Confusion about H-bridge Inputs

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ANS HAFEEZ

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SALAM
I m using this H-bridge Circuit
i am little bit confused about iinputs of bridge

WHAT IF BOTH INPUTS GOES HIGH ???????
some web sites say it will damag the H-Bridge and other says It will creat Break Situation ?? 8-O
 

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when all inputs are given to 1, the supply is directly grounded and their will no current flow through motors.

so the motor will not run

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in your circuit if both a and b switched on

the transistors Q3 and Q4 will turned on and the Q1 and Q2 will turned off. their will be no supply to the motor but the ground will be enabled for the favor of both sides so the motor will not run. their will be no damage to any thing until the poer is given to the favor of any one side rotation and ground is enabled in favor of both side rotations.
 

You need to do some additional research:
You should change collector and emitter for Q6, otherwise it will not work.
There is no current limiting in the driving section (Q5, Q6), so you will fry your transistors in real world: you need resistors in the collector or emitter of Q5 and Q6.
To have better (faster) off-switching, you would like to place resistors in parallel with BE for all 4 H-bridge transistors.
When you drive both base of Q5 and Q6 high, all 4 H-bridge transistors conduct and you will fry the transistors.
You need fast switching diodes across all H-bridge transistors to protect them against back EMF from the motor.
 

Q6 is incorrectly inverted, collector and emitter. You need resistors in the base and collector of Q5 and Q6 to limit the current. With those changes the circuit will work with one input on at a time but will short the supply when both are on. To avoid momentary shorting during the switching time you need a non-overlaping driver circuit.
 

nightmare :shock:

I also used UF4007 diodes for protection
10k resistor for base and 1k fir collector

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non-overlaping driver circuit. ????

I used AND gate gate ic for inputs
 

Audioguru said:
I see no schematic for your corrected circuit.
Why did you use the extremely slow 1N4007 mains rectifier instead of a fast diode.
Click to expand...

No no its NOT 1N4007 ITs "UF" 4007

this is my improved diagram
 

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What is the power of the motor and what type of load will be connected? This determines the values of the resistors (together with the HFE to force the H-bridge transistors into saturation). You probably know that you haven't any protection. When for some reason the motor is overloaded, or you try to accelerate a heavy load too fast, it will draw much more current and may fry your transistors.

If you have them in stock, use schottky diodes. This is not because of the speed of UF4007, but because of forward voltage drop. You need resistors across each BE junction of the H-bridge transistors.

You should make a drive circuit that assures that before one of the inputs goes high, the other is off for a certain minimum time (think of a 2..10 us). The reason for this is that a BJT requires more time to turn-off then to turn-on. If you don't do this, all 4 transistors will conduct when switching polarity, and that will go wrong.
 

WimRFP said:
........................

If you have them in stock, use schottky diodes. This is not because of the speed of UF4007, but because of forward voltage drop. ............................

.......................
Click to expand...
I don't see any significant advantage for using Schottky diodes for the reverse biased protection diodes across the transistors. They only conduct for a short period of time when the bridge turns off due to any load inductance and the diode forward drop for that has an insignificant effect on circuit operation.
 

@crutschow: This would be true for MOSFET, but he uses BJT, and BJTs have low HFE in reverse direction as far as I know.
 

i am using 12v motor with maxiimum currunt of 0.5A and use the resistors of 1 watt

and i didnt get my answer right now ???
 

WimRFP said:
@crutschow: This would be true for MOSFET, but he uses BJT, and BJTs have low HFE in reverse direction as far as I know.
Click to expand...
Okay. But I don't see how the reverse Hfe affects what type of CE protection diodes are needed. :?: The transistors are never significantly turned on in the reverse direction, even with standard Si junction protection diodes.
 

A motor that draws a maximum current of only 0.5A is tiny, like the vibrator motor in a pager.
Maybe it runs at 0.5A but draws 5A when it starts or when stalled.

Instead of non-overlapping you should stop the motor before changing its direction.
 

crutschow said:
Okay. But I don't see how the reverse Hfe affects what type of CE protection diodes are needed. :?: The transistors are never significantly turned on in the reverse direction, even with standard Si junction protection diodes.
Click to expand...

BJT's do not conduct well in there reverse direction so curent goes through the protection/freewheel diodes, even when the BJTs are turned on.

When the motor current goes from left to right and you turn-on Q2 (right-up) and Q3 (left-down), the current must go via the freewheel/protection diodes (because of motor inductance, or some series inductance for EMI reduction). So when having a good selection of inductance and switching frequency, the diodes conduct during a significant time and then schottky rectifiers reduce the dissipation.

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ANS HAFEEZ said:
i am using 12v motor with maxiimum currunt of 0.5A and use the resistors of 1 watt

and i didnt get my answer right now ???
Click to expand...

What do you mean with "maximum" current of 0.5A?

What type of load do you have and how fast would like to accelerate that load, as that determines the value and the duration of the inrush current. When you don't know, try to measure the inrush current of the motor with load when applying 12 DC to it from a large power supply. You need an real time sampling oscilloscope for that and a small resistor for measuring current (via voltage measurement across the small series resistor).
 

The inrush current of a DC motor is caused because it starts when it is stalled. THE SAME amount of current flows when you manually stop it from spinning because again it is stalled. Then the current is simply the voltage across it divided by the resistance of it.

The small 7.2V motors in my electric radio-controlled airplanes have a normal maximum operating current of about 5A when climbing straight up like a rocket. But just cruising around the current is only about 20mA. When stalled the current exceeds the 8A maximum allowed current of the speed controller so it causes the motor to pulse and SCREAM then it shuts down the motor.
 
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@audioguru: I fully agree with you, but I don't recommend fully stalled current measuremnet procedure to somebody that is not fully aware of the consequences (he may fry a motor winding). To assess the dissipation in the transistors it is also good to know how long the inrush phenomenon takes in time, and then you need to have/know the load.

Measuring the DC resistance is a good one. If possible, I would also like to know the inductance to select a nice switching frequency that is still fine for his BJTs.
 


0.5A is a steady state current not inrush current

i m using a small motor of 12v with maximum 140 RPM whose steady state current is 0.5A
actually i m making a light weight car which is controlled by voice
maximum weight of my car is 3Kg
 

We need to know the maximum current which is also the inrush current which is also the [/b]stalled current[/b].
The maximum current is simply calculated by the supply voltage divided by the motor's DC resistance (Ohm's Law).

140RPM is very slow. Like a quick heartbeat. Putt, putt, putt, putt. 2.3 revolutions per second.
 

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