SMPS soft start, why? - Page 1
Forum section: electronics / projects

author: Poe
author_rank: Frequent Contributor
timestamp: July 17, 2012, 06:02:35 pm
content: Thought I had a good grasp on SMPS theory until I discovered the "soft start" feature common on most BOOST controllers.Now I'm second guessing myself. They state this feature "gradually ramps the power in an effort to reduce stress during start up".Unfortunately I don't quite understand which parts are 'stressed' and why start-up is any different.Maybe I should back up. I'm designing a boost SMPS to simply light a 12Vdc string of LEDs from a ~2V supply (~7W). Currently the plan is to use a PIC's PWM and comparator modules. Voltage mode feedback only.I'm worried there's something I don't know and the lack of this 'soft-start" feature will bite me in the butt.Am I correct in assuming peak and average (switch/inductor/diode/capacitor) current during start-up is identical to peak current at full continuous power? What exactly is being protected during soft start?My first guess is that it protects devices which can not tolerate quickly rising supply lines. If this is it, I'm fine.My other guess is that when the output cap is near zero volts, the inductor will have a more shallow discharge slope resulting in residual inductor current.... or something along those lines. Since I don't have a current sense resistor in series with my switch, this might cause excessive current draw...hmm.. Is that right?Sorry. Long post. Any help would be appreciated and thank you in advance.

author: Simon
author_rank: Unknown
timestamp: July 17, 2012, 06:07:56 pm
content: it could be to stop overshoot on the supply's part an it damaging the load. it will be in fully on mode when first turning on and will hit the nominal output voltage so fast it would overshoot some before being brought under control. Yes I'd guess that having the caps all discharged means serious currents being drawn to charge it all up "instantly". This is all probably a problem with large supplies or delicate loads

author: Poe
author_rank: Frequent Contributor
timestamp: July 17, 2012, 07:11:10 pm
content: Quote from: Simon on July 17, 2012, 06:07:56 pmHello Simon. Thank you for the reply.Would there not be an identical amount of overshoot when the load transitions from full to nil? In my head, the same thing happens. i.e A 'fully charged' inductor is being dumped into a capacitor that was almost at the cutoff threshold. Once this happens the next cycle is skipped or truncated.My coworker mentioned 'in-rush current' as well, but the output capacitor and load are not being charged by a fixed voltage source. Instead they are being charged by an inductor which will supply a linearly decreasing current. The peak and average current into the capacitor is the same at start-up as during full load continuous use, right?I have a bad habit of second guessing myself and wasting massive amounts of time.

author: Simon
author_rank: Unknown
timestamp: July 17, 2012, 07:14:45 pm
content: no on start-up more current is needed to charge the capacitor as a discharged capacitor is effectively a short circuit. The overshoot can be dangerous because usually it is a buck SMPS and the input voltage is higher so with the switch full on the voltage could go well higher than the intended output. It is not usually possible for the output to drop under "0"

author: jerry507
author_rank: Regular Contributor
timestamp: July 17, 2012, 08:23:13 pm
content: From this datasheet:http://ww1.microchip.com/downloads/en/DeviceDoc/21876a.pdfWhen power is first applied to the MCP1650/51/52/53,the internal reference initialization is controlled to slowdown the start-up of the boost output voltage.This isdone to reduce high inrush current required from thesource. High inrush currents can cause the sourcevoltage to drop suddenly and trip the UVLO threshold,shutting down the converter prior to it reaching steadystateoperation.This inrush current comes from charging the output capacitor, and is in addition to the load. Especially as batteries start to get low you can easily trip the UVLO threshold.

author: Poe
author_rank: Frequent Contributor
timestamp: July 17, 2012, 10:10:55 pm
content: Thank you for the quote from that datasheet. Their explanation makes sense.To put in my own words...During start-up the INPUT capacitors are being charged causing a large in-rush through the source impedance and a relatively slowly increasing source voltage.If the controller attempts to charge the inductor the instant the source voltage exceeds the UVLO threshold, the increase in current draw through the source impedance might pull the input supply voltage low enough to trip the UVLO. This would be due to the source charging the INPUT capacitance AND charging the inductor. A condition that only happens at start-up and would not be a problem if the controller was disabled until the input capacitors were fully charged.Does that sound correct? If it's the only reason for soft-start, I might not have a problem due to a long power-up delay, low source impedance and low input capacitance.I don't think it has anything to do with charging the OUTPUT capacitor though. This UVLO issue would happen with or without a capacitor/load/short on the output. The output capacitor will never experience more peak current than the inductor since the inductor is essentially a current source charging the capacitor, right?Thank you both very much. It's nice to discuss these issues with someone.

author: IanB
author_rank: Super Contributor
timestamp: July 17, 2012, 11:04:05 pm
content: Remember that the output capacitor is going to look like a short circuit to the boost converter until it charges up. The boost converter is therefore going to try to draw the maximum possible current from the input source in response to that short circuit on the output in an attempt to get the output voltage up. This is what the soft start is designed to prevent. The output inductor is not current limiting in this scenario since it will take many operating cycles of the converter to charge up the capacitor.For an analogy, you can imagine the boost converter as a step up transformer. Suppose you connect a supply to the low voltage primary of the transformer and leave the high voltage secondary open circuit. The transformer will draw a little current in the primary, but not very much. Now imagine you short the secondary winding. As soon as you do this the primary winding will appear like a short circuit to the supply and will blow the input fuse.Although the operating mechanism is different, a DC/DC boost converter will behave in a similar way. Without overload protection in place a short on the output side will draw a huge current on the input side.

author: Poe
author_rank: Frequent Contributor
timestamp: July 18, 2012, 04:51:56 pm
content: The short circuit analogy and massive in-rush current only applies when avoltagesource is used, right? Like transformer, battery, or linear regulator output.As I understand it..This boost topology is more akin to a pulsed current source. In normal operation, the inductor is charged to a peak current (Ipeak). This is the largest current the output capacitor will ever experience. This peak current is limited by the current sense circuit and is no different at start-up than full power operation.Now... I do understand that during start-up the inductor will discharge more slowly due to the unusually low output capacitor voltage... resulting in an inadequately discharged inductor. This is the residual inductor current thing I was worrying about earlier. After thinking about it though.. The inductor current sense circuit would prevent this problem.err...Unfortunately I eliminated this circuit due to heat in the sense resistor...Maybe I can simply tie a capacitor across the top resistor of the feedback voltage divider? Wouldn't something like this create a pseudo soft-start?

author: Simon
author_rank: Unknown
timestamp: July 18, 2012, 05:07:18 pm
content: Quote from: Poe on July 18, 2012, 04:51:56 pmAnd at startup the current sense cutout circuit will keep cutting in as higher than expected currents flow to charge the output capacitor.

author: IanB
author_rank: Super Contributor
timestamp: July 18, 2012, 05:53:11 pm
content: Quote from: Poe on July 18, 2012, 04:51:56 pmRight. If you have a current regulated output (LED supply) then the circuit will be self limiting. The maximum input current will limited in proportion to the maximum output current. But you also have to consider how the boost circuit will behave when the output voltage is less than the input voltage. It will not be operating in its normal mode in that case and it may behave differently. The current sense resistor will still be there but the feedback regulation may not operate correctly. (I'm not saying your circuit will definitely have a problem, but these are things to think about when analyzing a given circuit.)

author: Poe
author_rank: Frequent Contributor
timestamp: July 18, 2012, 07:33:36 pm
content: Quote from: Simon on July 18, 2012, 05:07:18 pmI don't think this is so.Since this is a boost mode SMPS, there is no switch from input to output. The only switch charges the inductor. The current sense circuit releases this switch once inductor current (not output current) reaches Ipeak. When this switch turns off (from the current sense circuit or a time-out) the inductor wants to continue flowing at Ipeak. The only path is into the output capacitor. It starts at Ipeak and decreases. During that time, the output capacitor voltage slightly increases.No offense intended, I appreciate the conversation.I do see a problem during start-up, but that has to do with the inductor current building up due to my lack of current sense circuit. If someone has a possible solution to that I would be all ears.Quote from: IanB on July 18, 2012, 05:53:11 pmIt would definitely behave differently. I think the inductor current would continue to rise after the switch turns off due to the capacitor and diode acting like the switch. This actually doesn't happen since the output capacitors get charged through the inductor and diode along with the input capacitors as soon as the source is applied, right? The current sense circuit will protect the inductor...if I had one.I don't see why the feedback regulation wouldn't work properly... Unless you're talking about my pseudo soft-start idea. Yes, that would defiantly slow down the reaction time. Since I'm only powering LEDs with this supply I can't think how that could be a problem though. hmmm..

author: jerry507
author_rank: Regular Contributor
timestamp: July 19, 2012, 01:34:03 am
content: It would be quite a mistake to assume that the current limiting circuit inside a monolithic regulator works that fast. Also, the current limit value isn't exactly the maximum datasheet current. I have a circuit right now that uses a 175mA boost converter. The load pulses from about 140mA to about 190mA and the current limit NEVER kicks in during normal operation. These load spikes last about 500ms or longer. I'm actually limited more by the thermal cutoff in the device than by the current limit circuit.So no, inrush current will not be properly dealt with by the current limit circuit, hence the soft start circuit.

author: Poe
author_rank: Frequent Contributor
timestamp: July 20, 2012, 03:09:15 pm
content: How much would soft-start (a slowly increasing duty cycle?) help this simulation?Some input capacitance, a slight delay or a current sense circuit might help just as much if not more?The controller would wait until the input capacitor voltage exceeded some "turn-on" threshold. At this point Vout would be fairly close to Vin. Since a boost supply requires more input current than output, the input capacitor is not uncommonly larger.. making the above even more so. This is all assuming there's zero delay in demand. A little delay brings Vin and Vout even closer. Since the current sense circuit preventsthe switchfrom increasing inductor current beyond ~Ipeak, it alone would really help to eliminate this issue, no?I'm learning quite a bit just talking about this. Thank you. I suppose the general idea to pull from this simulation is that you don't want the output capacitance of a boost supply too large if you need your output to turn on instantly. Especially if you do not have a current sense circuit. It's a concept I never really thought about, but could see how it would really ruin your day.Luckily in my application the batteries are installed and the LEDs are not lit for at least one switch debounce period. By that time the ouput capacitor should be fully charged. I do not have a current sense circuit to protect the inductor from current build up, but I am planning on an additional inline circuit for linear current limiting after the output cap. That should cover all bases?Quote from: jerry507 on July 19, 2012, 01:34:03 amWould this be more of a comment on controller tolerances or some specific concept the soft-start helps with?

author: jerry507
author_rank: Regular Contributor
timestamp: July 21, 2012, 02:32:04 am
content: The soft start is intended to prevent an an event that lasts at most 100us. A current limit circuit on your average boost converter won't limit this current spike *at all*. However, your battery voltage from a large enough spike on a small enough battery will trip the undervoltage disconnect which will react that fast. That's the real reason for the soft start.

author: PChi
author_rank: Frequent Contributor
timestamp: July 21, 2012, 03:51:16 pm
content: Soft start can also be useful to limit the switch current.If the particular controller doesn't have soft start the duty cycle will be high at switch on (after the initial charge of the output capacitor up to the input voltage).When the duty cycle is high with the switch on for a long time the inductor current can get too high risking saturation and high switch currents and then smoke.Slow start can also help prevent boost converter instability where the switch isn't off for long enough for the energy stored in the inductor to be transferred to the output.

author: EEMarc
author_rank: Regular Contributor
timestamp: July 21, 2012, 05:06:46 pm
content: The overcurrent and instability problems can be solved by limiting the maximum duty cycle.The problem is when Vout is lower than Vin. For a simple boost switcher, there is a natural current that wants to flow from Vin through the inductor and then diode to the Vout and then to the output capacitor. The inductor resists the increase in current which will take time before it can build up. If you start by using a maximum duty cycle without delay, the current through the inductor will go through the transistor instead of to the output capacitor most of the time. The short off period charges the capacitor some. The switch turns back on and the inductor current continues to increase beyond normal levels and stresses both the inductor and the transistor. When the transistor turns back off, the current then stresses the diode and the output capacitor with above normal currents.If conditions are just right, the inductor can be driven into saturation and a runaway current will occur through the inductor and the transistor and maximum stress will happen to the diode and the output capacitor.A soft start is one method to prevent this problem since the low initial duty cycle allows the output capacitor to charge without the interference of a high duty cycle transistor hogging the current. It also can prevent an overshoot if the output capacitance is low.With a microcontroller, you can simply wait for Vout to increase through the natural current flow.

author: Poe
author_rank: Frequent Contributor
timestamp: July 21, 2012, 08:47:51 pm
content: Thank you PChi and EEMarc. That is exactly what I was talking about on the first page.Overall I feel much better excluding 'soft-start'. Due to my comparatively large input/output capacitance ratio and long delay, the output capacitor should be around 95% of Vin by the time my switch activates (worse case). This should protect me from any UVLO or inductor current build-up issues.BOM released to purchasing! Anxiously awaiting test phase!Thanks again everyone.