# Inductor boost regulator

Discussion in 'General Electronics Chat' started by KCHARROIS, May 25, 2013.

1. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
300
8
Hello,

So I'm looking to make a boost regulator circuit so I started looking at basic RL circuits and I understood how they got the formula

when t=0, voltage accross the inductor is E and current is 0 but then overtime the voltage accross the inductor is 0 and current accross the inductor is equal to V/R.

But when I look at a circuit like the one attached, how can you predict the output voltage when there is no load? When Q2 is ON, is it safe to assume that current through the inductor will eventually be equal to Vds/Rds after a certain time? But when Q2 is OFF the capacitors charge up to a certain voltage but to what value?

Thanks

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2. ### tindel Well-Known Member

Sep 16, 2012
620
206
If you simply turn the switch on and then off, the capacitors will charge to the input voltage (9V) as the inductor will act like a short at DC with no load.

What starts the output regulating is sensing the output voltage and turning the switch on and off quickly in an effort to regulate the voltage. Most Boost SMPS have a minimum current load they require to regulate properly. The minimum current is usually found in the sense resistor divider and is a function of the switching frequency.

3. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
300
8
So what your saying then is that this circuit here is incomplete?

Thanks

4. ### tindel Well-Known Member

Sep 16, 2012
620
206
Yes - this is just showing the topology of a inductor based boost circuit.

5. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
300
8
Ok great but how can you explain this, they say that
Vout = Vin + Vin(Ton/Toff) in a boost regulator. For example I would get the same output if Ton were 0.5ms or 0.5us and if Toff were 0.25ms or 0.25us.

Thanks again

6. ### tindel Well-Known Member

Sep 16, 2012
620
206
I think I gave you some bad information - I'm learning about these things myself.

After a little spice analysis:

Yes an inductor is a short at DC, meaning at steady state the output will be 9V. If you turn on the switch and turn it off, the reaction will be a bit different...

During the on cycle you will short out your supply and will charge your inductor, meaning you will have a large magnetic field. As soon as you turn it off the current will continue flowing through the inductor (inductors instantaneous change voltage, not current) and you will have a large voltage spike on the capacitor, which will in theory hold that voltage until you discharge the capacitors - the spice simulation shows it a little bit different though - the schottky will actually discharge the capacitor due to it's leakage currents - back to 9V.

The spice model also assumes that the inductor will not saturate and that the voltage source has infinite current.

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7. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
300
8
The spice model worked great thanks alot, but were you able to figure that the output would be about 50V?

Thanks

8. ### tindel Well-Known Member

Sep 16, 2012
620
206
I'm a bit too lazy and busy to solve the differential equation on a beautiful sunday morning in Colorado, but the idea is that the inductor resistance is 55mohm and the switch resistance is about 2mohm.

9V/57mohm is about 160A, which the spice simulation shows.

The question you're asking is to predict the output voltage after the switch has turned off. This is more difficult - not impossible - but does involve solving the differential equation. You can, however, verify voltage by using i(t)=C*dv/dt. If you trust that the simulation has solved the differential equation correctly (no doubt in my mind) then the spice simulation shows at 7s that dt is about 46us. This equates to dv being 73.6V at t=switch open. This dv/dt will fall as t goes on, but the result is 'on par' for the final voltage result.

Maybe I'll solve the differential equation tonight - but I'm not making any promises.

This is different from how a boost converter works though - we're solving for steady state values.

Wikipedia has good output voltage prediction calculations for a boost converter run in continuous and discontinuous modes. Though I haven't derived them myself.

9. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
300
8
Good prediction but I believe you need to take the average current over time and then use it in your formula i(t)=C*dv/dt. So for average I just used 0.636 * 160 and then plugging the values into your formula I got 46.8V, I know the average might be wrong but it is fairly close to the voltage accross the capacitor 47V.

Thanks again you've given out great advice and insigh,t the only real thing to really determine now is calculating why it took 46us to discharge.

Thanks

10. ### tindel Well-Known Member

Sep 16, 2012
620
206
It takes 46us to charge because that's what the differential equation solves too!

I'll take a closer look at solving the differential equation on paper tonight. (I'm at work today - no rest for the weary).

I too am enjoying the discussion, and you keep coming back - so I'll continue to help (or try to)!

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11. ### tindel Well-Known Member

Sep 16, 2012
620
206
Well, I did give it an honost shot today - And I have a formula for the short cycle, but with the input being a step function instead of starting at 9V.

The approach I was taking was to put things into the frequency domain (Laplace transform) and convert back to the time domain. I couldn't figure out how to make the system think that the input was at 9V for a long time. The Laplace of a DC source is Vsource/s -> which equates to a step function, not a long time function. I'll get it with some more time, but I will have family in town the rest of the week and don't know how much I'll get to work on it.

The function I derived was:
$V_{FET}(t)=\frac{V_{in}(t)*R_{FET}}{R_{FET}+R_{L}}\left[-e^{-\frac{R_{FET}+R_{L}}{L}*t}+1\right]$

$I_{FET}(t)=\frac{V_{FET}(t)}{R_{FET}}$

If you plot these you will get a rise from 0V to 9V*Rfet/(Rfet+Rl) in about 1ms...

Give me some time and I'll get the functions - it will just take some more head scratching with the Laplace transforms. Or you can do them yourself.

12. ### KCHARROIS Thread Starter Active Member

Jun 29, 2012
300
8
Woah, just a little out of my league lol although, I did find a good pdf that did help explain how to calculate Vout, current, ripple, etc. The only thing I dont really understand is how does Vin + Vl + Vo = 0.

Vin = 12
Vl = -12
Vo = 24

12 - 12 + 24 = 24???

I have briefly seen the laplace transform so not quite sure how to use that approach but I do appreciate the help.

Thanks

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13. ### tindel Well-Known Member

Sep 16, 2012
620
206
KCHARROIS - This post probably isn't going to be quite what you want, but I have spent a LOT of time studying this circuit (mostly because I too am developing a boost circuit - so I'm finding this interesting). I have discovered many things - things that can really only be learned by digging into the analysis yourself, but I will try to summarize my significant findings.

Note that a good electronics book will help with understanding these concepts better than what I will describe. I am using "Electronic Circuits" by James Nilsson and Susan Riedel, 7th ed. to help me understand these concepts. I consider this an excellent book for general electronics understanding. Suprisingly, the AAC book does not cover the step and natural response of RLC circuits. Not that I could find anyway.

One of the things that we were still questioning was the rise time to 50V and really even the 50V peak. There are some very long equations that predict these with great accuracy. I won't go into them here. However, consider the following equations for a series RLC circuit:

The damping factor (α) is α=R/(2L)|rad/s|
The resonant radian frequency (ω0) is ω0=1/sqrt(L*C)
The damped radian frequency is ωd=sqrt(ω0^2+α^2)

If ω0 > α then the system is underdamped. In this case, the circuit is underdamped. Underdamped means that you should see some ringing at the ωd frequency... But you don't! Why? The answer lies within the diode. Once the current goes to 0, current can't reverse (go negative) because the shottky is stopping that from happening. So why does this matter? Because it means that theoretically you can switch at a very fast speed and not have any ringing on the output capacitor!

There is still ringing in the circuit (on the Drain of the FET). I believe this is due to the very small currents in the inductor going through the FET capacitances. This may cause the FET to be damaged due to large transient voltage.

If you short out the diode, it does ring with a damping factor of ~800Hz and a damped frequency of ~5kHz, as predicted.

So, bringing this back to the timing: If you take ωd and divide by 2*pi then you get the resonant frequency of the circuit, but since it only resonates until the capacitor current reaches 0, in a quarter cycle (in this case about 50us.)

Since the ringing is sinusoidal, and we know that the sine wave is mostly linear except at the peak and troughs then, as you correctly suspected, we can average the current for the time duration and predict the output voltage to some decent amount. i=C*dv/dt => dv = dt*(Imax/2)/C+Vinit ~= 47Vish.

Finally note that if you set R to 0, the ringing will continue forever!

Something to note is that my book calls the damping factor α, many other sources use the ζ symbol for damping factor.

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14. ### JMac3108 Active Member

Aug 16, 2010
349
67
Hey guys, I don't have time to read all of your analysis, but I've designed LOTs of boost regulators and have derived the equations many times.

You do not start with the concept of an RL circuit when analyzing or designing a boost. You use V = L di/dt, the standard inductor equation. It can be rearranged and expressed as, I = (V/L) Ton.

When the switch is turned on, the inductor current is a linear ramp and...

I = peak inductor current (top of the ramp)
V = voltage across inductor (equal to the boost input voltage)
L = inductance
Ton = on time of the switch

When the switch is turned off, the current ramps down.

With this, and a google search, you should be able to find and understand derivations for the standard boost regulator.