Over driving a boost converter input - Page 1
Forum section: electronics / projects

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 06, 2019, 03:44:02 pm
content: Hi,I have designed a 91% efficiency Boost regulator based on UC3843, it would convert from 12V to 60V.The circuitThe simulated outputNow for some reason I want to over drive the boost input over the 60V, say to 340V, I wanted the design to be able to survive and workso the output should be around 340V in this caseThe problem is that when I simulate it, it would generate some 650V overshoots, I need a way to clamp this overshoot.I have changed the input voltage to 340V in hereWe have a 650V overshoot, How should I clamp it?Also I have built the above circuit and it works as expected with 12V input, I have not changed the Diode and MOSFET yet to be able to handle high voltages! so what part numbers for MOSFET and Diode do you suggest? I need them to be on the low cost side of things.I have not tried to over drive the input yet, for now I wanted to make my calculations solid and then over drive it in real world.

author: OwO
author_rank: Super Contributor
timestamp: February 06, 2019, 04:00:14 pm
content: Inductor too small/on-time too long for 300V. dI/dt is proportional to voltage. Try probing the current on the mosfet.

author: Yansi
author_rank: Super Contributor
timestamp: February 06, 2019, 04:02:02 pm
content: Mosfet too wrong.IRF540 for 340V input? lol. No way.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 06, 2019, 04:06:45 pm
content: QuoteLoool read the post completely! I said that I have plans for changing the MOSFET and diode!QuoteI do not want to blow up things, So should I increase the inductor? to what value?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 06, 2019, 04:14:07 pm
content: Actually now that you mention it, I'm very curious about what model TI used for the IRF540. IIRC most I've seen, contain diodes that include BV (breakdown voltage). That waveform shows it's clearly not using such a model!Addressing the transient is easy:a. Increase the source resistance to sqrt(30uH / 12uF) or greater, orb. Set C1 initial voltage to 340V or enable DC operating point setting, orc. Connect a TVS diode from GND to output, rated for say 400V. (Use a 550V+ rated MOSFET.)Tim

author: Yansi
author_rank: Super Contributor
timestamp: February 06, 2019, 04:16:53 pm
content: I have read all, unfortunately not the last sentence, sorryYou would need to replace not only the mosfet and diode, but the inductor and capacitors and resistors anyway.What you are trying to achieve does make only little sense, so you better start explaining what your intentions are with the 340V input capability, as this seems to me to be a classic X-Y problem to solve.

author: NiHaoMike
author_rank: Super Contributor
timestamp: February 07, 2019, 12:25:49 am
content: I'm going to guess he has a device with a common mains powered switching power supply, that will work down to 60V, but he wants it to work down to 12V.

author: Yansi
author_rank: Super Contributor
timestamp: February 07, 2019, 06:00:52 am
content: Yes, exactly same thoughts here.However he better makes a workaround to isolate the boost converter completely, when powered from 300+V.Using a 300V operation capable mosfet means pisspoor performance at 12V (very high Rdson) and the diode selection will become very problematic, as in a boost converter, you operate the diode in hard-commutation (The whole Qrr counts). SiC (Silicon Carbide) diode would be probably the only way to go here, due to the currents involved at 12V operation.All in all this will become rather nonsensical expensive mess with poor performance. Adding a bypass circuit, that would switch in and out this converter completely might become cheaper and simpler. But we do not know all the details.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 07, 2019, 06:47:31 am
content: QuoteI have used the built-in model in Tina TI for that.QuoteI plan to use a varistor (something like 7D471) on the Drain of the MOSFET, is it fast enough? I choose it because of it's price.QuoteExactlyI wanted to make my circuit work with 12V too, and also when apply 220VAC, it would work normaly too, there should be no setting from user input to mess around things,For the MOSFET I was thinking of 20N60, and for the diode I think some fast recovery ones like S5JBR5G would do the job? do you have any Idea?

author: Yansi
author_rank: Super Contributor
timestamp: February 07, 2019, 06:49:03 am
content: Using a varistor across the mosfet is just wrong.1) Capacity adds dissipation in the mosfet2) The mosfet itself is probably avalanche rated by itself.

author: TurboTom
author_rank: Super Contributor
timestamp: February 07, 2019, 10:02:19 pm
content: The UC384x variety is a current mode controller. In order for the control loop to work properly, it requires a valid signal on the current sense input (either sensed by shunt/current sensor or simulated by some R-C circuitry with an active switch). Just grounding the CS pin will result in something like 95...98% duty cycle all the time until the voltage setpoint is reached. No current limiting in the switch / inductor... Guess you know what that means

author: Yansi
author_rank: Super Contributor
timestamp: February 07, 2019, 11:46:38 pm
content: What are you complaining about TurboTom? The CS pin is correctly wired to the shunt.

author: aheid
author_rank: Regular Contributor
timestamp: February 08, 2019, 09:09:35 am
content: Silly question... You say you want the output to be 340V if input is 340V (220V AC). So couldn't you just bypass this whole circuit in that case?

author: TurboTom
author_rank: Super Contributor
timestamp: February 08, 2019, 05:07:47 pm
content: Sorry, now I noticed - I interpreted the right-angled line at the tab to the CS pin to symbol ground. Different conventions in drawing schematics... Please disregard my post above.Cheers,Thomas

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 10, 2019, 06:26:32 am
content: QuoteThere is no dumb questionhow do you boost 12V to 60v and also bypass 340VDC in a circuit automatically then?

author: Yansi
author_rank: Super Contributor
timestamp: February 11, 2019, 01:09:08 am
content: Use different inputs for each?

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 11, 2019, 08:04:38 am
content: QuoteIt's not viable for this product!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2019, 08:19:06 am
content: This product isn't viable.Tim

author: Zero999
author_rank: Super Contributor
timestamp: February 11, 2019, 12:50:03 pm
content: Quote from: ali_asadzadeh on February 10, 2019, 06:26:32 amHow about using an external power supply and connect it directly to 12V if necessary?What does the project do?How much power does it use?Are you sure the SMPS actually requires 340VDC? Quite often a much lower voltage can be used. I've run Rigol oscilloscope directly from 48V before and it works down to 32V, even though it's designed for 100V to 250V operation or is 60V the minimum voltage?If you could design a boost converter which will convert 12V to 60V, but will withstand 340V and do nothing when the output voltage exceeds the set output, then it's workable, but still not good design. It's better to use separate connectors or an external power supply.There should never be 12VDC on a mains connector or worse 230VAC on a 12V connector. One might think 12VDC into a mains appliance will do no harm and with many devices it won't, but it can destroy inductive loads such as motors and transformers, especially large ones if the 12V supply can supply a high current. Motors and transformers have a low resistance to DC, so will pass far more current at 12VDC, than 230VAC 50Hz, which can cause overheating and smoke.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 12, 2019, 05:58:42 am
content: QuoteI think this way is my best option.Please do not tell me to use separate connectors and why I wanted to do this in this way! it's the customer requirement, not my decision.

author: Zero999
author_rank: Super Contributor
timestamp: February 12, 2019, 09:38:48 am
content: Quote from: ali_asadzadeh on February 12, 2019, 05:58:42 amYou can run the UC3843 or a linear regulator suitable for 400V input and drop-out voltage low enough to ensure it will still work, when the input is 12V. The UC3843 feedback needs to be diode clamped to the Vcc pin to protect it and of course the MOSFET, diode and capacitors should be rated to at least 400V. Now if the UC3843 is set to 60V, it will maintain that voltage and be effectively bypassed if the input voltage rises beyond that.In future, you need to learn to say no to the customer when they want something too difficult or potentially dangerous.

author: Yansi
author_rank: Super Contributor
timestamp: February 12, 2019, 10:06:01 am
content: It is hard to give a specific suggestion, if the OP haven't even bothered to specify the output power required.Due to likely quite some high currents involved in the 12V operation, you will end up with excessively expensive input and output capacitor and mosfet and quite complicated inefficient power supply structure for the control circuitry.

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: February 12, 2019, 11:03:19 am
content: QuoteThe output power is in the order of 25-30WQuoteI have no Idea about running the UC3843 from 400V, would you explain more?

author: Zero999
author_rank: Super Contributor
timestamp: February 12, 2019, 07:31:30 pm
content: Quote from: ali_asadzadeh on February 12, 2019, 11:03:19 amI have no Idea about running the UC3843 from 400V, would you explain more?The idea is not to directly run the UC3843 off 400V but to use a linear regulator to get something like 9V to power the UC3843. I've done a quick Google and most HV linear regulators rated to 400V or more need the input voltage to be 12V above the output. One way round this could be to add a cascode to a standard low voltage regulatorSomething based on the circuit below could give 9V out, with 12V in, at the low current required to power the UC3843. Note that this circuit is not suitable as is for your application. It will need several modifications to work. R3 will need to be a constant current source and the LM317 replaced with a regulator with a lower minimum current requirement and drop-out voltage.