detail understanding of catch diodes role in h bridge - Page 1
Forum section: electronics / beginners

author: jf_NH
author_rank: Newbie
timestamp: August 29, 2020, 10:37:21 am
content: HiI am trying to acquire a detail understanding of catch diodes role in H bridge topology, most importantly in relation to voltage spike management during dead time.I have searched through google but all the pages or youtube video "tutorial" I can find don't explain what I think are important details such as for example what happen if the 2 mosfet of the same leg turn off at the "same" time ?I am looking for a reference "specialized text book" level that will explain this subject in detail in relation with design trade off when you designing an H bridge.I remember reading through an application note from a major electronic company ( unfortunately I don't remember which one) on the subject a while ago...So if someone here could propose reading material on this subject it will be very much appreciatedThank YouJF

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 30, 2020, 02:52:51 am
content: The best thing is to completely forget there such a thing as a "voltage spike" exists.For a purely resistive load, the diodes would do nothing.For a (partly) inductive load. very basic inductor properties are enough to understand it.When you apply a DC voltage to a (pure) inductor, then current through the inductor increases linearly.When you short circuit a pure inductor, the current will keep on flowing for all eternity. ( Unavoidable wire resistance will make the current decay, not the inductor itself).When you reverse the DC voltage, the current decreases linearly to 0, and then reverses and increments in the other direction.On a graph it's a straight line.The inductance of an inductor is the rate at which this current changes.U = L * dI / dthttps://en.wikipedia.org/wiki/InductorThe current through an inductor is always a continuous function. There is no discontinuity in the inductor current.The inductor current can not be stopped "instantly"So when you apply a reverse voltage the current decreases.If you try to stop the current by opening one of the H-bridge transistors, then the current will just keep on flowing, and decrease with a rate at which it finds resistance.Usually the current keeps on flowing through the catch diodes.If there are no catch diodes, the current will keep on flowing through the transistor. The inductor really does not care that you turned the transistor off.The generated voltage spike is just an artefact of the current that keeps on flowing, and the magnitude of the voltage is usually equal to the voltage that the transistor gives up it's magic smoke and stops working properly ever after.To understand the working of H-Bridges better, it is good to have a model of a simple brushed DC motor. These are (from a modelling viewpoint) the easiest motors.A usable way to model a DC-motor is to represent it as a series connection of 3 things.1). The self inductance of the motor windings.2). Some resistance (Wire resistance, brushes, etc)3). A dynamo, that generates a voltage dependent on the rpm of the motor.Then there is also a "torque constant" for the motor.Torque of a simple DC motor is linear with the current through the motor (idealized, fricion, saturation, etc)To make a long story short, the catch diodes seem to act a bit differently with the inclusion of the dynamo effect of a rotating motor.Sorry for braking it off but my attention is wavering a bit at the moment.Maybe I'll be back later if you have more direct questions.