells me that you have the disposition to become a really good designer. Of whatever you care enough about designing.  You are probably not happy with Skool As It Is, and that's good. I say this as a person who has taught at SUNY and IBM, and personally heard Buckminister Fuller say, "Dare To Be Naive"..  Keep on pushing until you understand it!
Regards, Terry King terry@yourduino.com  - Check great prices, devices and Arduino-related boards at http://YourDuino.com
HOW-TO: http://ArduinoInfo.Info

MarkT

Although it is trickier to drive an all NPN H-bridge properly, its relatively easy for all-n-channel MOSFET H-bridges because of the rich variety of cheap driver IC's available specifically for the purpose.

n-channel FETs and NPN bipolar transistors are inherently better-performing than p-channel and PNP devices so it makes sense to use them throughout an H-bridge for improved switching speed and power-efficiency.

For a little 5V H-bridge on a breadboard built from bipolar transistors though the NPN + PNP H-bridge design is the simplest DIY one (I've done just that a few times to drive small motors).   Bipolar transistors also have the ability to do crude current limiting (you provide just enough base current to saturate under normal conditions - fault currents are then limited by the current gain.  MOSFETs tend to pull a LARGE current if there's an accidental short and some sort of protection circuit or fuse is generally a good idea.
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Ischia

@terryking228: Thank you for your kind words. I guess I'll come back to thinking about the issue after I've made some progress in electronics, as I have difficulties understanding even half of the replies above...

Ischia

Actually, here's another way to ask the same question. Please bear with me and remember everyone was a newbie once.

Still following the first example circuit on this page, I assembled a circuit on a breadboard which represents only one diagonal pair, bottom-left to upper-right, using only NPN transistors. It looks like the attached diagram, the blue circles are just LED+resistor pairs which I've put to see if the switches work.

Obviously, the bottom-left part works (the switch connected to the base turns on the LED attached to the emitter) while the upper-right part doesn't, even while keeping the bottom-left switch in the closed position (and therefore creating a path to ground through the load).

Why is that, and how will a PNP transistor in the upper-right section make this work?

Thanks again.

nickgammon

This is how I think of it ... I may be wrong but it makes sense.

Transistors have 3 legs, where the collector/emitter pair are the "output" and the base is the "input". But we all know that you never control a circuit with one wire ... remember to connect the ground!

So you have to ask, to turn the transistor on, it must be current flowing between base/<something>.

So what is this something? For a BJT transistor it is the emitter. And since you connect the emitter to ground for the bottom two transistors, then it is the base/ground difference that turns on the transistor.

If we look at the image on the page you mentioned:



The bottom two transistors are doing that. The emitters are connected to ground, and it is the base/ground potential that turns them on. But the top two do not have their emitters connected to ground, so we do not have a voltage reference "to ground" there.

In fact the emitters are connected to 5V (or whatever). So we have to flip the whole thing over by using PNP transistors. Now it is the base/5V difference that turns the transistor on. It is still the difference between the base and the emitter, but with reverse polarity we can now turn the transistor off by putting 5V at the base (assuming the supply rail is 5V) and turn it on by putting 0V at the base. In effect, you could say that the transistor is turned on by putting -5V at the base, since the polarity of the silicon is the other way around.

If you use all NPN transistors, as in one of your sketches, the motor is "in the way" of the circuit between the base and the emitter, and thus your attempt to turn on the top transistors will be thwarted (partly or completely) by the resistance of the motor coil.
Please post technical questions on the forum, not by personal message. Thanks!

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Ischia

#12
Jun 09, 2012, 12:13 am Last Edit: Jun 09, 2012, 12:17 am by Ischia Reason: 1
@Nick Gammon: Thanks to your explanation I'm beginning to see the light here. One last thing, though: suppose there was no motor or any other load "in the way" as you say. Even then, the top right NPN transistor still doesn't work *although* the bottom-left transistor is kept "active" (i.e. +5V to base, current flowing from C to E) and therefore opening a clear path to ground for the top-right one.

EDIT: actually the top-right NPN works with no load, I was silly enough to assume that a single LED as a load (in the place of the motor) would be insignificant, but it wasn't. If you remove the load and keep the bottom-left switch closed (i.e. transistor active), then the top-right NPN works just fine.

I appreciate your taking the time to explain all of this as it finally makes sense to me.

P18F4550

Nick, your arrows are pointing towards you pnp transistors, I think they should be pointing away from them as you sink the base on the pnp's not drive them but you've explained it well,
Here is a circuit i like, it uses a small npn to sink the pnp base and drive the npn base at the same time requiring only switching between forward and reverse, I have tried and tested this circuit and it works well


nickgammon

That wasn't my diagram, but do you mean the blue arrows pointing to the resistors? Yes I see what you mean there.
Please post technical questions on the forum, not by personal message. Thanks!

More info: http://www.gammon.com.au/electronics

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