Which one of these 4 circuits do I need?

I have a Max773 adjustable DC-DC controller. I want to input 6V and output 8V at 1Amp, I'll be using 2 lithium batteries that can source 3V and 1A each. None of the typical circuits in the datasheet will give me exactly what I want so I'm not sure which I should start to adapt.

Which one is best?

Thanks,

Antknee.

You can supply 6 volts at 1 Amp and want 8 volts at 1 amp? That's a conversion efficiency of 133% and it will not happen.
You need more/bigger batteries.

You need to do the calculations to work out the values. That's why the data sheet is so long. You need to use the feedback connection (FB) like the 16 V and 24 V examples use, as the voltage you want isn't one of the presets on the MAX773 (5, 12 or 24 V). You need to adjust the ratio of R1 to R2 to give the correct output voltage.

Wade Hassler said:

You can supply 6 volts at 1 Amp and want 8 volts at 1 amp? That's a conversion efficiency of 133% and it will not happen.
You need more/bigger batteries.

Click to expand...

Lithium Manganese batteries can supply up to 3A so its not a problem.

Diver300 said:

You need to do the calculations to work out the values. That's why the data sheet is so long. You need to use the feedback connection (FB) like the 16 V and 24 V examples use, as the voltage you want isn't one of the presets on the MAX773 (5, 12 or 24 V). You need to adjust the ratio of R1 to R2 to give the correct output voltage.

Click to expand...

I spent 2 hours going through the datasheet last night, it was a fair effort, I need diagram 3d to be adapted. I wasn't sure what bootstrapped means, I'm still not sure to be honest but I don't want the current limited so that is the way to go. I have made a parts list, the only value I can't work out is C4 the output bypass capacitor, how to calculate that value doesn't seem to be mentioned in the datasheet.

Hi antknee,

for a variable output voltage you must use bootstrap mode.

"Setting output voltage" is described on page 13 of the data sheet. There are also formulas to determine the values of R1 and R2 for the F/B input.

Boncuk

the bypass cap isn't a value that needs to be calculated, use what they have there..... it's a bypass cap, all that's required is that it's a low impedance at the switching frequency of the converter.. if you need the formula for that it's C=1/(2πfZ) (remember to convert C to microfarads)

I've updated my parts list with values for R1 and R2 to give an 8V output. I've also put the output bypass capacitor at 10uF, that should have an impedance of around 0.15ohms.

The only thing I'm not sure on now is the PCB layout. The datasheet says "proper PCB layout is essential, protect sensitive analog grounds with a star ground configuration"

I know what a star looks like...but what is a star ground configuration? And how big does it need to be?


• C1 – input bypass cap – 150uF
• C2 – output filter cap (low esr) – 300uF tantalum?
• C3 – reference bypass – 100nF
• C4 – output bypass – 10uF
• L1 – 20-22uH
• D1 – 1N5817
• Transistor - N Mosfet - Si9410DY
• R1 – feedback determining output 10Kohm
• R2 – feedback determining output - 43Kohm
• Tie Low Bat Indicator to gnd
• Leave Low Bat Output floating (no contact)
• R_sense ~ 100mohms to 200mohm (from graphs).

Hi there,


Bootstrapping is a way to supply the chip with a higher voltage than is available with a low voltage battery like 2 volts. It makes the MOSFET switch better and lower resistance when 'On'. If your input is 5 or 6v you probably wont need bootstrapping. If your input is say 3v then you probably do. It's all about how much input voltage you have to start out with and what the chip requires to run efficiently.

Anyway, i would choose the simplest circuit there. That's about it. For the diode D1 though i would choose something with a current rating a little higher than 1 amp. For the inductor current i would go with something that could handle the peak current, which you can determine experimentally with a circuit simulator for all input/output states you might encounter with this circuit. Estimating your inductor current to be about 8.6/6 amps, a 3 amp inductor should suffice, might get away with a 2 amp device, and the ESR should be low like 50m ohms or lower.

A star ground is simply a single physical connection point where all of the lead to be grounded gets connected to. Choose one solder connection as your star center and connect everything up to that single point rather than daisy chain some of the grounds.

MrAl said:

Bootstrapping is a way to supply the chip with a higher voltage than is available with a low voltage battery like 2 volts. It makes the MOSFET switch better and lower resistance when 'On'. If your input is 5 or 6v you probably wont need bootstrapping. If your input is say 3v then you probably do. It's all about how much input voltage you have to start out with and what the chip requires to run efficiently.

Click to expand...

I get it. I would prefer to use just one Lithium battery at 3V but I decided I probably couldn't get enough volts on the chip at the current I need. The bootstrapped graph did have lower volts than the non bootstrapped. I will have to build the circuit and see if I need one or two batteries. I also investigated using Lithium Thionyl Chloride batteries which give 3.6V (which would be fine), they don't seem to offer much current in comparison to Lithium Manganese though. Perhaps I misread the datasheet for those I'm not sure?

MrAl said:

A star ground is simply a single physical connection point where all of the lead to be grounded gets connected to. Choose one solder connection as your star center and connect everything up to that single point rather than daisy chain some of the grounds.

Click to expand...

I found this pic which gives the idea. I hadn't thought about ground paths before and probably have daisy chained some of them.

Thanks,

Antknee.

Hi again,


Well, if you use only 3v input that means you'll be drawing roughly 3 amps from the battery. At 6v input, only about half of that or around 1.5 amps.
The efficiency will also go down somewhat with 3v input rather than 6v input. It's up to you though as long as the battery cell can provide the current.
The runtime for 6v input of course will be nearly twice that of the 3v input, using the same kind of cells for each...one cell for 3v and two for 6v.

Yeah. I can't really tell exactly how the circuit is going to perform, so it is a case of build it and see. I'm also going to be using the Max774 that is an inverter/DC booster from the same family to provide a dual supply on the same board with the same batteries. I will have to have a think how that will affect the layout and earthing. I didn't mention that at the outset because it would confuse matters and they are very similar chips, now I understand one I also understand the other. So I'm probably going to have to use 2 Lithium batteries, they are expensive and it would be better to use just one.

Regards,

Antknee.

========Duplication===========
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for the star ground, all you need is to pick a common point where it's convenient to put a large area of copper, the best place would be either the ground of the output filter or the ground side of Rsense. the idea is to keep from developing voltage drops on ground runs that would cause errors or instability in the circuit. the "common" of the star should be the ground for a circuit element that has the highest total current passing to ground. this would either be the ground for the FET (as i said, on the ground side of Rsense) ot the output filter cap ground (because of the high charge/discharge currents from the switching action of the FET. if one of these is used as the common, the voltage drop across the other ground traces will be minimized. point-to-point grounds can cause a lot of problems, especially in circuits where there are high peak currents. these currents create voltage drops along the trace. if the ground is point to point, the voltage drops interact, as some are in series with each other, especially along the long center trace in the picture. as the extremities branch in, the currents add, increasing the voltage drop along the trace. with a star ground, all of the currents sum at a single point, and the voltage drops are isolated from each other if the current along one branch increases it does not add a voltage drop to any other ground point.. think of it as a resistor network, any trace line is a resistor, with point-to point, you have several resistors in series with each other, with a star ground, no resistor is in series with another, they all tie at one common point,