Originally Posted by Styx

Unfortunately this precludes the easy use of the hardware PWM.

PWM is just a ON/OFF signal. A pic is a digital device dealing with ON and OFF so what is soo special abt the PWM pin, that another pic cannot do? If it is just sourcing current, nothing a BJT to boost that cant fix

The PWM pin is connected to specific hardware inside the PIC, you simply set a few registers and it runs without any further intervention. To alter it's output you simply change the value in one of the registers, it makes PWM so very easy!.

You can do PWM in software (and so use any pin), but it's fairly complicated to write, it doesn't run anywhere near as well as the hardware one does, and it takes lots of clock cycles - the PWM hardware takes no processor clock cycles at all!.

There are various PIC's which have two PWM outputs, this makes them VERY useful for motor control in a small robot or model.

Originally Posted by samcheetah

i dont want to interrupt the ongoing discussion here but ive found a H-bridge circuit and its driver in the August 2004 issue of EPE. what are your comments about it

It's a crude little circuit, but fine for what it was intended. My biggest problem with it is driving the motor from the emitters, you lose so much battery voltage by doing that, and the transistors run hotter because you can't get enough base current into them.

The selected C on the gates (the value of about 10nF was needed) of the N devices may solve the spike problem at the rising edge but there's still a problem on the falling edge.

Makes no difference even with a 20K. You most probably need a big cap at the top as well.

Cap's at the top seems to work.

Originally Posted by TheOne

Makes no difference even with a 20K. You most probably need a big cap at the top as well.

I found this strange, so after trying another FET model I re-run the simulations. This time the delay was enough with a much smaller value of C (100pF) and the 7k resistor suggested by Styx originally.

That specific model must have been flawed as it made no real difference if I had a 1k gate resistor or 100k!

Originally Posted by samcheetah

actually we did a simulation in our lab in which we had to build a H-bridge circuit and drive a motor with it and measure the voltages and the currents. the circuit that our teacher gave us was simillar to this one. the motor was being driven by the emitters. uptil that day i had only seen H-bridges in which the upper transistors were PNPs (or P channel MOSFETs) and the lower ones NPNs (or N channel MOSFETs) so the motor was driven by the collectors. i told the teacher that the configuration that you have given us is wrong and it should be the other way round. he said that it doesnt matter, you just have to make a path for the current to flow. could you you explain how the base current is low and the voltage drop is higher

In the existing configuration (with the emitter the output), the emitter has to be 0.7V lower than the base - so you're losing at least 0.7V straightaway, plus losses in the transistor.

Then the base current, this is governed by ohms law, and is equal to voltage across the resistor divided by the value of the resistor. Because the base has to go high, you've got no voltage to work with - so the base can't go as high as 12V (again restricting the output voltage).

Assuming the base can get within 0.3V of the 12V rail, and the transistor has a gain of 100, and it's driving a motor requiring 1A. This gives a base current of 10mA. So to calculate the base resistor, 0.3/0.01, giving only a 30 ohm resistor feeding the base. The values used in the circuit shown will limit the output current considerably, resulting in a much lower voltage to the motor, and the transistors running MUCH hotter than they need to. There's also the obvious disadvantage of two 30 ohm resistors permanently across the power supply, wasting lots of your battery power! (one from each side of the H-bridge, either top or bottom, depending how it's switched).

My 'best case' calculation above would only supply a theoretical maximum of 10V from the 12V supply - but the values in the circuit would be far less under load.

With the transistors the other way round, things get MUCH better - for a start the voltage drop in that configuration is much less - 0.2V or 0.3V may well be possible, depending on the transistor. You've also got MUCH more voltage to play with for your base current, instead of the paltry 0.3V I suggested above, you've now got a whopping 11.3V instead! - so you don't need 30 ohm resistors anymore!.

If we assume a worst case of 0.5V drop across the transistors, we now have a reliable 11V for the motors, from the 12V supply, with the transistors turned hard ON, and running a great deal cooler.

Originally Posted by Styx

But With Common-Collector there is chance of shoot-throughs (common-emitter doesnt have this)

In what way?.

Originally Posted by Styx

you get them during a HIGH to LOW input to the leg - TheOne has already shown that with FET-based bridge leg. he same stands up true for BJT phase leg.

There is a finite time it takes to switch a device ON and OFF. It is worse for switching OFF the P-type/PNP since it takes alot longer than the N-type/NPN to turn-on.

I see what you mean, but I wasn't suggesting simply changing the transistors over, but using a correctly designed circuit for the configuration I suggested. I would tend to cross-couple the bottom transistor of one leg with the top of the other - this avoids any potential problems like that. It obviously introduces the possibility of turning all four devices ON at the same time (which isn't a good thing!) - but you can counteract that either in hardware, or by carefully written (and hopefully bug-free) software.

In any event, S/C failure of any one device in an H-bridge is likely to be fairly catastrophic in a similar fashion :lol: