[resending]

Hi - thank you Olin for the very detailed reply and workthru starting
with fundamental equations.  In my design I am using a FET, and there
are very few choices of nfets with both 80V+ ratings and low Ron @ 3V
gate voltage, so using a bipolar as you've suggested makes sense, and
I'll look closely at your design example.  And my spec of "miniscule
current"... the epaper is a prototype display from the R&D arm of a
japanese company with a datasheet with obvious errors and poor english,
and the spec on the 70V requirement was a scope picture showing
allowable dip on the rail in the several usec the display takes to
erase.  So... "miniscule" is what it seems to be, and a 2.2uF 100V ceram
cap on the 70V works.  (Using an electrolytic is not possible due to
very low profile requirement of all the electronics.)

Thank you again for spending the time to write it all up.

J



Olin Lathrop wrote:
> Jesse Lackey wrote:
>> This all worked, but I had to manually "tune" the PWM signal for best
>> efficiency and ensure it could boost to 70V @ the 3V minimum input
>> voltage.  By adjusting duty cycle and frequency one could see when the
>> inductor was saturating, and how efficiency was poor at low frequency,
>> then got pretty decent within a range, and then poor again as the
>> frequency went up.
>>
>> All kind of fun, but I'd certainly rather have a set of equations to
>> come up with duty cycle and frequency given voltage in range/out
>> range/output current range/inductor value & current spec, etc.  At
>> least
>> know good starting points for duty cycle/frequency and around what
>> inductor specs are going to be required for a given design to know if
>> cost/performance is going to be reasonable prior to building a board.
> 
> I'm not sure what you're asking since the equations are the same whether a
> microcontroller produces the switching pulses or some other electronics
> does.  The guiding equation for inductive switching power supply is:
> 
>          Volt Sec
>   Amps = --------
>          Henries
> 
> In other words, if you know any three of applied voltage to a inductor, the
> length of time that voltage is applied, the inductance, and the resulting
> change in inductor current, you can find the fourth.
> 
> For a micro to control a switching power supply directly from firmware, we
> want switching edges to be a few uS apart.  Just to see where we're at,
> let's see what size inductor would charge up to 1/4 amp from 3V in 10uS:
> 
>             Volt Sec   3V 10uS
>   Henries = -------- = ------- = 120uH
>               Amps      250mA
> 
> OK, that's not way out of line so a 100uH inductor should be quite usable.
> That's a common value and finding one with at least 500mA saturation current
> will be no problem.  So now that we've tentatively decided on 100uH 500mA
> inductor, let's see what the longest guaranteed not to saturate switch on
> time is.  Obviously, this needs to be computed at the maximum input voltage,
> which you said previous was 4.2V:
> 
>         Amp Henries   500mA 100uH
>   Sec = ----------- = ----------- = 11.9uS
>            Volts          4.2V
> 
> So this tells us that as long as the firmware never leaves the switch on for
> more than 11uS, and the inductor current starts out at zero, we are safe and
> the inductor won't saturate.  Let's say the controller is a 10F204 where
> each instruction cycle is 1us, so that would be 11 instruction cycles.
> That's actually a fairly long time, but then again that is the max
> saturation limit.  Nothing says we have to charge up the inductor to 500mA
> each pulse, and there are some reasons not to.
> 
> We'll likely use a NPN switch, since FETs that can handle 70V aren't going
> to switch well with only 3V on the base.  The downside of a bipolar is you
> have to keep feeding it enough base current to stay in saturation.  Let's
> say you can get a NPN that can withstand the voltage with a guaranteed gain
> of at least 40 (I haven't checked, but that sounds about right).  If you
> wanted to max out the inductor at 500mA, the would require 500mA / 40 =
> 12.5mA base current to keep the transistor saturated.  At least that's
> easily doable from a PIC pin, since those are rated for twice that current.
> I wouldn't actually run the transistor at the limit, since there will be
> some rise in C-E voltage.
> 
> I don't want to write a book here, so I'm going to cut short a few thought
> processes.  Let's just pick 6uS as the inductor on time, and we'll use a
> pulse on demand system.  Whenever the output voltage goes below its
> regulation threshold, the micro will produce a 6uS (6 instruction cycle)
> pulse.
> 
> Note how each time we pick a parameter, we can feed that into the equations
> and nail down something else more tightly.  At this point we've decided to
> use a 100uH inductor, a NPN grounded emitter switch with a minimum
> guaranteed current gain of 40, and a fixed size 6uS pulse on demand control
> system.  Now we go back and find what our highest inductor current can be:
> 
>          Volt Sec   4.2V 6uS
>   Amps = -------- = -------- = 252mA
>          Henries      100uH
> 
> 1/4 amp is going to be no problem for finding a inductor or transistor, so
> that all looks good.  To keep the transistor saturated, it needs 252mA / 40
> = 6.3mA base current.  Obviously we want a little more for margin and to
> keep the saturate voltage low.  10mA sounds like a nice number given the
> circumstances.
> 
> Here's something that might not be obvious.  The limiting case for base
> current is the lowest input voltage, not the highest.  You might think it
> doesn't matter because the inductor current will scale linearly with input
> voltage, so proportionately less base current is needed as the input voltage
> goes lower.  However, there is a larger fixed offset subtracted from the
> base current drive voltage than from the inductor drive voltage.  This is
> because the voltage accross the base resistor is the supply voltage minus
> the fixed B-E drop of the transistor, whereas the inductor charge voltage is
> the supply voltage minus only the transistor saturated on voltage.  The net
> effect is that the base current scales down more quickly as the supply
> voltage drops than the inductor current does.  So we need to know what
> current the transistor must sustain at 3V:
> 
>          Volt Sec   3.0V 6uS
>   Amps = -------- = -------- = 180mA
>          Henries      100uH
> 
> 180mA / 40 = 4.5mA minimum required base current with 3V supply.  Let's say
> 700mV is lost between the B-E drop of the transistor and the high side
> output pin driver in the PIC.  That leaves 3.0V - 700mV = 2.3V accross the
> base resistor.  2.3V / 4.5mA = 511 ohms.  We of course want to build in some
> margin and keep the saturation voltage down.  360ohms sounds like a nice
> number.  That provides 30% margin.
> 
> Now we need to do a sanity check at the maximum supply voltage.  Again let's
> assume 700mV is lost that is not applied accross the base resistor.  (4.2V -
> 700mV) / 360ohms = 9.7mA.  That's real close to the 10mA we thought sounded
> nice earlier, so that checks fine.
> 
> I'd add another minor detail to speed up the switching transitions a little.
> At 6uS on it's not a big deal, but a $.01 part can help things just a
> little.  Turning the transistor on isn't a issue since the inductor current
> starts out at 0 anyway.  The base current will build up to its steady value
> much faster than any appreciable inductor current will.  However, we want to
> make the transistor switch off as fast as possible.  Grounding its base via
> a 360 ohm resistor is actually not that bad, but we can still speed it up
> with a small capacitor in parallel accross the base resistor.  The resistor
> alone will probably turn off the transistor within about 100nS.  To see
> where we're at, let's see what capacitance would result in a 100nS time
> constant, meaning about 2/3 of the effect is over in 100nS.  100nS / 360ohms
> = 278pF.  Hmm, that sounds too big.  Make it 100pF.  That results in a time
> constant of 100pF * 360ohms = 36nS.
> 
> Normally you have to consider the inductor discharge time.  In this case
> it's obvious from casual observation it's not a issue given the 70V output,
> but let's do the calculation anyway just to show it:
> 
>         Amp Henries   252mA 100uH
>   Sec = ----------- = ----------- = 383nS
>            Volts      70V - 4.2V
> 
> which is well under a single instruction cycle.  Or put another way, the
> inductor current is guaranteed to reach 0 with the switch off for just one
> instruction cycle.
> 
> Of course the above is only true during normal operation when the output
> voltage is already around 70V.  Startup is a whole other issue.  This is one
> of the advantages of having a micro.  You can easily use a different
> algorithm and different parameters during startup than during normal
> operation.  I don't feel like getting into startup here, but it's something
> that has to be considered.
> 
> So the 10F204 code for the normal operating loop looks like this:
> 
> loop
>          skip_low            ;output below regulation threshold ?
>          goto    loop        ;no, go back and check again
> 
>          switch_on           ;turn on the switching element
>          nop
>          nop
>          nop
>          nop
>          nop
>          switch_off          ;turn off switch 6uS later
>          goto    loop        ;back to wait for output low again
> 
> SKIP_LO, SWITCH_ON, and SWITCH_OFF are macros that expand to single
> instructions.  It's a good idea to add that layer of abstraction.  It
> separates the control algorithm from the details of how that particular
> project is wired to the micro.
> 
> Now we can calculate the maximum output current the supply can maintain
> voltage regulation with.  From the code above it should be obvious that
> there is a minimum of 4uS between the fixed size 6uS pulses, for a maximum
> pulse repetition period of 10uS, and a maximum pulse rate of 100KHz.
> 
> At the lowest specified input supply voltage, we calculated earlier the
> inductor charges to a maximum of 180mA.  This was assuming no switch loss,
> but let's see what the best possible case is and scale back from there.  1/2
> * (180mA)**2 * 100uH = 1.62uJ, x 100KHz = 162mW, / 70V = 2.3mA.  Obviously
> we're not going to get 100%, but certainly this supply will be good for at
> least 1mA.  Since the spec said "miniscule" (rant about bad engineering
> suppressed), we can claim we're done.
> 
>> I have to admit I spent only 1 or 2 evenings online reading and
>> looking
>> at dc/dc design info before just tuning it for the particular design I
>> had in front of me.
> 
> The calculations above took way longer to describe than to do.  A couple of
> minutes with a calculator is all it should take to get to this stage.
> 
> 
> ********************************************************************
> Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
> (978) 742-9014.  Gold level PIC consultants since 2000.

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