No! Not in this context.

When you say dt in th equation:

dI/dt = V/L

It is referring to an infinitesimal delta time, that is calculus.

Because the solution is linear, you can use any dt and the slope if the current vs time remains the same.

Not in any context in this context. The dt is in regard to the change of current through the inductor over time. The mosfet is regarded as an ideal switch with effectively zero transition time, and the formula you query is describing the behaviour of the INDUCTOR as time passes.

 

This is the part that is getting lost in communication I'm trying to get at.

So when we say dt, what dt are we referring to?

dt mean the rise and fall time of the mosfet. Aka, how long does it take for the mosfet to be fully on and or fully off.
That is the dt i thought is generally discussed when we talk about V = L di/dt. However i am thinking that is my confusion??

Or is dt the amount of time the mosfet is actually turned on that is in the V = L di/dt equation?


is dt in this context something else.
View attachment 297704

First - a bit of pedantry.
Δt is a time interval. dt is part of dI/dt which is a slope. dt is an infinitesimally small amount which really doesn't make much sense on its own.
What you have shown appears to be the MOSFET gate waveform. Your first dt is part of a dV/dt which is the gate voltage risetime, which is not relevant here, only to say the steeper the better.
Your right hand side dt (really Δt) , is the MOSFET off time. PW is another Δt and is the MOSFET on-time.
V/TD gives you the downwards slope of current when the MOSFET is off. V is the difference between Vsupply and Vout for a boost converter.
V/PW gives you the upwards slope of current when the MOSFET is on. V is the power supply voltage.
In both cases is the the voltage across the inductor.
If you drew the inductor current waveforms, then you would have some slopes and dI/dt would make sense.

 

This is the part that is getting lost in communication I'm trying to get at.

So when we say dt, what dt are we referring to?

dt mean the rise and fall time of the mosfet. Aka, how long does it take for the mosfet to be fully on and or fully off.
That is the dt i thought is generally discussed when we talk about V = L di/dt. However i am thinking that is my confusion??

Or is dt the amount of time the mosfet is actually turned on that is in the V = L di/dt equation?


is dt in this context something else.
View attachment 297704

Hello,

I think you need to read ALL the posts after yours not just the last one. I had explained this fully in post #13.

The 'dt' is the time increment. For that expression 'dt' is said to go to an infinitesimally small value. To understand the basic operation though, you can make 'dt' some small value like 10us or something. If the current changes by 1ma in 10us that means di=0.001 and dt=0.00001 and the division results in di/dt=100, and so the voltage is L*100 for that example.
In a switching converter, 'dt' may be taken to be the pulse width if the output is nearly linear, which it usually is. It will not hold for large values of 'dt' because there are often other considerations such as the output capacitor.

To add to that, once the inductor has some current flowing through it, it starts to act like a current source and when the switch turns off the impedance of the circuit goes up and so the current source produces a much higher voltage. In theory, if an inductor has current flowing though it and it is suddenly open circuited, the voltage rises to a value that approaches infinity. In real life there are losses, but the voltage can still go up very high.

 

if an inductor has current flowing though it and it is suddenly open circuited

Which it is not in a boost converter.

 

Which it is not in a boost converter.

I suppose, technically, it will be open circuit for those few hundreds of nanoseconds after the MOSFET stops conducting, before the output diode conducts, whilst the output voltage slews charging up the stray capacitance.

 

Which it is not in a boost converter.

Hi,

What is your point?
That's part of why a boost circuit boosts

 

Hi,

What is your point?
That's part of why a boost circuit boosts

So the TS would not get the wrong idea.

 

The mechanism that turns the transistor on and off is the error amp. You ask the converter to give you 20 V with an input voltage of 5V. The error amp inside the IC gets feedback Vfb from the output and compares this voltage with a reference voltage Vref.
Vref >Vfb keep the pulse on. Vref<Vfb, shut off the pulse. Thats what turns the FET on or off. The rest of the IC is other stuff.
The transfer function of a BOOST is Vo/Vi = 1/(1-D). D = duty cycle.
If you dont have a load eventually the Vfb will be > Vref and no more On pulses..
A boost converter is more than just an inductor and a switch. you need the diode, cap and the load. (Perhaps not the cap). Without the load you are energizing the inductor and not giving the energy anywhere to go. without the cap and the load, your transistor will blow up due to Vdss > the FET rating.

 

The mechanism that turns the transistor on and off is the error amp. You ask the converter to give you 20 V with an input voltage of 5V. The error amp inside the IC gets feedback Vfb from the output and compares this voltage with a reference voltage Vref.
Vref >Vfb keep the pulse on. Vref<Vfb, shut off the pulse. Thats what turns the FET on or off. The rest of the IC is other stuff.
The transfer function of a BOOST is Vo/Vi = 1/(1-D). D = duty cycle.
If you dont have a load eventually the Vfb will be > Vref and no more On pulses..
A boost converter is more than just an inductor and a switch. you need the diode, cap and the load. (Perhaps not the cap). Without the load you are energizing the inductor and not giving the energy anywhere to go. without the cap and the load, your transistor will blow up due to Vdss > the FET rating.

Hi,

The basic physical operation though is based on the inductor being energized with a lower voltage, then when more or less open circuited, produces a higher voltage. That's back EMF and it's how we get true energy conversion. You don't need an error amplifier to construct a boost converter. An error amplifier is used to get line and load regulation, which falls under the category of control theory.

 

Hi,

The basic physical operation though is based on the inductor being energized with a lower voltage, then when more or less open circuited, produces a higher voltage. That's back EMF and it's how we get true energy conversion. You don't need an error amplifier to construct a boost converter. An error amplifier is used to get line and load regulation, which falls under the category of control theory.

His question was " I am asking what the electrical mechanism is that changes the dt in V = L * di/dt equation. "
now tell me how you would answer that?
Instead of starting an argument with me, if you have a better explanation then give it.
The guy was getting exasperated with the answers he was getting and wrote his question in BOLD.
Did you understand his question?
I read all the answers and there may have been on that did answer what he asked.

 

His question was " I am asking what the electrical mechanism is that changes the dt in V = L * di/dt equation. "
now tell me how you would answer that?
Instead of starting an argument with me, if you have a better explanation then give it.
The guy was getting exasperated with the answers he was getting and wrote his question in BOLD.
Did you understand his question?
I read all the answers and there may have been on that did answer what he asked.

Hello there,

I am not arguing with you, I was just stating some facts about boost converters.

You were asking me if I understood his question maybe because you didn't think my answer was good enough, yet. Did you attempt to answer his actual question? If you did I'll read it over.

I think you brought up an interesting point in any case. Did anyone actually answer the question that was asked. This leads me to believe that another approach to the view of how a boost converter works is in order. I'll answer the question more directly in my next post. Thanks for bringing this up.

 

That is what i am asking.
I am asking what the electrical mechanism is that changes the dt in V = L * di/dt equation.

Hello again,

There are a couple different ways of looking at this that's probably why you may or may not have gotten the answer that you were looking for yet. I'll try to explain from another perspective.

We start with the equation again:
v=L*di/dt

This equation happens to be the solution for the voltage across the inductor. If we assume that L and di/dt are constant, then the voltage 'v' will be constant. There are four variables there though, so we could solve for any one of them, or even consider di/dt to be a single variable.
Since you were interested in 'dt', we can solve for that first:
dt=di*L/v

What this tells us is that the change in time is dependent on the change in the current times L/v. This isn't really a practical view though because we don't usually have to solve for 'dt'. What we usually have to solve for is 'di', or di/dt together as a rate. So, let's solve for di/dt:
di/dt=v/L

Now this tells us that the rate of change of current is equal to the voltage divided by the inductance. If the voltage is constant (a usual assumption to start out with) and the inductance is constant (also a usual assumption and often not changed at all) then we see that the current will increase at a given rate and that, combined with the pulse period, tells us how large the current will become at the end of the pulse period. The pulse period is the time the switch is either on (inductor charging) or off (inductor discharging).
Since when the switch is closed the inductor is charging, the voltage is the power source voltage Vin. Thus, we know what the rate of change of current is when the switch is closed:
di/dt=Vin/L

Now we know what the rate will be and so after a given pulse period (which is NOT 'dt' BTW) we know the current will be just so large. After the switch opens, the voltage is now the difference between the source (Vin) and the output (Vout) or simply vL:
vL=Vin-Vout

Now the rate of change of the current is:
di/dt=vL/L

and this is simply the voltage across the inductor (vL) divided by the inductor value L.
This tells us how fast the current will decrease as it discharges into the output capacitor.

BTW, in this view of the theory behind the converter, we assume that the initial transients are gone we just have the cycle by cycle transients left to deal with. This is the time when the current ramps up and then down, then keeps repeating, and the ending current is always the same and the starting current level is always the same. So it looks like a sawtooth to a first order approximation (which is usually good enough for most purposes).

What this means is that 'dt' is sort of an ancillary variable where we don't really pay too much attention to it directly. As mentioned about, this has nothing to do with the pulse period that the switch is 'on' for, nor the 'off' period. It's just part of the rate of change and this may be even more apparent if we use a single variable for the rate of change of inductor current:
r=di/dt

and now we see that when the switch is on we just have this:
r=Vin/L

and it becomes clear we are not that interested in 'dt' anymore.

I think this makes it a little more clear where that 'dt' fits in. It's not really worth mentioning too often except in some other basic theories that are really related to just an inductor not a converter. It's going to continue to appear in formulas as di/dt, but it is better to look at that di/dt as a single variable when it comes to the theory of these converters.

Sometimes an example helps to drive the point of the theory home.
Say we have a 12v input and we want a 36v output, and our inductor is 0.2 Henries. The charging rate of change of current di/dt is:
di/dt=12/0.2=60 amps per second.
Now if we leave the switch closed for 0.1 seconds, the current at the end of the pulse period would be higher by:
di=60*0.1=6 amps.
If the current was 30 amps to begin with, it will now be 30+6 amps which is 36 amps.
Now when we turn off the switch the rate of change of current will be:
di/dt=24/0.2=120 amps per second.
If we leave the switch off for 0.05 seconds, the current will decrease by 120*0.05=6 amps. The current at the end of the cycle will be:
i=36-6=30 amps.
The cycle will then begin again and the inductor will start to charge again.

Notice we never really looked at 'dt' alone, we just looked at it as being an integral part of di/dt.
If you must know what 'dt' is responsible for, then if we look at the rate as a numerical value then we would have some units to consider:
di/dt=60 amps per second
and this means that when we multiply both sides by the time we get the change in current:
di=6

We can look at some waveforms if you like.

 

You can't change dt on its own. dt is part of the symbol dI/dt, which is the symbol for the derivative of I with respect to t (or the slope of I over time).
If you put it in Newtonian notation it would be I (upper case I with a dot above it), or Iʹ in Legrangian. It's like trying to define the ʹ on its own.
You can change Δt, but that is a different matter

 

You can't change dt on its own. dt is part of the symbol dI/dt, which is the symbol for the derivative of I with respect to t (or the slope of I over time).
If you put it in Newtonian notation it would be I (upper case I with a dot above it), or Iʹ in Legrangian. It's like trying to define the ʹ on its own.
You can change Δt, but that is a different matter

Hi,

That's a good way to put it too.
The 'dt' in di/dt does have some relevance though, it does pertain to the units of di/dt. It is amperes per second. That's about it though I think.

 

Hi,

That's a good way to put it too.
The 'dt' in di/dt does have some relevance though, it does pertain to the units of di/dt. It is amperes per second. That's about it though I think.

Yes. Messrs Newton and Lagrange don't make it clear what variable is being differentiated with respect to.
I think that the TS is implying that it is t(on) or t(off) that is being varied.

 

Yes. Messrs Newton and Lagrange don't make it clear what variable is being differentiated with respect to.
I think that the TS is implying that it is t(on) or t(off) that is being varied.

Hi,

I got that impression also, that's why I tried to make it clear that 'dt' was different than the pulse period. We could say that in the first order approximation that dt=pulse period, but in general that is not really true. For example, if the pulse period is 10us and during that time the current changes by 30 amps, we could state that the rate of change of the current is 30A/10us which of course equals 3 amps per microsecond which is 3 million amps per second, and this would fit in with the inductor expression v=L*di/dt, with di=3 amps and dt=1us, or di=3000000 amps and dt=1 second, but that would be stretching it unless the inductor current was a perfect ramp, and that means there would be no filter capacitor. In many calculations though we do use the approximation of a perfectly linear ramp, so maybe we can think of 'dt' as being the pulse period in the first order approximation.

 

Not sure if I am misunderstanding, but the original poster seems to be getting confused between the rise and fall times of the FET (which he is calling “dt”) and the switching period of the FET. These of course are different things. The rise and fall times of the switch are substantially faster than one would set the switching frequency. The point of using a high frequency (lower dt) is to reduce component size or to minimise either current/voltage ripples within the circuit. Reducing the FET ON time (dt) may also demand that you reduce the FET rise and fall times by choosing a FET with lower input capacitance, internal resistance, or increasing gate drive strength.

Therefore, the answer to the question “what the electrical mechanism is that changes the dt in V = L * di/dt equation" is both the switching frequency and the duty cycle. It has nothing to do with the FET rise and fall-time, but to increase FET switching frequency you may need to speed up the rise and fall times too for optimal operation.