# Can HFC0300 startup at 17V

Since the VCC startup voltage is 12V, does HFC0300 flyback IC start at 17V input (HV=17V)?

Excellent question, this doesn’t look like it is covered in the datasheet. The HV current source is specified at 400VDC. A purist reading that might conclude that at 399VDC the current source might not work. Certainly from the datasheet the part is useless for a 120VAC circuit.

This feels like you got to try it and see. Can the current source provide reasonable current with only 5V across it?

In fact per the datasheet the minimum current of the startup HV fet is allowed to be zero so really even at 400V the part might not work and still technically meet the datasheet.

Thank you for your reply, but what do you mean with 400VDC? in the datasheet, it says the universal main supply voltage could in the range of 85VAC to 265VAC. At 85V AC input, the DC voltage is around 100V to 120V. therefore, it does work below 400V DC. I was wondering if it can still work under 100V DC, down to 17 V. I would not want to design a circuit and try it, if I does not work, it wastes time.

Well I am being a bit silly, but. The HV startup device that provides power for this thing to get going is extremely poorly specified on the data sheet. It is "typically " good for 2mA off a 400V rail.

However no minimum is specified so you could theoretically get a part with a minimum of zero, i.e useless won’t start.

In a 85VAC application the device has to deliver current from a 100V rail will it do that?? Can’t say from the specs in the datasheet. We don’t really even know if it will work at 399V.

A more reasonable spec would show what minimum current you can expect at say 85V.

You are probably going to have to get some parts and do your own investigation of startup in the very low DC area you are trying to operate in. I would imagine you can just short VCC to ground with an ammeter and crank up the voltage on the HV side and see what current you get vs what DC you have. I suspect the part will start successfully at 17V all kidding aside

Hi haiguo.li
It is quite likely that the device works down to 17V HOWEVER you must first understand what it is you are asking about…
There is more than a simply IC in the circuit and unless all the components are considered, then it doesn’t matter what the chip does, it won’t work. Ultra wide input designs are quite specialized, and even if you could confirm the chip, it will take several attempts to finalize the circuit.
Also you will note… that Vcc (from secondary winding) is expected to be in the 8.2V to 11.7V range, so the chip is fully functional at that voltage. (see figure 8). It is possible that the Vcc can rise above 12V, but no more than 24V or it will go into OVP (that datasheet doesn’t have figures of this). Also since there isn’t a boost capacitor… this will also be the DRV voltage (the absolute max is the point of destruction not operation) since this pin is an output, not an input.
Mid point 8.2 - 11.7 is approx 10V which would seem about right for a MOSFET, going above 20V would damage most MSOFETS so that’s the limit.

The company I work for previously had a very specific PSU designed for them with ran from 20V to 110V AC or DC That’s a 5.5 to 1 ratio, and well beyond the usual 2:1 ratio these mains converters are use to. In our case the frequency is constantly adjusted so that the minimum/maximum on time isn’t violated. You also have power concerns… our PSU was 5W… so 250mA @ 20V, 45mA @ 110V, may not seem much, but it changes the requirements for the FETs involved. This whole PSU was designed by a specialist PSU designer, Worked on 3rd attempt which was pretty good.

My question is quite simple, actually. In the figure on page 2 of the datasheet (https://www.monolithicpower.com/en/documentview/productdocument/index/version/2/document_type/Datasheet/lang/en/sku/HFC0300/document_id/59/), if the input voltage is 17V DC, and I design the transformer, feedback, etc., accordingly, will this circuit work?

Cant say from the datasheet, you have to try it, or maybe get some aps support rather than internet randos. Have you tried via my MPS?

Okay. Answering this question requires detailed internal design. Not sure why this is not included in the datasheet. Thank you!

So I then ask the Question… Why??
Features:
• Variable Off-Time, Current Mode Control
• Universal Main Supply Operation (85VAC to 265VAC)
They are quite specific about the design criteria.

So if you only want to run off 18V… Why not the MP6005? This is specifically designed for 8 - 80V DC operation unlike the HFC0300 which is designed to work in the 85 - 265V AC range.

And as noted… the chip starts at 8.2V … so well below the 17V you ask for. Just size R2 to ensure there is at least 2mA flowing into the HV pin. Alternatively, you could put a regulator & diode in to regulate the 18V down to 12V and bypass the HV current source. there are lots of alternatives here. But I would really suggest you use something designed for the job if aren’t experienced in fly-back design.

Also note: All MPS testing is done at 110 - 300V … so I wouldn’t expect 86% efficiencies, and most of their test data is invalid if you operate at 18V. You will simply have to build it and test it.

Thank you. I know there are many controllers. However, I need to run it at a super wide range from 17V to several hundreds of volts.

In which case I would suggest this design is too simplistic. To achieve > 4:1 input ratio, you will need active frequency control. You will also need to understand how the chip starts-up, and you will have to simply try the chip as no datasheet will tell you what you want to know. 5.5:1 was pushing the technology at the time, and is a very specialized design. The guy who designed it was a specialist PSU designer and knew what is capable of being done. We were unable to get to 12V due to the transformer/MOSFET limitations.
20V - 200V would be 10:1 which is incredibly difficult. I would suspect you wont get this to work as the frequency range will be beyond the capability of the device (regardless of whether the device will start at 18V).