What is the output switch saturation voltage when making a boost converter? - Page 1
Forum section: electronics / projects

author: WarFreak131
author_rank: Contributor
timestamp: October 09, 2020, 12:25:40 am
content: I am reading this datasheet to try and make a boost converter.https://www.ti.com/lit/ds/symlink/mc33063a.pdfOn page 13, it shows the sample layout for a boost converter, and it says the following:QuoteWhat is that? It lists the output voltage later, so they are not the same thing. I've seen this saturation voltage in many datasheets from many manufacturers, but I haven't been able to find a source on what it means.Thanks

author: aheid
author_rank: Regular Contributor
timestamp: October 09, 2020, 12:57:03 am
content: I interpret it as the voltage drop over the switching transistor when fully on. This is a typical parameter you find in BJT datasheets. For a MOSFET it'll be V_DS which you'll find a graph of based on V_GS used to drive it and I_D current.

author: MasterT
author_rank: Frequent Contributor
timestamp: October 09, 2020, 01:03:32 am
content: Page 5, 7.6 Electrical Characteristics—Output Switch.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 09, 2020, 01:11:41 am
content: Page 13, and referring to Fig.10?See 7.6, Vce(sat), non-Darlington, and associated notes.You saturate a sponge when it has no more capacity to absorb. Electronically, saturation is defined as the point where an amplifier cannot amplify any more, it has run out of operating range. The transistor Q1 is amplifying the current from Q2, drawing load current in the inductor and pulling the voltage down; the voltage can't pull any further than saturation. (Conversely, when Q2 turns off, Q1 also turns off, and the voltage saturates up to some maximum, defined by the diode's forward drop plus Co's voltage.)33063 is... maybe okay, for getting started, but it's rather mysterious to understand. It's not a great lesson, and it's not a smart design. It's an ancient relic, from the... what, 70s? The sticker price is tempting, but the sheer size of inductor and capacitors shows just how expensive these things are to actually use in production.I would strongly recommend a UC3843 or similar device to learn with: it uses current mode control, and is much faster. It can be used at higher power (external switch), and at higher efficiency (MOSFET switch). Current mode control prevents destructive operation, even under source and load extremes -- provided you've designed for worst case operation, which is pretty easy.Tim

author: WarFreak131
author_rank: Contributor
timestamp: October 09, 2020, 01:41:31 am
content: What determines whether I'd use Darlington or non-Darlington configuration? Would that just be a function of how much gain I'm looking to get?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 09, 2020, 02:05:59 am
content: Using the two internal transistors, a Darlington configuration gets you somewhat higher drive efficiency (drive current is drawn from the load), at the expense of saturation voltage.The gain is about the same either way, since you're using them at the same hFE(sat). Defining gain as load current over internal drive current, the Darlington case will be higher by about 10%, due to that sharing of current.In the non-Darlington case, the supply voltage dropped by the current-limiting/drive resistor dissipates excess drive power.Presumably, you'd calculate the losses both ways, and choose the more efficient option.Either way, efficiency above 70% isn't easy to get. The examples claim mid-80s % at full load, which I'll bet tanks quickly at lighter load.Note their choice of Darlington for the buck example, with its high (25V) supply voltage: the ~1V drop is a relatively small fraction of the total supply, making this a better option (it's also the only option, without adding a bootstrap or auxiliary supply).The boost example burns probably ~0.4W in the drive resistor; but this is probably worthwhile compared to the extra ~0.6W (1.3 vs. 0.4V, at 1A, times some duty cycle) a Darlington switch would cost.Both have pretty awful ripple...Tim

author: magic
author_rank: Super Contributor
timestamp: October 09, 2020, 06:24:46 am
content: Using this chip for boost is not even the worst thing one can do, because one could be using it for buckThen the darlington works as emitter follower and drops almost two volts rather than just oneI learned it by dissecting a Chinese battery charger which was getting unreasonably hot during operation.

author: Pawelr98
author_rank: Regular Contributor
timestamp: October 15, 2020, 08:57:09 pm
content: Unless you want to run on low input voltages just use UC3843 with some decent mosfet.In boost configuration it can easily achieve 90+% efficiency.Or just use some 1$ chinese module.They also offer excellent performance for the price.34063 cannot even reach 90% because it uses a bipolar transistor, a source of noticeable conduction losses.