how this H-brigde is working?**

Please help me out to understand *how this H-brigde is working?
i tried but i couldn't make any sense.
i mean i know the general working of the circuit...but i don't know the reasons why one transistor goes to Saturation and other goes to cuttoff...
i found it through google.it is attached in the form of image..
thanks

It may help to check up PNP & NPN transistors.
Then you will see that the top transistors will work with a positive signal and the low transistors when there is a connection to ground.
The small arrow in the symbol can help you

Hi there,

To keep it simple and not take all the fun of researching yourself from you:
- treat transistors as switches: if base-emitter voltage is >0.7V and its polarity is the same as the transistor arrow than the switch is on = saturation, else the switch is off = cutoff.
- take what I said above and the fact that the 2 sides of the bridge are identical, let's examine the part with Q1, Q2 and Q5. If input A is high, than Q5 is on, and the bases of Q1 and Q2 are pulled low through Q5 stronger than they are pulled high through R1. For Q1 this means off and for Q2 this means on. If input A is low, than Q5 is off, and the bases of Q1 and Q2 are pulled high through R. For Q1 this means on and for Q2 this means off.
- do the same for the other side and you have your explanation.

If this explanation is homework than this is a starting point in getting to a complete explanation... simply taking this and copying it will not get you a passing grade. If you are genuinely interested there are tons of information about BJTs on the internet and you can understand by yourself if you understand how BJTs work. (trust me it sounds more complicated than it actually is).

Also the schematic is done in Proteus Isis, try to resimulate it yourself and probe the different points in the design to see what happens for different A and B input values.

Best of luck!

Also, read up on switching transistors and why they are biased to stay out of their "active region" (i.e. flip from saturation to completely off).

Actually, what difference does it make if he/she (don't know the gender of the name "wiki", but I'll call it "he" from now on so i don't go nuts!) reads it here or somewhere else...and if he chooses to "cheat" and merely copy this, then that's his responsibility--karma will take care of that ;)

So...

The reason a switching transistor is biased to avoid it's active region, is because the idea is to send as much of the power, as possible, to the load that is being switched. When the transistor is, say, half on, it will be dropping half the voltage, and thus for the current being drawn, only half the power will reach the device. The other half will be dissipated across the transistor. Why do you suppose that is not such a good thing?

Now, as a thought experiment, using the diagram, below, consider what happens when the transistor is at saturation, and then when it's at cutoff. How much power is lost in the transistor and how much reaches the load (hint, in one case, neither dissipates any power). And then, think about what happens if the transistor is only partially on.

Regarding the H-bridge, look at the current paths through the transistors for each of the states in the truth table using what you learned from Dorin's post. In the case of the H-bridge, there are two switching transistors in series to drive the load. See if you can relate the current paths and their affect on the motor, to the label in the truth table (e.g. "Stop", "Forward", "Backward" and "Off").

I actually didn't forget to add the diagram, it just didn't upload for some reason (even though it's a PNG!!) And, because there's no @#$%! preview, there was no way to catch it before going live--just sayin' in hopes the webmaster is watching

Thanks for the explanation and considerations.
If a switching transister is that vulnerable for overheating, what should I do to work around that or better to avoid that.
Are the characteristics of an IGBT better, i.e. are they switching quicker, so that there are not so many losses. Or is this completely the wrong track (as I think, I'm quite unexperienced)?
Thanks for any further consideration. (currently I'm sticking with chopper circuits to drive my high current motors, is that perhaps the right track - i.e. keeping the switching transistors' count as low as possible?)

Oh, I see, you mentioned a switching transistor is "biased to avoid its active region", that explains the solution to the heat problem - unless they still get very hot.
Would the use of strongly overrated Transistors fix the heatsink requirement? (only out of interest, considering finding the transistors at same price)

Do the analysis:

Consult the datasheet and determine the Vce(sat). Then knowing the current, you can compute the power that would be dissipated and knowing that and once again consulting the datasheet, determine if the transistor is up to it by looking up the Ptot.

A MOSFET tends to be a better choice than a bipolar transistor when large currents are involved, but the manner of driving MOSFETs is different. IGBT are only viable for high voltage applications.

Thanks For the Reply dear...And don't worry this is not my assignment or something based on which i would be graded, so there is no chance of copy paste ...

i have understood a lot..and i am very thankful to you that you gave me a nice advise and a concept of pulling voltages(it was new to me).but there are still some confusions.

1)i have learned that for a PNP to be in saturation Veb>=0.7V ... due to some saturation resistance of Q5 the base of Q2 would not be at exactly 0V... then how are you so sure that *Emitter Voltage* of Q2 is at-least 0.7V greater than its *base voltage*??

NPN's turning ON and OFF is okay but i couldn't understand how PNP is getting saturated??..

thanks for the Reply..i will edit my profile asap. :p

your reply added a very good piece of information to my little knowledge.

and i have already predicted all these things....but i couldn't co-relate them with what i have learned in books...
and my circuit analysis is weak....

To finish my thought (which I forgot to do), a switching transistor can still get hot, but that will happen at MUCH higher currents than if the transistor were biased in the middle of it's active region. A switching transistor, when fully on, is still not a perfect switch. There is still some "resistance" (I put resistance in quotes, because it's not likely to be a linear resistance, at least, not in a Bipolar Transistor--a MOSFET, on the other had, will exhibit a linear resistance [as long as the temperature doesn't change]).

But, the bottom line is, there is still a voltage drop and thus a power loss (a perfect switch, when "on" would have zero voltage drop and thus would dissipate NO power, and thus not get hot). What happens to that power? It is converted into heat by the resistance in the Vce path.

MOSFETs are purely resistive devices and thus can be made with very low channel resistance (on the order of milli-ohms)--thus they are better for high current applications, as their low channel resistance (i.e. Source-Drain resistance) means they can control very large currents with little or no power loss and thus little or no heat.

See why it's difficult to explain these things? There are so many variables.

Maybe this will help. One thing that used to puzzle me about bipolar transistors was the question of how could the base voltage be greater in magnitude than the Collector-Emitter voltage? The Vbe on a transistor can be .7 volts and yet the Vce can be as low as 0.1V [PN2222A]. If you consider that the Collector-Base junction is reverse biased this becomes more plausible.

exactly.....

actually this was me 2nd question.but i forgot while replying.

i have also learned that for a transistor to be in saturation both the EB-junction and CB-junction should be forward biased..

so if we consider Q1
when A is low.the bases of Q1 and Q2 would be pulled to the +Vcc .. so i can say that the Base would be at higher potential than emitter and so EBJ is Forward biased.... but when it comes to CBJ... here shouldn't the base be higher than collector as well,for it to be a forward biased??
but from the cct base can't become greater than collector!!
so according to theory the Q1 should never come in ON condition. !!

OK, now I learned something. In this arrangement, none of the transistors can ever reach saturation. Why? Because, the emitters can never go higher/lower than around .6-.7 volts below/above the base voltage.

Here's an example:

Consider Q1 -- if the "A" input is low, then R1 is free to pull the base of Q1 high. The highest the base can be pulled to is V+ (lets call it "Vp" so the math is less confusing). Thus, the highest that the emitter of Q1 can be pulled to is around Vp - 0.7volts. Because, if the emitter were to go any higher, the transistor would begin to turn off. Instead, it finds an equilibrium point between "shutting down" and "turning on", and that equilibrium point settles at around 0.6 to 0.7 volts across the emitter collector junction. Thus, this circuit may not be the best in terms of efficiency. It appears to be a compromise between simplicity (and thus cost) and efficiency.

Below is a circuit where the transistors _can_ go to saturation, but as you can see, the trade off is greater complexity.

Now, I'm not sure, but I suspect that this isn't an issue for high power transistors that typically have an Vce saturation voltage of a volt or more, so this may only be an concern for small current switchers such as the 2N2222.

I didn't catch this, because I primarily work with MOSFETs which saturate with such low resistances that they behave, nearly, like ideal switches.

Addendum:

bq). Now, I’m not sure, but I suspect that this isn’t an issue for high power transistors that typically have an Vce saturation voltage of a volt or more, so this may only be an concern for small current switchers such as the 2N2222.

Because of the availability of FETs, it seems counter productive to use bipolars for this application in all cases but a small power load, and in that case, the slight loss of efficiency is tolerable.

Also, this arrangement of complementary push-pull bipolars is often used to drive the gate of a MOSFET where the only current involved is charging/discharging parasitic capacitance. Thus, the fact that the transistors never reach saturation is not a concern, and in fact is a boon, because they will switch faster.

Well...for some strange reason my circuit diagram won't display, so here's a second attempt...