H-Bridge flyback

Hi there!

I hope this won't be a duplicate topic, but I wasn't able to find an answer to two simple questions about h-bridges that I have.

  1. I am building an h-bridge with tip122 / tip127 darlington transistors, following the ideas in http://letsmakerobots.com/node/27578. Why do some h-bridge designs (see http://letsmakerobots.com/node/20466) put a "small" transistor (like 2n2222) between µc pins and the power transistors?

  2. I've understood the use of a single flyback diode for clamping motor reverse emf, but can't understand how the diodes in the h-bridge can be useful. I think that the back emf will make a current flow backwards in the power supply with that design. Isn't it a problem for the supply?

Thanks

ik

1) The reason for the "small" transistors is to allow the input to control the H bridge with a smaller voltage swing than the full 24V motor voltage. The H bridge using the four emitter follower output transistors will have to have an input voltage swing equal to the full 24V motor voltage. The arduino output voltages are not high enough to control the H bridge without the "small" transistors.

2) If protect diodes are not included to sink the flyback current then the output voltage can get high enough to damage the output transistors.

Sorry, but I can't understand. For example, the designer of the h-bridge I attached says "The circuit requires the inputs to be about 12v" . Why? Isn't it just good to turn on the transistors with some voltage above 2.5v (that is the Vbe(on) of the tip127/122)?


Firstly that circuit's a bad circuit, forget it. You never use emitter-follower configuration for a switching device.

Each circuit has its own specific drive requirements - understanding transistor and MOSFET function is pretty much a prerequisite - a non logic-level MOSFET for instance needs 10 to 12V of gate drive, that's just how it works.

I am building an h-bridge...

An interesting read on DIY h-bridges.

http://forum.arduino.cc/index.php?topic=53425.0

MarkT: Firstly that circuit's a bad circuit, forget it. You never use emitter-follower configuration for a switching device.

Can someone explain me why the emitter-follower configuration isn't good for switching? Thanks :)

Yes, it wastes more power than the usual common-emitter configuration when switched on. Good switching circuits are power-efficient.

A good switching BJT will have Vsat from 0.05V to 0.4V or so, this is the voltage "wasted" across it when switched on and carrying load current. The power wasted = Vsat x I

With an emitter follower (we'll go with that circuit's darlington devices) you have a Vsat of 2V (typical for Darlingtons) and also there is some voltage across the base resistor needed to provide the base current, again this might be a volt or two, leading to 3 to 5V lost across the device. With a simple BJT, not darlington, you'll see Vsat = 0.9V or so (we are talking power devices, Vsat is higher due to emitter IR losses). Add a volt for the base resistor and you're up to 2V lost.

In either case the common-emitter circuit is more efficient since there is no voltage lost due to a base resistor and the base voltage can be higher than the collector voltage (the definition of saturation really).

There are circuits using what looks like emitter-follower, but the base drive voltage is bootstrapped up to above the supply voltage to allow the device to saturate properly - these perform OK.

Thank you MarkT! Now I've got it.

My second question, instead, is still there: let's suppose I've this good-designed h-bridge with flyback diodes. In my case, I want to power a dc motor with a modified atx power supply. When I switch off the motor, I will have back emf forcing a current from ground to the supply "high", am I right? I attached an image I found on this forum, describing this. So I'm wondering: won't that be bad for my power supply? In this post http://forum.arduino.cc/index.php?topic=50931.0 MarkT pointed out that

… Some [power supplies] can [sink current] and some can't.

This solution from evading seems good: http://forum.arduino.cc/index.php?topic=50931.msg1252094#msg1252094. The freewheeling current is kept "in the h-bridge" by switching independently the four mosfet/transistors. If using that approach, can you suggest me how proper timing for the "back-emf discharge phase" can be calculated? :sweat_smile:

Thanks!