Recovery Time of the Diode in a DC/DC Converter? - Page 1
Forum section: electronics / beginners

author: rentner
author_rank: Regular Contributor
timestamp: April 30, 2013, 12:43:55 am
content: What is the minimal Recovery Time of the Diode in a DC/DC Converter? I found a nice Diode for my purpouse, but is 35ns good enough? Will the losses be high or won't it even work? Or is that perfect for such a purpouse?Buck Converter?Boost Converter?Which Recovery time is recommented for 100khz Operation?

author: Kremmen
author_rank: Super Contributor
timestamp: April 30, 2013, 06:26:07 am
content: "Good enough" is a relative thing. If it is good enough for you circuit, then it is good enough. That of course is less than helpful so let's see:While 35ns recovery is not even close to the fastest there is, it is a reasonable value and will itself probably not pose a problem. However, that is not at all the only thing you need to worry about.In all switchmode converters the diode is biased in reverse and forward once per cycle. That means at switch turn-on you have to recover the stored (minority) charge accumulated in the PN junction once per cycle as well.To characterize the losses associated with recovering the diode charge we can reason as follows. Assume a simple buck converter and the phase where the diode conducts and the switch transistor is off:- initially he diode is forward biased (Vf) passing the inductor current (Ii). There is a diode specific minority carrier charge present at the PN junction.- the transistor turns on. Now the supply current starts to neutralize the junction minority carriers, and the carriers are neutralized by recombination as well. The magnitude of the recombination current is typically defined by the inductances of the actual circuit where the diode is located. Potentially the recombination current can be quite high, approaching a short circuit as seen by the switch.- The recovery current appears as a transient single shot half wave, the area of which corresponds to the diode reverse recovery charge Qr.During the recovery process, 3 points in time can be distinguished.At t0the turn-off starts by initiation of the charge recovery current in the PN junction. At this time the full supply voltage is effective across the switch transistor. Diode current rapidly rises to peak value which it reaches at t1.At t1the junction charge has been neutralized. Now the diode biases in the reverse direction and starts to block voltage. The transistor voltage starts do decrease rapidly towards its saturation value. The diode current decreases to the blocking current (~~0).At t2the diode is blocking the full voltage with effectively no reverse current.If the time interval t2- t1is long(ish) compared to interval t1- t0, the diode is called soft recovery. If the reverse is true, we have a snap recovery diode.Now to the point: transistor switching losses. In addition to all the other losses, the transistor has to dissipate that caused by the diode recovery as well. The exact form of the loss is:Wd=t0INTEGRALt2Uds(t) * ids(t) dtFor a snap recovery diode this can be approximated byWd= Uds* iL* tr+ Uds* Qr, and from therePd= Wd* fswwhereWd= energy in joules, dissipated during one switching event,Pd= dissipated power,Uds= (supply) voltage across the switch transistor,ids= transistor current,iL= load current,Qr= diode reverse recovery chargetr= diode reverse recovery time, divided into intervals t0, t1, t2fsw= switching frequencyMany high voltage rectifier diodes are doped P-N--N+effectively resembling the so called pin diodes with wide lightly doped central region providing the high breakdown voltage. It also brings a high amont of stored charge. Thus as a rule of thumb a fast high voltage silicon diode will also be one with a tendency for spiking during reverse recovery.There are soft recovery diodes that limit the current spike, but a preferred solution, voltage limits permitting, would be to use a Schottky type diode. While normal silicon diodes are bipolar, i.e. the current is the sum of movement of both majority and minority carriers (electrons and holes), in a Schottky diode the junction is a metal-semiconductor one and only majority carriers are present. That means there is no recovery phenomenon as such at all.If you want to minimize both recovery losses and forward conduction losses in a diode, then substitute a synchronous FET switch in its place. A FET transistor will not exhibit a recovery charge and its forward voltage will be significantly lower than that of any diode. The downside is increased complexity in controlling the synchronization but that is price that is worth paying in many cases.

author: rentner
author_rank: Regular Contributor
timestamp: April 30, 2013, 04:02:50 pm
content: Wow, that's, what I call an answer! Thank you for that scientific explenation. I'm sure, this will help me many times in the future.Thank you, that's all.

author: Ghydda
author_rank: Regular Contributor
timestamp: May 02, 2013, 08:02:50 am
content: To Kremmen:You sir, deserve a medalThat was one excellent piece text describing a very non-noob subject in a very noob-friendly way without loosing the fine details.This will get scissored into my notebook at once.