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I am trying to work out the math of what is the highest cap voltage or capacity I can have to charge from a boost topology in a given time from a source of 5V, 500mA. Here is a rough sketch of what I am describing.

schematic

simulate this circuit – Schematic created using CircuitLab

SW1 will be on for 90mS to charge the cap, then will close, then SW2 will open for 10mS to dumb current to a load. So 10Hz duty cycle.

I started to see if I have enough power: (Please correct me if I did this wrong)

PART 1

I am going to look at this from joules side first.

My source has a 2.5W which is 2.5 joules/s. There for for my duty cycle of 90mS I have 2.5 joules/s * .09 = 0.225 joules/ 90ms.

A cap of 30V 300uF with has energy of 1/2*C*V^2 = 135m joules.

So from energy point of view knowing switching loses and what not, that I have enough energy to charge my capacitor. Is this correct?

PART 2

This is where it all falls a part when I try to calculate it using constant current capacitor charging equations. The numbers don't add up.

My source is 2.5W, lets assume switching losses of 0.5W to make the math easy.

So I have 2W at the cap. Using P=VI with V=30 and P=2W, I will be 66mA.

Using the equation of charge of the capacitor V = It/C assuming I is constant and solving for t=dVC/I (Note: dv=25V since my 2V buck converter will most likely not work below 5V on the cap. I know this will make the math a bit different but I don't think it should change by that much)

So we solve for t = 25V * 300 / 66mA = 113mS. So it will take me 113mS to charge the cap even though I have almost double the amount of energy needed? Perhaps I am looking at this all wrong and just want someone to point out why my numbers are not adding up and that I am just mixing up two different thing together.

Thanks

EDIT: The load is a laser of 10A for 6mS 10Hz

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  • \$\begingroup\$ You are going to lose >50% of your energy. this way Why do it? \$\endgroup\$ – Tony Stewart EE75 May 30 '19 at 18:23
  • \$\begingroup\$ Why are you doing this the wrong way? WHat are you really trying to do? \$\endgroup\$ – Tony Stewart EE75 May 30 '19 at 18:36
  • \$\begingroup\$ @SunnyskyguyEE75 I need to deliver a 10A load in 0.004seconds, how would you suggest doing it? \$\endgroup\$ – J. Jones May 30 '19 at 18:42
  • \$\begingroup\$ WIth a Li Ion cell and FET ?? what is the actual load? a Diode which? repetitive? \$\endgroup\$ – Tony Stewart EE75 May 30 '19 at 18:43
  • \$\begingroup\$ From a Solar panel? Please define everything.. also since we have Siemens change mS to ms Don't you want optical or emitter feedback? \$\endgroup\$ – Tony Stewart EE75 May 30 '19 at 18:48
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If you're pulling constant energy from the input, and putting all of it (or a constant proportion of it) into the cap, then of necessity the capacitor current will change over time as it charges.

You can calculate this, and find that the initial capacitor current is -- inconveniently -- infinite. But you can certainly take your 135mJ and divide it by 2.5J/s to get a lower limit on the capacitor charge time.

To really minimize that charge time you'd have to give some love to your boost converter. This is basically what photoflash and capacitive-discharge ignition systems do, but even that has limits -- basically you end up being current-limited in your coil and switching element, and while the upper limit isn't infinity, because of finite switching speeds, it can still be arbitrarily large.

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  • \$\begingroup\$ Can you elaborate the comment regarding 135mJ divided by 2.5J/s? How is that the lower limit on the capacitor charge time? \$\endgroup\$ – J. Jones May 30 '19 at 18:44
  • \$\begingroup\$ 2.5W = 2.5J/s. Conservation of energy says that if you're going to end up with 135mJ, you have to wait for 135mJ/2.5W (or 2.5J/s) to get it. \$\endgroup\$ – TimWescott May 30 '19 at 19:04
  • \$\begingroup\$ What formula are you using to come up with this expression that the energy needed divided by the total energy equals time? \$\endgroup\$ – J. Jones May 31 '19 at 21:56
  • \$\begingroup\$ Your confusion confuses me. It's simple conservation of energy. 1st-year college physics, if not high-school physics. Power is the rate of energy change. Energy is energy. Time = amount / rate. So time = (energy change) / (energy rate of change). \$\endgroup\$ – TimWescott May 31 '19 at 22:33
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You have only stated 10A 2V? 6ms = 120mJ

But with what sag? ? 100mV? How does sag affect current i.e. Rs of Laser diode 10mOhm?

if C= I dt/ΔV= 0.1V/6ms = 10A 6ms /0.1V = 600 mF
ESR loss ?? est. < 10mV = ΔV/I=0.1V/10 = 10 mOhm

Meanwhile to charge up 600mF to 2V , E= 1/2CV^2 = 1.2J so only using 10% thus 10 Hz is max rate using ideal parts. But you have 5V 0.5A or 2.5W = 2.5 J-s so you have enough energy, just need to define how to regulate the laser current better.

Standby current = ? 40mA
Lasing current = 10A CC
Rise time ?
Fall time ? 10 mOhm Half bridge?

Your specs could be improved. on VI impedance of everything.

or start with a NiMH or Li Ion cell and pulse 10A from here.

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