What DC-DC topology should I look for? - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 01, 2021, 02:36:56 pm
content: I built an adjustable power supply using two buck converters. It worked well, but I found some limitations with it, specifically it only stepping down the voltage. Now, I'm looking at making a new version that would either step up or step down the voltage, and I need some help.My previous design took a DC voltage (13.2 V) as an input and used two separate buck regulators to step down the voltage to a positive and negative value. This worked extremely well, so I'm hoping to accomplish the same thing. Considering how I want to either step-up and/or step-down my output voltage, I'm looking at the following topologies: buck-boost, SEPIC, and Cuk. Flyback topology appears to require a transformer; as both the positive output and negative output will be referenced to the same ground, a transformer appears to be detrimental to this goal. Plus, looking at Digikey, an isolation transformer seems rather large, and I want to keep this as small size as I can.Given this, how do I search for a positive regulator? I'm having trouble searching for one that can output 2 amps. Searching for a negative regulator is straightforward (I was led to the ADP5074), but I was wondering what buck-boost converters would be good? What would you guys recommend, or are there any topologies that I should also have a look at? Looking at a roughlinkon digikey, all of these seemed overly complicated with pins for control signals and whatnot. Plus, they seem way more expensive than what I was thinking.

author: Benta
author_rank: Super Contributor
timestamp: September 01, 2021, 03:53:00 pm
content: What kind of output voltages and power levels are we talking about?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 01, 2021, 03:55:47 pm
content: Just use a boost/flyback controller and SEPIC.Buck-boost is inverting, or do you mean the four-switch "flying inductor" kind (literally a buck into a boost, sharing the same inductor)?If you need tracking bipolar outputs, SEPIC and Cuk can even be combined. You need a multiwinding inductor to do it, which can be something like Coilcraft Hexa-Path, or just a couple dual-winding inductors with the primaries wired in parallel and the winding ends all cap-coupled together. Then only one boost/flyback controller will suffice (though you should add an op-amp to get joint +/- regulation; the -out can't simply be summed into the feedback node by itself).Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 01, 2021, 07:40:04 pm
content: Given a 13.8 voltage input, I would like to be able to have voltage outputs ranging from -15 volts to +15 volts minimum. So, the positive voltage output converter would go from 0-15 volts while the negative voltage output converter would go from 0 to -15 volts, at minimum. That way, I can use this power supply for a wide variety of applications, specifically when I need such voltages to power op-amp circuits. It was my mistake for thinking that a 13.8 V DC supply would be good enough for this, so I'm trying to rectify that by going with buck-boost rather than buck converters.Looking at the available buck-boost converters, I don't think I can't go with the current output I previously had. So, I'm looking for a max output current of 2 amps minimum from each IC. Previously, I used the LM2673 (3 amp output) for positive voltage. For negative voltage, I was planning to use the same IC, but I heard somewhere that rearranging the components to get a negative output voltage would reduce the max output current. Thus, for the negative voltage, I went with the LM2679 (5 amp output) to compensate.

author: Benta
author_rank: Super Contributor
timestamp: September 01, 2021, 08:56:14 pm
content: So, +/-15 V output at 2 A. OK. Balanced voltage or independently adjustable?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 01, 2021, 09:00:35 pm
content: Quote from: T3sl4co1l on September 01, 2021, 03:55:47 pmThanks for your reply. Yeah, looking at the types ofconverters, I guess I was thinking of the flying inductor that you mentioned or the four-switch buck-boost converter. Basic buck-boost is inverting, but the flying inductor keeps the same polarity as the input.If I may ask, what do you mean by tracking bipolar outputs? I don't believe my application doesn't require tracking both output voltages together. I would like to have them independent of each other rather than have both be the same magnitude. For some applications, I need both voltage magnitudes to be different, so they don't need to be tracked, if that is what you're referring to.Having another look atDigikey'slist of parts, these were the list of components that seem to fit the bill. Their topologies are all different, though I think it means that the IC can be used in various topologies rather than be limited to one (correct me if I'm totally mistaken). What are the pros/cons of all these topologies here?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 01, 2021, 10:35:47 pm
content: Independently adjustable.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 01, 2021, 10:49:29 pm
content: OK.Just as well then, use two boost/flyback controller/regulators, one for each channel, SEPIC for positive, Cuk for negative, and now the added op-amp will be compulsory, to invert the negative output voltage to a positive sense voltage for the regulator.No need for fancy controllers or anything, bog standard like UC3843 or newer will do.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 02, 2021, 03:34:06 pm
content: Thanks for your reply. I'm learning a lot of interesting stuff through this discussion. If I may, I'd like to discuss this with you a bit more. (Boy, this all could have been avoided had I thought about the input voltage a bit more)So, I've attached my first design below as an image. Putting aside op-amp comparators for now, given the voltage goals I'm trying to achieve, I think my negative output voltage might be okay for now. I was researching for alternatives for both the positive and negative ICs. When I saw that the LM2596 can be reconfigured as a negative voltage, it made me think about what I've done. The datasheet for the LM2596 stated that the absolute sum of input/output voltages cannot exceed 40 volts. Assuming that the same principle applies with the LM2679, with an input voltage of 13.8 volts, I should theoretically be able to reach -15 volts. If this is the case, then maybe I could be fine with what I have for the negative voltage. (certainly a lot better than what I was considering, the ADP5074; max switch voltage also seems to have a difference of 40 volts with PVIN, so I would still be facing the same output voltage limitation as with the LM2679) {Hypothetically, if that's the case, if the input voltage is 5 volts, could the LM2596 output -35 volts?}Now, all I have to worry about is the positive voltage. Going by what you said, the LT8364 seems like a good IC to use as a SEPIC converter. Going by this though, I know all ICs are different, but for SEPIC converters, what is a rough limitation on the output voltage given the input voltage? Similar to the constraint above, would a SEPIC converter be limited in the same way?EDIT: Sorry, but this thought occurred to me. On the other hand, if my problems all stem from the input voltage being limited.......The reason I went with a 13.8 fixed supply is because of current issues. Start-up current was too much, so I got a fixed supply that outputs a lot of current. If my issue can be solved with just increasing the supply voltage (and adding a soft-starter circuit at the input), I could just stick with my old design...LM2596 as inverting buck-boost, page 18Guide for inverting voltageLT8364

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 02, 2021, 05:48:56 pm
content: SEPIC is just like any buck-boost: the output voltage equals the peak switch voltageabove the supply voltage. So, switch rating must be greater than the sum magnitude of input and output.Same for the bootstrapped version above, with the difference that the output must handle sum magnitude currents as well. So it's not great in terms of capacity, but it's nice in a pinch. I suppose that means lower efficiency is a downside.Did you mean to have a separate "+5V" regulator referenced to the negative rail? The INA might not be too happy with so much voltage across it, though it looks like it should still work fine.Can remove almost all the 6.8u's on the 5V supply, nothing here needs them, aside from perhaps the regulator itself. The 0.1's can be shared between adjacent opamps too, probably only two or three needed in total.Q1 isn't going to be happy with potentially -15V on its base; a clamp diode from GND to B will do (1N914 etc). Some positive feedback (hysteresis) around U4, U7 would also help to limit oscillation.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 02, 2021, 06:36:18 pm
content: To be fair, I initially designed this circuit to provide +5V and -10V for an external circuit. It worked well (at least, not counting the inrush current issue, but that was solved by limiting I_ADJ for each of the ICs). For the op-amps, I was worried about powering them using +5V from the positive voltage output (moreso since the positive output can be adjusted to something higher), so that drop-in regulator was used to solve that issue. INA293 can handle a 22-volt supply voltage, so with 15 volts across it (5-|-10|), it worked.I'm not sure about the sum magnitude currents, but if I recall correctly, my circuit assumes 3-amps max on each regulator. They were designed around that limitation, and most of my applications don't even draw that much current at all. Ripple-wise, I measured it with an oscilloscope, and it doesn't appear to be a lot of ripple from what I can measure. If I may, what are you using to determine efficiency here?I keep on thinking that I'll have to reinvent the wheel, but if the only limitation is the input voltage (which I'll have to test with my bench supply), then that's all I need to do. With a 13.8 voltage supply, I could output +10 and -10 volts. If I go with an 18-volt input, the circuit should be able to output that (well....maybe not with those feedback resistor values). I've been running a simulation of it, and it seems to work. Technically, it should work with 20-volts, and if that works in hardware, then all I'll need to do is to just add a soft-starter to limit the in-rush current.Then, yeah, the INA293 is really close to its limits there. Regarding the negative current measurement, how would you redesign it? Plus, you mentioned the use of an op-amp for the sense voltage for the regulator. Are you referring to a current sense op-amp like what I have? I just have it so it turns on an LED when it reaches a certain current limit, but are you talking about feeding that alert voltage back to the feedback node?Thanks for the notes about hysteresis. That would definitely help in this case. I'll have to research that more.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 02, 2021, 06:59:41 pm
content: Would look something like this,Example values are given, say V(FB) = 2.5 so that's what the opamp output will also be. Resistor ratio being as shown, output will then be -12.5V. Adjust accordingly for desired output. Which can be varying resistors, or adding a voltage offset to the FB pin -- this is a good way to connect an MCU + DAC to a power supply for example.A boost/flyback controller is used, while the topology is Cuk. SEPIC is simply flipping around the diode, and swapping GND/VOUT around the transformer/diode. (So you see, simultaneous SEPIC and Cuk is possible, and it's a convenient way to get bipolar (tracking) supplies from a single regulator, if that were what you were after.)About efficiency, I just mean that, since part of the bootstrap's power is spent pushing itself below GND, and that power does not go to the output, it should be less efficient overall compared to a direct method. Basically, it's why you need a 5A instead of another 3A regulator.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 02, 2021, 09:45:29 pm
content: Quote from: T3sl4co1l on September 02, 2021, 06:59:41 pmRight, I do recall that from the document now that I think about it. That's why I used the LM2679 for the negative regulator instead of another LM2673.So, an op-amp here to control the feedback. Let me see if I get this. The feedback voltage at the IC has to be 2.5 volts. The resistor ratio is fixed to have an output of -12.5 volts. IC1 acts as acomparator, with the reference being the divider output voltage (fixed so that you should get 0 volts ideally given the feedback voltage). Roughly speaking, any change at VOUT will cause a voltage change at IN- of IC1. As a comparator, if -IN gets more negative, IC1 would output V+; if it gets more positive, IC1 would output ground. So, if the output voltage becomes more negative, IC1 will induce a voltage to FB, causing the regulator to adjust VOUT to the set value, causing IC1 to output 0 volts again. I think I saw something like that before in a video where someone did this in the form of a current limiter, measuring the voltage across a sense resistor and feeding it back into the feedback pin. Shouldn't you put a diode between the output of IC1 and the node of R2?Going back to what you said about some other stuff, with the limit of the INA293, assuming -15 volts, having a 20-volt supply is coming awfully close to its rated values. Nevertheless, would it still work, or would you say that's too close for comfort? Also, you mentioned the following about Q1. Isn't it still within the operating range of the transistor? Worst case, I can find a different transistor. That component is rather trivial in the grand scheme of things.Quote

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 02, 2021, 11:41:35 pm
content: Quote from: LoveLaika on September 02, 2021, 09:45:29 pmNooo not a comparator. Plain old inverting amplifier. Vout = -Vin*(R3/R1) or something like that.Simply:The regulator expects a sense voltage at FB, proportional to the actual output (namely, growing with increasing throttle).But the output is inverted.So the op-amp inverts it back.It's important that the op-amp GBW be higher than that of the error amp, so it doesn't introduce undesired phase shift. But that's easily achieved, pretty much anything will do. Obviously it needs to be either "single supply" or RRIO type, which is also easily done (LM321 for example).QuoteYeah, it still works.Actually the transistor is fine too, but you wouldn't know that from a datasheet:Its base voltage ranges from a bit under +5V, to -15V.The LED is rated for 5V reverse (I don't have to look that up, I've never seen a datasheet show anything else). The transistor is rated for Vebo = 6V. Thus the minimum safe base voltage is -11V, but the op-amp can apply -15V.In actuality, the LED will probably break down anywhere from 20V (high efficiency green, cyan, blue, white; breakdown is fatal) to over 40V (GaP green, alloy yellow) to over 100V (GaAlAsP red). This is an "undocumented feature", it will work but it isn't actually guaranteed to.Adding a clamp diode, sinks excess current from the op-amp, keeping the transistor and LED "technically" safe; cheap insurance.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 03, 2021, 01:58:09 am
content: I see now. Sorry about that. It didn't occur to me how R3 closed the loop. Boy, I would have failed my test for sure. With the way the LM2679 is oriented to give out a negative voltage, is it necessary to invert the voltage to a positive value? In the app note, the output voltage is connected to ground, while the regulator's ground is thought to be the negative output. So, with the regulator output being more 'negative' than ground, using that as the point of reference, you would still get a 'positive' voltage seen at that feedback node.I think I understand now. Boy, it's a good thing I used through-hole parts for the current sense circuit. It's a little McGyver-ish, but I can solder a simple 1N4004/1N4148/etc. to the resistor leads (from base to ground node) to achieve the clamp diode function. So, in that sense, when the LM321 outputs -15 volts, the diode will will be forward active, and it will pull the base to 0 volts (or some value close, but a lot closer than -15 volts). When it outputs 5 volts, the diode is reverse bias and doesn't affect the circuit.Thank you for mentioning this. I never thought about adding something like this. Given the potential outcome, I'm rather surprised I didn't break my LED already. Though, knowing that I'm within my tolerable range for my IC (and the need to add the catch diode), I feel a bit more confident testing this at my desired voltage range. If that's the case, and it works, all I need to do then is redesign my board.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 03, 2021, 04:42:36 am
content: Sorry, I've been thinking of the op amp a bit more, and I want to challenge myself a bit to learn better how to use it. I see how it works, but now come the resistor choices.R1 and R3 form the gain. Using an LM2596 as an example, its feedback network form a resistive divider, such that the output voltage is divided down to a certain voltage that is fed back to the feedback pin. Here, R1 and R3 need to do the same while inverting it. If so, either R1 or R3 can be a variable resistor. Then, is the purpose of R2 then to limit the current into the feedback pin? That would be based on the max voltage the op amp would provide given the max output voltage?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 03, 2021, 06:55:50 am
content: Right.Your configuration works because the regulator acts to push itself more negative (namely, the sense input is between FB and GND). So the feedback is always the right way around, and it knows to stop when it gets there.R2, is relevant when the FB pin is a summing node, including compensation elements. Like if the module has external voltage-mode compensation, like a UC3843 or TL494. For internal-compensation or gm-amp types like this, it doesn't matter.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 03, 2021, 04:32:57 pm
content: I see now. The inclusion of R2 would allow other elements as well to control the output voltage of the regulator. With the ICs that you mentioned, looking at their block diagram, I see how they have a pin (COMP for the UC3843 and and maybe FEEDBACK for the TL494) that would allow you to implement your own error-feedback compensation mechanism. I'll admit, I'm not familiar with PWM controllers, but this is fascinating to learn about.Thank you very much for your help and your patience. I have a lot to consider now regarding the next design. I hope I can salvage my current design. Funny thing about the current limit on these ICs. The current resistor limiting values were too much, so I had to make adjustments to them by using larger resistors. This reduced the overall switch current, but my current usage doesn't really draw as much current as to reach the minimum values of 1 amp and 3 amps for each of the ICs.I hope you don't mind me asking, but what do you think of this circuit for an external soft-starter? It seemed like with a simple module like that, incorporating it as an external module before powering my supply would solve a lot of issues. I was planning to make my soft-starter like this.Link to Soft Starter Circuit

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 03, 2021, 05:06:46 pm
content: Indeed the TL494 has two error amps, outputs wired-OR together, inputs free: you could implement the inverter circuit right there without the external op-amp, just by swapping the inputs +/- relative to a normal (positive output) circuit.BTW, have you tried just longer soft-start timing? That affects the current draw due to charging the output caps. Does nothing for the input side, but you usually don't need as much capacitance on that side.The linked circuit, is a very rough and basic implementation of the best-case function; what it's missing, is:- What if the source can't supply as much current as is limited by the resistor? (The resistor should be set based on source capability, and load demand to some extent.)- What if the load is a low resistance, so its voltage never rises to full output? (The resistor is a voltage divider against DC load current.)- The relay is hard timed, so pulls in regardless of whether the load reaches full charge or not.- The precharge resistor is hard wired, so is present regardless of whether the load is a short circuit.Consider mitigation to each of these steps: maybe you use a series of resistors, switched in as needed; or a linear or switching circuit to drop a current (might be programmable). Of course the linear circuit can only do that for so long (some ms maybe) if it's compact, or it has to be big enough to handle ~continuous power, so probably doesn't scale well; but a switching circuit somewhat passes the buck, as it can't work without some bypass, either.We might make the relay conditional, so it only pulls in once the output is within some margin of the supply, say 5 or 10% below. But if this never happens (because of a shorted load), then the precharge just sits there cooking forever. So the precharge itself should be conditional, probably timed to turn off if the above never happens.So, if you're only ever using it with the average case -- the load is never shorted, and the resistor is suitable for the particular supply and load -- yeah, it works well. Outside of that case, it's not very general, so be careful.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 04, 2021, 10:29:31 pm
content: According to the datasheet for the LM2673/2679, depending on the soft-starter capacitor value, the output may overshoot during turn-on. Also, with no-load or low load, it may not even be effective. So, they recommend avoiding using soft-start caps between 33 nF and 1 uF. I believe I tried larger values (10 nF), but it didn't solve the problem.Well, here's how I'm overcoming my current power issues for now. Initially, with my board the way it is, I was drawing 4-amps with my use load at startup. In order to get it to work, I used my two outputs of my bench power supply. Each output had a current output of 3 amps; setting it to a 'tracking' mode and combining them together, I can output a single voltage while outputting 6 amps. Thus, that's how I powered my board at the bench. Now, since I don't have access to it on the field, I had to find an alternate method. Thus, for my own use, I used a single-output fixed voltage power supply capable of 4.5 amps at 13.8 volts. It works for my current needs (putting aside the start-up current, my load draws 390 mA).So, in order to make this capable with my bench top supply, the current at start-up will have to be limited to 3 amps or less. In order for the positive output to be 15-volts, the input voltage would have to be 15 volts at least. Looking at it in the worst case scenario, max voltage would be 20 volts (though we would definitely set it lower than that).To answer the questions, from this, I believe the source can supply enough current for the power supply. Perhaps this is a matter of figuring out why it's drawing so much in the first place. I did try to remove one of the input and output caps, as well as lower the capacitor values, but it still was drawing a lot of current at start-up. Wouldn't the load reach the full, desired output voltage after the relay has switched over? The RC circuit closes the relay after a certain amount of time to form a short circuit through the relay. Even if the load is present, it shouldn't have an effect on the circuit right? If that's the case, we could just use a different relay system to have the circuit not see that precharge resistor at all after the relay switches.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: September 06, 2021, 08:48:53 pm
content: Sorry about asking so much, but going back to what you said, about picking a SEPIC converter, I'm rather confused by the example layouts, and I wanted to get some insight.Take this, theLT3436. Typical application has it as a boost converter, but you can reconfigure it as a SEPIC converter as shown on page 12 and 13. Now, the first figure shows a "3V to 20VIN 5VOUT SEPIC with Either Two Inductors or a Transformer" while the second figure shows a "4V-9VIN to 5VOUT SEPIC Converter".It seems pretty basic, but what I'm confused about is the use of inductors/transformer. Figure 1 uses the same 2 inductors while Figure 2 uses a coupled inductor to act as a 1:1 transformer. SEPIC converters seem to use two inductors in practice, and I've seen proposed layouts on other products showing the inductors close together. I guess they would have to be using a coupled inductor, but do you have to keep the two inductors close together usually in a SEPIC configuration?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 06, 2021, 10:29:36 pm
content: It works fine with separate inductors, but performance is increased (lower ripple, core loss) when coupled.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 06, 2021, 03:35:59 am
content: From what I see, isn't the cuk topology good for small currents? I'm trying to get 3 amps on each channel max

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 06, 2021, 06:57:57 am
content: In what way would it vary with current, that's different from any other switched-inductor topology?Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 10, 2021, 03:58:41 pm
content: Right, that's a fair point. I didn't think of that. I was looking at IC topologies, specifically Cuk, and all I got were ones that had low current ratings. I was looking at them the wrong way.If I may ask your opinion, which IC do you think would be more advantageous in terms of features: theLT1370or theLT3579? Both seem to work in terms of my needs, and I like how simple the LT1370 is. It's more expensive, but it has its own dedicated pin for negative feedback. On the other hand, the LT3579 looks to be more...versatile, in terms of protection against in-rush current (as shown in HOT PLUG near the end of the datasheet).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 10, 2021, 05:22:50 pm
content: It's not a common topology, so you'll get a lot of oddball, special case things. Or LT, they have everything... if you're willing to pay for it.3579: I don't mind the hot plug feature, it's one of those quirky additions that LT has a lot of (again, oddball special case things, are one of their specialties), and I can't say I've needed that particular feature, but if you do, sure, it's handy having it all in one chip (plus support components).And the 1370, fine I guess. It loses points for me because of a peculiar set of errors:1. "Switch Current Limit vs Duty Cycle", why is this truncated? If it's because of slope compensation just added in, evidently they added a clamped sawtooth wave (see the ramp waveform and "Σ" (summer) in the 3579 block diagram -- this representation is a common sight in block diagrams of peak current mode controllers), so that it doesn't compensate anything at low duty.2. The block diagram is very basic, and seems to have current sense go directly into the comparator, no slope compensation at all. Is this in fact only good for DCM (discontinuous conduction mode)?3. The paragraph on page 8,Quoteseems to confirm the truncated-ramp hypothesis. To be clear: it's an oft-repeated claim that subharmonic oscillation occurs over 50% duty, but this is only true when the voltage ratio is the same. That is, a boost to 2*Vin, or SEPIC to Vout = Vin, etc.What'sreallymeant is, the necessary consequence of operation at that point: for the voltage ratio to be exactly 1, AND duty to be over 50%, the inductor current must be continuous (CCM).And this is the fact that actually matters, current being left in the inductor from cycle to cycle. That is, when the switch turns on the next time, the current will have decayed an unknown amount, and so the on-time (until the comparator ends that cycle) is also unknown. It turns out, at first the duty cycle alternates every cycle (half subharmonic), then at a little higher load current, a cycle of 4, and so on; the number of states diverges rapidly (period doubling) until full chaos occurs. In fact, if you plot the states (plotting a point for every pulse width seen, at a given load current), you get what's called a bifurcation diagram (as the number of states doubles every time it splits up like that). The figure is identical to that of the logistic map -- it's actually an analog implementation of it. (Cool, huh?)Under chaotic operation, inductor current is always bounded, as is duty cycle, so it's not unsafe to operate, as such; but since exact operation is impossible to predict, there is higher output ripple, and usually audible hiss from the inductor. Switching loss may also be increased (which if it causes overheating, could lead to failure).And, ironically -- they suggest higher inductance, but this acts to reduce the ripple fraction, pushing the system closer to CCM, and subharmonic oscillation or chaos. And, I assume the comment about "reduced slope compensation" at high frequencies, is just due to the truncated ramp being also delayed by internal circuitry, a delay which is relatively significant by that frequency. That's fine.So, that explains that situation.I don't know what exactly they did in the design of this chip, but it's just... weird. Not weird enough to be like "eww no", but weird enough I would keep looking for something better.They also discuss tantalums and ceramics, which haven't been an issue for YEARS; ceramic and polymer won, and most chips support them by design.Just how old is this datasheet, anyway? 1370 must be a pretty early part, if they're just numbering sequentially... ah there it is, very corner of the last page, 1998 it seems.Which I guess agrees with the price: the D2PAK version is $18 in singles at Digi-Key, clearly they don't want you designing it in. If you want to keep buying it to support a legacy product, sure, it's up to you how much you want to spend doing that...What would I use?Probably something like a TI Simple Switcher of comparable ratings, or a controller of whatever sort (probably something nicer than ye olde UC3843, I'd have to go shopping to see what), and just tack on the op-amp to invert the feedback (which is exactly what LT1370 does, they simply set V(FB) = -0.5 V(NFB) with an op-amp). Indeed, I've done exactly that before.Or even more random, probably the flashlight board I made, which indeed I've used for aCuk converterbefore. But that's not exactly helpful here.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 10, 2021, 10:42:55 pm
content: Thanks for your input. If only TI provided good simulation models for their products. Partly, that's why I was drawn to the LT ICs, as LTspice already has most of the models built in that actually work.Since you mentioned the SIMPLE SWITCHER family from TI, I think it gave me some ideas. Initially, I was looking at a converter with an integrated switch, like the LM2588 even though it's kind of old. Other converters I found were those rectangular integrated modules that had strange footprints which I wanted to avoid. But then, looking at controllers (without the integrated switch), those have the advantage of allowing higher currents via the external switch. Thinking about it like that really helped me with search results.