H Bridge for DC motor(+schematic)

[thedude92]

Hello guys! I am participating in a project that involves building an H bridge to control the speed and direction of a DC motor ( with brushes).

The problem is I must use an already given schematic and I am having difficulties in understanding the way it works sinces it`s quite different from all the schematics that I have seen before. For example usually in an H bridge you have your ground at the bottom of the schematic whereas in my case the motor is at the bottom

main problem : NO place for the current to drain after passing through the motor. all ground connections are before the power mosfets and as far as i know current CANNOT pass backwards through them .

Triangular hexfets : TC 428 http://ww1.microchip.com/downloads/en/d ... 21415c.pdf

Square mosfets: IRF7425 P type power mosfet http://www.irf.com/product-info/datashe ... rf7425.pdf

Diodes : 30BQ015 http://tec.icbuy.com/upload/database/20 ... b52198.pdf

ORIGINAL SCHEMATIC : h-bridge-upb _original.jpg

my idea of modification : punte H modificata+observatii.jpg . I inverted the mosfets that I circled and added Ground to each of them at their drain point . I don`t know if it was a good idea but that`s my best shot.

So can anyone tell me if the original is functional? And if so explain were the f*** is the current draining??

P.S. the bridge is controlled by a Microchip Micro controller : PIC18F4XK20 ( if it has any relevance) it`s connectors i believe are at the top left and right of the schematic

[pebe]

Hi,
The circuit comprises an H bridge formed by the top four FETs, with a second H bridge using the bottom four FETs, overlaid on to it – I think to increase the current handling. There is no 0V rail shown so I assume it is at ground.

Let’s just look at the top bridge, because the second bridge operates identically. U1A is an inverter and (in the absence of any signal from J2) with pin2 tied to ground by R1 its output will be high, keeping the P channel FET U5 turned off.

U1B is a buffer and with its input tied to ground by R2, its output will be low, turning on the (unmarked) FET below associated with D2. So with the motor connected between the pins of J1, the LH side of the motor will be at 0V

There are identical conditions for the two RH FETs so the RH side of the motor will also be at 0V. The motor is not running.

Now take both pins of J2 high (fed from a PIC at5V). The outputs of U1 will both change states, turning U5 on and turning off the unmarked FET below it. Current will now flow from +12V supply through U5, RT1, the motor, RT3 and the unmarked FET below U11, to ground. That will make the motor turn in one direction.

Taking both pins of J3 high, instead of J2, will turn on U11 and turn off the FET below it, so reversing the direction of current through the motor.

I have never before seen a circuit where the complimentary FETs are switched separately as they are here. There is a possibility that if only pin 1 of J2 is switched high, then U5 and its complimentary FET will both be switched on – with disastrous results! The same goes for J3. But judging by the missing numbers in the sequence of IC labels, this may only be a part of a larger circuit, and the problem may be taken care of elsewhere in the circuit. But if not, beware!

I hope that helps.

[thedude92]

hi there pebe sorry for answering 2 weeks later but I`ve been through a series of exams with next to no sleep. However I`ve managed to get my project supervisor to take a look at the schematic and she admitted there were some mistakes and gave me the original schematics designed by the guys at Microchip. If you are still curious how it works here`s a screenshot

it was designed to be used with the PIC16F684 microcontroller

[pebe]

Hi thedude,
Thanks for that. But it's the same circuit as the original except the FETs are shown by their symbols - rather than by their pinouts.

It does not explain why P1A and P1B are taken out to separate pins - rather than being connected together. Same applies to P1C and P1D.

It would greatly help if your project supervisor could provide a link to Micxrosoft's writeup on the circuit. I mean the application note number, or whatever.

[thedude92]

I think this might work

http://www.microchip.com/stellent/idcpl ... e=en012134

http://ww1.microchip.com/downloads/en/A ... 00893a.pdf << entire circuit

[pebe]

Microchip’s Application Note makes it all clear.

Speed control is done by PWM. They include a PGA (not shown on your circuit) to enable back EMF measurements to be made when the motor is cut off, to measure the motor speed.

The micro will turn on Q2 and Q4 to ground the left hand side of the motor. The voltage on the other side of the motor at ‘RPM’ is then measured whilst all other FETs are switched off.

[thedude92]

One minor thing I`m not sure of: shouldn`t Q2 and Q4 ( respectively Q6 and Q8) have pins 1 and 5 reversed in the circuit. According to the IRF7425 datasheet 1 is the source pin and 5 is the drain the and current cannot pass from 5 to 1 to drain as far as I know, right?(without destroying the component)

Therefore lets say P1A and P1D are both OFF and do not allow current to pass and P1B and P1C are ON allowing the passing of current. The current flows from the +12V line through Q5 and Q7 through the engine and when it reaches Q2 and Q4 it enters through the drain pins according to the schematic. Is that alright??

Am I thinking it wrongly? As far as I know current cannot pass both ways through the fets(might be wrong on that one).

[pebe]

But there are two different FETs used - a 'P' channel IFR7425, and an 'N' channel IFR7457, and they handle current in opposite directions.

With Q1 and Q6 switched on, current will flow from source to drain of the 'P' channel Q1, then through the motor, then from drain to source of the 'N' channel Q6.

Does that clear up the problem?