12V to 85V Buck Boost Regulator Needed

[Frosty_47]

Hello, I need to boost 12V from parallel array of batteries to 85V. The desired output current must be around 1.5A - 2.0A. I want to use a buck-boost topology for space efficiency. My search for national parts was unsuccessful, as I could not find anything that suits my needs. What are my options for this project?

Thanks

 

[BuffaloEngineer]

You could always build your own - it wouldn't be easy though, that's a huge output current requirement for such a large voltage boost. What kind of batteries are you using?

This might work:
https://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1003,C1042,P38055,D25148

 

[Frosty_47]

You could always build your own - it wouldn't be easy though, that's a huge output current requirement for such a large voltage boost. What kind of batteries are you using?

This might work:
https://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1003,C1042,P38055,D25148

They are all SLA batteries (9 of them to be exact). Wired in parallel, to give a total of 45Ampere/hour rating. These batteries can supply a lot of current if that’s what you worry about.

 

[Claude Abraham]

Don't use a buck-boost, but rather just a straight boost. Since the output is never less than the input, a boost is the right topology. A buck-boost has 2 transistors & 2 diodes, vs. 1 of each for the boost.

Keep in mind that for 2.0 amp output at 85V, the input current at 12V must be a little more than 14.2 amp average.

I would use a UCC28C43 peak current mode controller. It operates with a supply of 8.4 to 18 volts, perfect for a 12V application. It has a totem pole FET driver, low side, perfect for a boost topology. The FET & rectifier must withstand 85V, so I'd use 100V parts for each. I'd operate the inductor in the continuous conduction mode. All in all it looks pretty straightforward.

 

[Frosty_47]

Don't use a buck-boost, but rather just a straight boost. Since the output is never less than the input, a boost is the right topology. A buck-boost has 2 transistors & 2 diodes, vs. 1 of each for the boost.

Keep in mind that for 2.0 amp output at 85V, the input current at 12V must be a little more than 14.2 amp average.

I would use a UCC28C43 peak current mode controller. It operates with a supply of 8.4 to 18 volts, perfect for a 12V application. It has a totem pole FET driver, low side, perfect for a boost topology. The FET & rectifier must withstand 85V, so I'd use 100V parts for each. I'd operate the inductor in the continuous conduction mode. All in all it looks pretty straightforward.

Thank you, I will look into that. Sometimes I wish for the school curriculum to at least introduce me to switching regulators, but instead our provincial government forces everyone to take useless subjects such as psychology, economics, workplace safety (biggest joke ever), and other useless crap that have no relevance to electronics. So my only option for side project is to do my own research and pay a little visit to this place.

Thank you for your help.

 

[colin55]

If you have 9 twelve volt batteries, why don’t you connect 7 of them in series and get the voltage you need. You can then charge them individually and this will cost you NOTHING.

 

[Ubergeek63]

at low battery you are talking about 17A average in so you are looking at something on the order of 150A peak from the battery which makes for a 200A inductor and 400A FET.

Dan

 

[Hero999]

You'd be far better off connecting all the batteries in series and using a buck step down converter.

If you must use 12V in then use an autotransformer as a flyback rather than a simple inductor.

 

[Claude Abraham]

at low battery you are talking about 17A average in so you are looking at something on the order of 150A peak from the battery which makes for a 200A inductor and 400A FET.

Dan

But one can *section* the design so that each section carries a portion of the total current. Current mode control readily lends itself to this practice. Using 8 to 10 sections makes the design quite feasible.

At low battery, however, it depends on whether or not a full 2.0 amps is needed. If a reduction in max output current can be tolerated, maybe it can't be as I don't know the function, then it gets even easier. At 12.0V input, with 85V/2.0A output, the power is 170W, which when divided by an efficiency of 0.90 gives 188.9W. At 12V in, this gives 15.74A average current at the input, but the average inductor current is 123.9 amp, quite a lot. With 8 sections, it is 15.5 amp per section.

Not easy, but feasible.

 

[Ubergeek63]

But one can *section* the design so that each section carries a portion of the total current. Current mode control readily lends itself to this practice. Using 8 to 10 sections makes the design quite feasible.

At low battery, however, it depends on whether or not a full 2.0 amps is needed. If a reduction in max output current can be tolerated, maybe it can't be as I don't know the function, then it gets even easier. At 12.0V input, with 85V/2.0A output, the power is 170W, which when divided by an efficiency of 0.90 gives 188.9W. At 12V in, this gives 15.74A average current at the input, but the average inductor current is 123.9 amp, quite a lot. With 8 sections, it is 15.5 amp per section.

Not easy, but feasible.

sure you can. you can parallel components or use multiphase converters.

I can not count the times I have seen hobbyists ask for directions to do things that are not practical for any other than power or RF engineers and that wonder why they can not get a complete schematic, step by step instructions, and hand holding when their dreams collapse into rubble at their feet.

And then I get scolded for trying to explain life in the real world to them! So I usually just do what I did here: point out the component requirement and let them go find the price on the part, if they *find* it. Quite often what they are asking for requires custom parts or parts from a MFG that laughs at my ordering a single 1K reel, never mind a hobbyist looking for just one or two parts.

In this case it would be the custom. At any rate, the best bet would be a very high frequency multiphase switcher with custom magnetics: transformers of at least a 1:10 turns ratio capable of 15-20A peak on the primary running full bridge.

Dan

 

[colin55]

If someone has to ask about a problem like this it is way beyond their capability.
That's why I suggested connecting 7 batteries in series and using a simple $25.00 battery charger to charge each battery in turn. This is about the maximum capability of the poster.

Whenever you get currents greater than 3 amp, you need to be an engineer to deal with the heating effect.

 

[Ubergeek63]

If someone has to ask about a problem like this it is way beyond their capability.
That's why I suggested connecting 7 batteries in series and using a simple $25.00 battery charger to charge each battery in turn. This is about the maximum capability of the poster.

Whenever you get currents greater than 3 amp, you need to be an engineer to deal with the heating effect.

better watch out: I get yelled at pretty regularly for pointing that out

 

[Oblong]

So I guess this isn’t the place to ask for detailed design files on a 2kw inverter ,with 120% conversion efficiency that operates off a single 9Vdc alkaline?

 

[Frosty_47]

at low battery you are talking about 17A average in so you are looking at something on the order of 150A peak from the battery which makes for a 200A inductor and 400A FET.

Dan

How did you come up with such calculations ? Last night I used one of Ipk formulas from some book on switching regulators, and the peak current came up to be 45Amps - 49.8Amps.

 

[Frosty_47]

If you have 9 twelve volt batteries, why don’t you connect 7 of them in series and get the voltage you need. You can then charge them individually and this will cost you NOTHING.

I do not want to do that because if one battery drains, the whole chain fails! even if the rest of the batteries still have plenty of charge on them.

I experienced this issue before with one of the 6V battery being more drained than the other 6V battery in series.

 

[Frosty_47]

You'd be far better off connecting all the batteries in series and using a buck step down converter.

If you must use 12V in then use an autotransformer as a flyback rather than a simple inductor.

Originaly I didn't want to use the flyback topology due to the size and wheight of the transformer. However, it seems that a flyback boost circuit is more realistic for this aplication... Oh and no I don't want to connect batteries in series for reason explained above...

Thanks for your suggestion

 

[colin55]

We really can't help you.

 

[k7elp60]

Another solution might be to buy an off the shelf 12V/120V inverter and then use a rectifier and buck regulator to get the 85V. The current would be less and with a 200W continous type you may be able to get the 2A you need at the 85V level.

 

[Ubergeek63]

How did you come up with such calculations ? Last night I used one of Ipk formulas from some book on switching regulators, and the peak current came up to be 45Amps - 49.8Amps.

Well I thought I answered that briefly... but I was rushed and distracted as I was answering from work.

85V*2A=170W

170W/10V=17A@10V

If you have a high enough inductor value to have 0A inductor ripple current you end up with relatively high switching losses and a minimum peak current of 19A.

At critical conduction (not realistic in normal circuits) you ramp linearly up and linearly down requiring twice the peak current or 38A minimum.

Normal most switchers will be below that since most either in one mode or the other since the crossover tends to cause stability problems with the two modes requiring different compensations values.

Dan

 

[Claude Abraham]

Well I thought I answered that briefly... but I was rushed and distracted as I was answering from work.

85V*2A=170W

170W/10V=17A@10V

If you have a high enough inductor value to have 0A inductor ripple current you end up with relatively high switching losses and a minimum peak current of 19A.

At critical conduction (not realistic in normal circuits) you ramp linearly up and linearly down requiring twice the peak current or 38A minimum.

Normal most switchers will be below that since most either in one mode or the other since the crossover tends to cause stability problems with the two modes requiring different compensations values.

Dan

I made the same mistake. I think I divided by the factor (1-D) twice instead of once.

Anyway, at 85V/2.0A out, P = 170W, divide by 0,9 gives 189W. At 9V input the average input current is 21A which must equal the average inductor current as the inductor is in the input. Using continuous conduction mode places the peak current at more than 21A and less than 42A.

It's not as bad as I first erroneously thought.