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Can anyone teach me how this H-brige is working??

wiki

Sep 28, 2012
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Please help me out to understand how this H-brigde is working?
i tried but i couldn't make any sense.
i found it through google.it is attached in the form of image..
thanks :)
 

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Harald Kapp

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If all transistoirs are Off, the motor stops.
If Q1 and Q3 are On, the motor turns in one direction.
If Q2 and Q4 are on the motor's direction is reversed.

For a complete explanation of the workings of an H-bridge see e.g. this wikipedia page.
 

wingnut

Aug 9, 2012
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Q3 and Q6 supply an amplified signal to the H bridge.
This signal (+ or -) is taken to the bases of Q1 and Q2 which are joined together. Thus if a + signal goes to a pnp (Q2) it turns it on, while simultaneously turning off the npn (Q1).

Thus either Q1 or Q2 are turned on at any given time, but not both.
Suppose a + voltage comes to the bases of Q1 and Q2, Q1 turns on, and Q2 off. This supplies current to the motor from left to right.

The exact same logic applies to Q4 and Q5. Suppose a negative (0V) voltage is applied to their bases, Q5 turns on, but Q4 turns off.

Thus current flows from the top (+ power supply), through Q1, through the motor, to Q5 and ground. Reversing the bases causes current to flow from the + supply, through Q4, through the motor, through Q2 to ground, reversing the motor.

The diodes and resistors are there to protect, so ignore them for now.
 
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wiki

Sep 28, 2012
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Thanks for the reply...but i have still some confusions..

1)if +ve voltage comes on the bases of Q1 and Q2,then Why does Q2 goes to Cuttoff region...why not Q1 goes to Cuttoff??
2)How someone had Selected R1 to be 1kohm??why not some other value??
3)....(i just forgot the third confusion....... :D)
 

BobK

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The Q1 and Q2 bases will be pulled to ground if input A is high, or pulled to V+ (by the 1K resistor) if input A is low. The value of 1K was chosen to provide enough base current to saturate Q2.

Bob
 

wingnut

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Do you know that npn transistors like Q1 require positive bases for them to conduct and pnp transistors like Q2 require negative voltage for them to switch on.

By positive or negative we mean .7V more or less than the emitter.
 

wiki

Sep 28, 2012
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i have also learned that for a transistor to be in saturation both the EB-junction and CB-junction should be forward biased..

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!!
 

wingnut

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Suppose you are using a 6V power supply.

The collector of Q1 sits at 6V
The base of Q1 and Q2 sit at 6V
The emitter of Q1 sits at 0V because it is shorted to ground by Q5 being switched on. Thus the collector voltage is above the emitter voltage and the base is above the emitter, so it switches on.

Here are the rules for an NPN bipolar junction transistor

" 1. In order to conduct, the voltage difference between the collector and the emitter (VCE) must be above a certain threshold (say, VCE > 0.2V).
2. In order to conduct, the voltage difference between the base and emitter (VBE) must be above a certain threshold (say, VBE > 0.7V).
3. Once it is conducting, the current flowing from the base (IBE) causes a current to flow from collector to emitter (ICE), amplified by a factor \beta \equiv h_{fe}. "

There is no rule that the base voltage must be above the collector voltage
 
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