yes, its alright but you said that you were going to use an H-bridge driver. so why would you want to leave one of the base terminals floating?

 

I intend to use 2 PNP transistors at the top of the driver connecting it directly to the power supply. However, to turn off the PNP transistors, the base voltage has to match the emitter voltage. Obviously the PIC voltage to the base would be much lower than the power supply.

Therefore, just to confirm, leaving the base floating wouldnt cause any negative effects would it?

 

I don't really understand what you mean by this?.

You would normally have a resistor between base and emitter of the PNP transistor (to ensure the transistor turns OFF), with the base fed via a resistor from the collector of an NPN transistor - with it's emitter to 0V, and it's base fed from a PIC pin via a resistor. When the PIC pin goes high, it turns the NPN ON, this in turn pulls the base of the PNP low (through the resistor) and turns the PNP ON.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

i think you havent studied the pages i referred you to.

http://www.dprg.org/tutorials/1998-04a/

and ive found a couple of other pages for you

http://www.micromouse.co.uk/micropic/hbridge/
http://www.barello.net/Papers/H-Bridge.pdf

 

Actually i did read through the websites. Maybe my question was abit unclear. So what i was actually asking was about transistor Q1 and Q2. If Q5 and Q6 are OFF, wouldnt Q1 and Q2 base be connected to nothing(basically an open circuit)? Therfore, is it alright to leave it like that?

By the way, i require 4 inputs because i intend to use dynamic braking. IF i apply HIGH to Q5 and Q6 it that shold do the trick.
Anyway, just to reconfirm, is my circuit alright or did i left something out

Attached Images

hbridge.jpeg (18.5 KB, 313 views)

 

As I said above, it's advisable to have resistors across the bases and emitters of Q1 and Q2 - overwise any slight leakage in the NPN transistors could bias the PNP ones ON - which could well be disasterous!.

It 'may' work OK without them, but it's not guaranteed, and under some circumstances it could all go badly wrong - it's not good practice to leave them out!.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Base-emitter resistors will also allow the PNPs to turn off faster. When a BJT saturates, excess charge is stored in the base region, and it has to be cleared out before the transistor will turn off. Resistors will speed that process. Without the resistors, the charge has to drain off through the base-emitter and base-collector junctions.

 

This is fundamentally wrong. BJTs are current control devices.

j.

 

You can look at it either way, since voltage and current in the base-emitter junction are tied together in the diode equation, but I agree that, unless you are using a transistor in some sort of log converter, current control is a lot easier to understand intuitively. I would like to meet the guy who designs a saturating switch with the diode equation in mind. :roll:

 

dont mind me asking this, but what are darlington transistors used for? some h bridge drivers seem to be using it

 

okay lets talk about fundamentals.

why is a transistor called a transistor???

well, that because a transistor "transfers resistance"!!!!!

okay so what is resistance??

from Ohm's Law we have R = V/I which means that when you change the voltage across a resistor the current changes proportionally. so a change in voltage results in a change in current.

okay now what is transfer of resistance??

resistance is transfered when a change in voltage between two points results in a change in the current at some third point. so its basically just like resistance but the change in voltage and current is happening at different places.

thats exactly what happens in a transistor. a voltage across two terminals controls the current through the third terminal.

so does that mean that the equation Ic = BETA * Ib is wrong??? no, it isnt. it isnt wrong but it is just an approximation. it works in explaining alot of applications of transistors and thats why it is used so much. but that doesnt mean that a transistor is a current amplifier. to understand differential amplifiers, logarithmic converters, temperature compensation and other important applications you must think of a transistor as a transconductance device - collector current is determined by the base emitter voltage. and according to the Eber Molls model of the transistor

Ic = Is [ exp( Vbe/VT ) - 1 ]

where Is is the saturation current and VT is the thermal voltage.

the quantity Ib is just a by-product of the whole transistor-action. its something like who came first? the chicken or the egg? :lol:

and if you remember the voltage divider bias amplifier (VDB) you would remember that Ib and BETA are eliminated from the calculations. when designing a good amplifier Ib must be zero. so now i ask you that lets say you have a control panel consisting of Ib, Is, BETA, VT etc and lets name that control panel a "transistor" and the manual of this control panel says that keep Ib = zero for an optimum performance then how would you say that Ib is a controlling factor. either it is the controlling factor or it is an un-necessary factor in the whole process. thats up to you to decide.

now i know that it Ib can never be zero in a BJT. but thats the defficiency of the transistor. if it had been ideal there werent be any Ib. so if you see that Ib is always present then that doesnt mean that Ib is the controlling factor.

one thing more, do you know why Ib should be kept zero. because if Ib is zero then all the electrons entering from the emitter will go to the collector and become the collector current (in an NPN transistor) but if some of the electrons combine with the holes of the base they wont become the collector current. thats why Ib should be zero.

for further reference read the book "The Art of Electronics" from page 79 and onward.

and read the article http://amasci.com/amateur/transis.html

i hope that helps

 

Im thinking of using this Hbridge cct. Actually this belongs to PIKE. And i hope you dont letting me make of it as a reference.

As Nigel mentioned previously, a resistor is required in between the PNP base-emitter junction. Any value would do right? Just as long as it keeps the transistor OFF.
As shown below in this diagram, there are 2 10K resistors connected at the NPN emitters, why are they required ?

Attached Images

h-bridge-.jpg (24.6 KB, 252 views)

 

1. Lower value base-emitter resistors (Rbe) will allow the PNPs to turn off a little sooner, at the expense of requiring more current from the driving transistors. The current through those resistors when the PNPs are on is Vbe/Rbe. Even with 1k, that current is less than a milliamp. Since you are driving about 10ma into the base nodes, that is pretty insignificant. On the other hand, since you have to build in generous dead time at switching time, turn-off speed is probably not critical. For the circuit you have drawn, my personal choice would be 1k base-emitter resistors.
2. Notice that the resistors at the NPN emitters also connect from base to emitter of the lower H-bridge transistors. Without the resistors, those bases would be floating when they are off, just like the PNPs.

 

i have forgotten to ask about this. Is it alright to place a motor in parallel with another motor in the H bridge?
Im facing this problem where I dont have enough PWM channels to drive the hbridge. What im planning to use is the PIC16F876 which has 2 PWM channels. I'll also be using 2 motors as my forward and reverse(2 motors required due to heavy load).
If i build 2 separate hbridge for each motor, i would require 4 PWM channels to control the forward and reverse. Therefore, isit possible to place the motors paralel with each other?

THX for any suggestions!!!!!

 

why not use a single motor capable of driving the load you have???