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Author Topic: 5V,2A H bridge help  (Read 13309 times)
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junaid_766
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« on: September 27, 2008, 03:30:40 15:30 »

I want to make a H bridge with discrete component (Mosfet) which operates on 5V, 2A can anyone post a circuit
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fernandodiaz
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« Reply #1 on: September 27, 2008, 08:35:28 20:35 »

Hi my friend.
dont lost time and money  buy
L293E   
chip dc H bridge

http://www.datasheetcatalog.com/datasheets_pdf/L/2/9/3/L293.shtml
and
LMD18200
LMD18201
LMD18245   
http://home.cogeco.ca/~rpaisley4/HBridge.html


I like more L293E 

Can make regulation curent add resistor, and thermal protection IC (read PDFs) to make connections pins
but two work fine...

Best Regards..
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junaid_766
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« Reply #2 on: September 28, 2008, 06:54:28 06:54 »

thanx
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fernandodiaz
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« Reply #3 on: September 28, 2008, 07:07:46 07:07 »



L6203    H bridge  surface mount


http://www.datasheetcatalog.com/datasheets_pdf/L/6/2/0/L6203.shtml
read this
Is special for build robotics projects
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Diramo
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« Reply #4 on: September 29, 2008, 01:07:08 13:07 »

MCZ33887EK - low cost H-bridge for motor application, 5V, 28A max. Datasheet here: http://www.freescale.com/files/analog/doc/data_sheet/MC33887.pdf
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junaid_766
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« Reply #5 on: November 02, 2008, 03:29:51 15:29 »

i had made an inverter (hbridge) and an inverter logic(2x4)  both work perfectly in simulation but when i had join them together an error occur an anyone help me ...im atttach its DSN file
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dennis78
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« Reply #6 on: November 14, 2008, 11:27:29 23:27 »

Schematic is good, but i think that you have problem with dead-time. Most of electronics simulators have problems in simulating H-bridge circuit. Ignore it for your real plans. This schematic principile work in real world, but without generating dead-time you can have problems with overheat or destroy transistors. In small time steps (when  bridge switching) you can have situation that Q5-Q7 or Q6-Q8 are ON im same time. This situation provide short high current peak through Q5-Q7 or Q6-Q8. It peaks is reason for overheating  or destroying transistors. For detailed information you can find on net (e.g. http://www.powerdesigners.com/InfoWeb/design_center/articles/Inverters/inverters.shtm).
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sohel
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« Reply #7 on: November 15, 2008, 09:50:24 09:50 »

u may read pic ref  from microchip AN967, AN877, AN843
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tAhm1D
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« Reply #8 on: January 07, 2009, 01:52:38 13:52 »

Hi Junaid,
I have seen your simulation. You have to understand the difference between simulation in Proteus and Real Life Application. Proteus simulates the Circuit without High Side Drive requirement but in Practical Circuit you require High Side driver for 2 High Side Mosfets which will provide Virtual Ground to those 2 High Side Mosfets. Otherwise all the Mosfets will conduct at the same time and all the Mosfets will be burned and the whole circuit may also be destroyed. You must have adequate knowledge regarding High and Low side Mosfet Driving procedure before attempting the H Bridge Circuit on ground. I will try to provide you a Practical Circuit
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tAhm1D
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« Reply #9 on: January 07, 2009, 06:49:38 18:49 »

Hi Junaid,

You can try with the following Circuit. Hope, this will solve your problem. You can use this circuit  upto 10-15A and 25V for low frequency application. Good luck.

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« Reply #10 on: January 13, 2009, 02:24:57 02:24 »

hi tAhm1D,

How to make 200mA constant current output from your schematics ?
I want to build 24v with 200mA constant current bipolar square wave at 50Hz.
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caveman508
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« Reply #11 on: January 13, 2009, 03:45:14 03:45 »

Free

You can use PWM from a PIC (or other controller) to control current, based on a current sense system and a flip-flop for a steering device an a bit of simple AND gate type-logic.  The "steering" FF needs to control the output of the PWM so that each side (polarity) sees the same pulse width.

That was a short answer, sorry.  I am just not clear what hardware and topology you have in mind.   

With this circuit (I am not sure, I am a firmware kind of guy with a fair knowledge of hardware), you will have to filter the output and set the table values for the PWM to do a square wave, I suppose you could do it with that sort of system.

I do know (been there with this stuff "H bridges" for many applications) that a fast PWM, say 50KHz can do this job.

I can be more helpful when I know what topology you plan to go with.

200mA seems small, but the same problems and control remain.  In my past, fun, but not easy.

Caveman
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tAhm1D
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« Reply #12 on: January 13, 2009, 04:56:08 16:56 »

Hi free,

To use this circuit for 24v and 200mA, change the values of R5, R6, R7, R8 to 1k. Do not connect the sources of Q2 and Q4 directly to ground, instead add a resistor in the path as a sense resistor. Using the formula,   V=IR,   there will be a voltage drop of 600mV across this resistor if its value is 3ohm. This 600mV will turn on 2 transistors which can ground the gates of Q2 and Q4, turning these off and therefore stopping the bridge from operating when output current becomes >200mA. The heat across the sense resistor will be:
P=VA or P=I*I*R or P=(V*V)/R, thus P=120mW. So you can choose a 1/4W resistor, or for more safety a 1/2W resistor.

You can use IRFZ44N MOSFETs as they can handle 24v and 200mA with no problems, but you can replace them with less powerful MOSFETs if you want.

For input signal, you can use SG3526 or 4047 or a 555 followed by Flip Flop stage or any oscillator that produces complementary outputs at 50Hz.

If you are still not clear, I can post a circuit.
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« Reply #13 on: January 14, 2009, 02:07:03 02:07 »

hi tAhm1D,

Please post the circuit....
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tAhm1D
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« Reply #14 on: January 14, 2009, 05:59:04 17:59 »

hi tAhm1D,

Please post the circuit....

Hi free,

I have made the circuit for you. Hope this will fulfill your requirement. This circuit is made for currents around 200mA, but if you remove R9, R10, R11, Q5, Q6 and replace fuse with appropriate amperage, you can use this circuit to upto 24v, 5-10A max with appropriate heat sink arrangement.

« Last Edit: January 14, 2009, 07:56:46 19:56 by tAhm1D » Logged
tAhm1D
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« Reply #15 on: January 14, 2009, 06:03:43 18:03 »

Hi Junaid,

For a working H-bridge circuit (I have posted it today), you can find it here:

http://www.sonsivri.com/forum/index.php?topic=15212.new#new
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caveman508
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« Reply #16 on: January 14, 2009, 09:24:36 21:24 »

This circuit should work, but be aware that during current limit, the two lower FETs (Q2 & Q4) will be forced into the linear region and begin to dissipate a fair amount of heat.  Make sure they are properly mounted to a heat sink that is capable of dissipating 10 or more watts.
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tAhm1D
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« Reply #17 on: January 14, 2009, 10:22:57 22:22 »

This circuit should work, but be aware that during current limit, the two lower FETs (Q2 & Q4) will be forced into the linear region and begin to dissipate a fair amount of heat.  Make sure they are properly mounted to a heat sink that is capable of dissipating 10 or more watts.

Hi caveman508,
If you look at the Circuit, you will find that all the Fets are always in Linear Region as the Circuit itself is Linear Type and the Frequency is 50 HZ. If you look at the Oscillating IC 4047, you can find that with Variable Resistor RV1 and Capacitor C3, it is producing 100 Hz and supplying to pin 10 and 11  opposite phased 50HZ to drive the Mosfets.

The Fets will handle max 5-8 watt load and hence for irf z44 this is very low amount of load and even then it is wise always to use heat sink with Mosfets.

Whenever load exceeds 200 mili amp, the transistors Q5 and Q6 will make the circuit off.
Very simple circuit but one has to understand clearly. I understand your concern and assure you that your concern has been addressed properly in the circuit.
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« Reply #18 on: January 14, 2009, 11:24:39 23:24 »

hi caveman508 and tAhm1D,

Your posts are very helpful to me, I will try the circuit.
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ajmain
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« Reply #19 on: January 18, 2009, 02:23:22 14:23 »

Hi,
The sense is not really essential. It can be left out as irfz44 is a powerful mosfet and can handle currents of >200mA if necessary and the input fuse will suffice.
But if your requirement is a necessary precise and limited 200mA, then the circuit is fine.

Just a question to caveman508, why would the mosfets be heated? Aren't they going to be turned off?
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caveman508
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« Reply #20 on: January 18, 2009, 05:42:05 17:42 »

This circuit will limit the current by dragging down on the gates of the lower FETs, not turning them off, but rather in that tricky area called the "linear" region.  Normally, in an H bridge, the FETs are biased such that they are saturated ON or totally OFF.  When a FET is driven into saturation, the only drop it has in voltage from Drain to Source is the rated specification called Rds-on that is the Resistance Drain to Source when ON, which can be in the 10's of milli-ohms easily.

So, in the case of current limit, and supposing that the top FET is fully on, the current limit is being done by the lower FET.  the power dissipated by the lower FET will be 24V * 200mA = 4.8 Watts.

I am not an analog engineer, so I could be wrong, but that is what I see.

Maybe Tahmid can better explain the biasing during both normal mode and during current limit.

Hope that helps,

Caveman
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tAhm1D
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« Reply #21 on: January 18, 2009, 08:44:01 20:44 »

Hi,
Heat dissipation of 4.8 watts is no problem for all the fets as each irfz44N can withstand unto 140watts. Yes, even the sense Transistors are not required for 200mA current,which is very less in comparison with the capability of the H Bridge Circuit. However, it is provided to show a precise circuit.
This is a low power linear circuit, so much calculation is not required. For High Frequency, High Powered circuit, lot of parameters should have to be calculated. Since my good friend caveman508 has touched regarding structural linear region of Mosfets, I would say something regarding that in brief. Mosfets fall in the structural linear state when the Drain voltage becomes less than the gate voltage and then Drain to Source Channel becomes very narrow and every possibility that the Mosfets become hot and get damaged as it fail to dissipate the internal heat created in access of Rds On. Here, the case in not like that as assumed by caveman508. Here, with access load, the gates of the lower fets will be off and hence circuit will be off till the required condition prevails.
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caveman508
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« Reply #22 on: January 18, 2009, 09:41:58 21:41 »

I will leave the rest of this discussion to all of you.

Best wishes,

Caveman
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sohel
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« Reply #23 on: January 18, 2009, 11:33:02 23:33 »

((caveman = digital) anding (tahmid = analog)) % yesno

Posted on: January 18, 2009, 10:47:19 22:47 - Automerged

please cotinue. it will answer a lot of quary.
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sohel
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« Reply #24 on: January 19, 2009, 03:57:47 03:57 »

A transistor drive pulse of less then 12v could result in excessive heating which occurs when the transistor being driven are operating in the linear mode of operation. which by the physical transistor construction, happens to be the most resistive regions of operation.,  by the way could u please upload another design for H-BRIDGE and Full-BRIDGE ?  SORRY IF I AM WRITEEN WRONG. NEED TO SLEEP.
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