Electronics > Beginners
Inductor Discharge Rate and Other Questions
TheDood:
Perhaps worrying about the wrong stuff here?:
Given a conventional single inductor boost topology, will the filter cap voltage impede a magnetic field collapse? Will the filter cap voltage effect the inductor discharge rate?
If I've stored [0.5L×(I^2)] J, how long will it take to dissipate? Does a charged cap in the inductor discharge path effect the rate at which the inductor discharges? Or how much energy discharges?
If Tau = L/R and 5Tau seconds equals full charge/discharge, then the amount of discharge time would be dependent upon cap V?
If cap V was known, ending charging inductor I was known, and total J stored in inductor was known, can I determine discharge time? Is it important to determine exact discharge time or will it be less than charge time if impeding V is higher than input voltage?
PFC:
I'm trying to measure cap V at zero cross, then calculate remaining J's needed to achieve desired constant V of filter cap, then set PWM based on the needed J's to achieve the desired filter cap V. If Ω is kept constant during inductor charging then current will model a sinusoidal function as input voltage does, but the amount of current needed per switching period to arrive at the sum total of J needed to fill the filter cap per switching cycle will be time dependent. So I'm trying to determine exactly how the energy stored in an inductor is released so that I can provide enough discharge time for the inductor to fully dissipate the stored energy onto the filter cap.
Will ALL the energy deposit onto the filter cap? Or is some left behind in the form of hysteresis, or maintaining residual mag field?
Is this an electrically efficient way of achieving high PFC, or boosting V? Ie will the amount of power consumed during charge up equal the amount of power discharged onto cap? I've seen some graphs showing PFC, but it almost looks as if they aren't discharging the inductor completely. Is discharging completely not advised?
Curious:
Also, if your inductor charging V is 170V, and your filter cap is at 170V, and if current doesn't immediately drop to 0 during initial discharge event, but rather drops to 0 along some function originating from the initial current down to 0, then right after the charging event has concluded and you begin to discharge, the change in current will be almost 0, but the filter cap is at 170V, so in order to maintain the current flow, or in order to have a change in current = 0, your voltage needed to flow that current will be 170V more than the impedeing voltage (cap charge level) due to the fact that it took 170V initially to flow the current, so you'd effectively have to be generating 170V + any impeding voltage to flow the same current? Is this correct thinking?
Final thoughts:
If I were to integrate a J/s curve based on sinusoidal V and I and constant load R to arrive at a maximum J's per cycle possible, then I should be able to use a static switching Hz and calculate my duty cycle % based on a [(Joules needed to fill cap) ÷ (maximum joules possible)] figure?
So perhaps choose an inductor with enough inductance that at 50% duty cycle of a certain Hz, the time constant is such that at peak Vin, the inductor I = desired Irms × 1.41.
So instead of using a higher R in the charging event to reduce peak current in the inductor, use a bigger inductor to reduce peak I through the inductor, given a set Hz and maximum of a 50% duty cycle?
If this "Final Thoughts" section is sound then I can continue on with determining core specifics, wrap #, skin effect, ect, per the Hz desired.
Please give any insight you see fitting. Thanks.
T3sl4co1l:
You need something more basic:
V = L dI/dt
If you've fixed the voltage on the inductor (say by clamping that voltage with a diode and a relatively large value capacitor), and L is constant, then dI/dt is also fixed.
For approximately-square wave voltage, dI and dt become ΔI and Δt, trivial to integrate as the ramp is linear.
If you know initial and final current, and V, you can solve for Δt, the time required to discharge the inductor.
I never liked the "magnetic field collapse" phrasing. It implies something different has happened. The inductor is linear (for the most part), nothing different happens. The same equation applies, regardless of whether its arguments have pluses or minuses. Apply a positive voltage drop (given a convention for which end of the inductor is positive and negative, and its current flow is positive when flowing from positive to negative), and dI/dt is positive. Apply a negative voltage drop and dI/dt is negative. Integrate V/L dt and you have inductor current.
Tim
TheDood:
--- Quote from: T3sl4co1l on January 30, 2020, 12:45:09 am ---You need something more basic:
V = L dI/dt
If you've fixed the voltage on the inductor (say by clamping that voltage with a diode and a relatively large value capacitor), and L is constant, then dI/dt is also fixed.
For approximately-square wave voltage, dI and dt become ΔI and Δt, trivial to integrate as the ramp is linear.
If you know initial and final current, and V, you can solve for Δt, the time required to discharge the inductor.
I never liked the "magnetic field collapse" phrasing. It implies something different has happened. The inductor is linear (for the most part), nothing different happens. The same equation applies, regardless of whether its arguments have pluses or minuses. Apply a positive voltage drop (given a convention for which end of the inductor is positive and negative, and its current flow is positive when flowing from positive to negative), and dI/dt is positive. Apply a negative voltage drop and dI/dt is negative. Integrate V/L dt and you have inductor current.
Tim
--- End quote ---
Thanks Tim,
You've been helping for awhile now, I wish I was catching on quicker, sorry, I appreciate your insight.
I thought that the voltage across the inductor reversed polarity upon the event at which the energizing voltage was "instantaneously" removed? Im getting confused on how to clamp the voltage if it's reversing?
For a cct with a 12V source, single inductor, and switch:
State 1:
Open cct
Energizing V = 12V
Voltage across inductor = 0
State 2:
Closed cct
Energizing V = 12V
V across inductor = 11.99 V
State 3:
Closed cct
Energizing V = 12V
V across inductor = 5V
State 4:
Closed cct
Energizing V = 12V
V across inductor = 0V
State 5:
Open cct
Energizing V = 12V
V across inductor = -12V
Is this incorrect thinking?
If the energizing V is 12V but I'm trying to push current onto a 13V cap, does discharge time drop in duration compared to the charge time? If I charged a coil for 1s at 12V and placed a 13V cap (cap charged to 13V) in its discharge path, the time of discharge is reduced from the time of charge up, correct? So that if 50% duty cycle were my max ON duration, and I was boosting V, that I shouldn't be worried about discharge time as it'll be less than the charge time?
Also is my assumption that, the total energy stored in the mag field is being discharged completely or almost completely (depending on hysteresis) true? That if V = J/C, and I = C/s, that the energy will dissipate completely regardless the resistance/impedance in its path? Like a transformer maintains power but the V,I of the primary is different that the V,I of the secondary but total power is conserved (ideally)? The inductor charges up with whatever V, but it reaches a certain energy before cct is opened, and if a cap charged to a higher V is put in the inductor discharge path, that the energy discharged will still be equal to the energy stored, but less current will flow due to the higher V cap in its path? IE it will reduce its discharge time because current is maintained initially at switching event, but then would drop faster as more J/C were needed to discharge the energy?
Attached is what I think you were saying to me, but I'm not certain. If it is what you were indicating, it looks like an additional buffer for the transient(?), but I'm still not certain on exactly why or how we get such transients if there's a route for the current to take that doesn't require such a high V. For example, when I simulate in LTspice, I often get large spikes at switch off events. This would mean that dI/dt were large. I'm not sure how you can have such a large change in current if the inductor opposes it? I feel like I'm going in circles lol... In fact this is where I thought Tau played a role? I don't see any talk about Tau ever, and its not seen in any of the equations posted surrounding my PFC commentary, does Tau not factor in as much as I think?
Extra:
I get confused and this is not as heavy on my mind but figured you could elucidate. I've read that current in an inductor wants to remain that current, even after energizing voltage had been removed, but then I've read that it acts as a spring in such a way that the current flowed through wants to reverse back the way it came upon voltage removal? Which is it?
Also, attached is an example of the equation previously posted. How do they know that the current drops to 0 in the time they've decided to switch off? I see that the time off is 10ms, but what if it was 5ms or 100ms? How do they know it drops to 0 in 10ms? They integrated V/L(dt)?
T3sl4co1l:
--- Quote from: TheDood on January 30, 2020, 02:38:49 am ---I thought that the voltage across the inductor reversed polarity upon the event at which the energizing voltage was "instantaneously" removed? Im getting confused on how to clamp the voltage if it's reversing?
--- End quote ---
This phrasing suggests a circuit where the driving impedance rises suddenly. Like switching a relay coil. Where you don't care what happens when it turns off, you just want the current gone.
Of course we don't want unconstrained voltage, that damages transistors. So even in this simple case, we use a diode to clamp the voltage, which keeps the driving impedance low. The inductor voltage is always under control, always well defined.
In a switching converter, the inductor voltage is always well defined: either Vin when switched on, or Vin-Vout (for the boost case shown) when switched off (and discharging). There is a third state, if the inductor has finished discharging: voltage rings down to zero.
--- Quote ---For a cct with a 12V source, single inductor, and switch:
State 1:
Open cct
Energizing V = 12V
Voltage across inductor = 0
State 2:
Closed cct
Energizing V = 12V
V across inductor = 11.99 V
State 3:
Closed cct
Energizing V = 12V
V across inductor = 5V
--- End quote ---
Where's the voltage dropping? There's no resistor shown in the circuit.
--- Quote ---State 5:
Open cct
Energizing V = 12V
V across inductor = -12V
--- End quote ---
This is missing the output voltage. Is it 24V then?
--- Quote ---If the energizing V is 12V but I'm trying to push current onto a 13V cap, does discharge time drop in duration compared to the charge time? If I charged a coil for 1s at 12V and placed a 13V cap (cap charged to 13V) in its discharge path, the time of discharge is reduced from the time of charge up, correct? So that if 50% duty cycle were my max ON duration, and I was boosting V, that I shouldn't be worried about discharge time as it'll be less than the charge time?
--- End quote ---
Assuming you're still talking about the boost configuration, no, discharge time is inversely proportional to discharge voltage. For a 13V output, V_L(disch) = -1V. The equal flux condition is V_L(chg) * t_chg = V_L(disch) * t_disch. If flux is 12Vs then discharge is 1V and 12s.
If the switch turns on during the discharge, current begins at whatever it left off at. If the inductor is 12H, the peak current is 1A (12Vs / 12H = 1A; henries have units of flux per ampere). Say you turn on after 6 seconds: current starts at 0.5A, and an on-time of for example 0.5s is needed to raise it back to 1A. This is CCM.
--- Quote ---Also is my assumption that, the total energy stored in the mag field is being discharged completely or almost completely (depending on hysteresis) true?
--- End quote ---
If I_L = 0 at the end, yes.
--- Quote ---That if V = J/C, and I = C/s, that the energy will dissipate completely regardless the resistance/impedance in its path?
--- End quote ---
Energy only dissipates if given a path to dissipate through. Inductance is lossless. You can short an ideal inductor at 0V forever and it will hold its magnetic charge (current and flux and energy). Superconductors show this.
A regular conductor will eventually dissipate its energy through the voltage dropped across its resistance.
In switching converters, we choose DCR to be small compared to L*Fsw, so that converter efficiency is worthwhile.
--- Quote ---Like a transformer maintains power but the V,I of the primary is different that the V,I of the secondary but total power is conserved (ideally)?
--- End quote ---
An ideal transformer stores zero energy. A nonideal transformer is a transformer in parallel (and series) with inductance. The elements can be analyzed independently.
--- Quote ---The inductor charges up with whatever V, but it reaches a certain energy before cct is opened, and if a cap charged to a higher V is put in the inductor discharge path, that the energy discharged will still be equal to the energy stored,
--- End quote ---
It won't, actually. The inductor is in series with the DC source, which boosts its energy. You get some energy for free. Which is intuitive, as in the complete absence of switching, the output is still at least Vin.
A buck-boost configuration (in essence, the output cap is connected between IN and OUT, not GND and OUT) does satisfy this condition.
--- Quote ---Attached is what I think you were saying to me, but I'm not certain.
--- End quote ---
You've shorted the inductor in reverse with a diode, so it can never discharge into the capacitor.
--- Quote ---If it is what you were indicating, it looks like an additional buffer for the transient(?), but I'm still not certain on exactly why or how we get such transients if there's a route for the current to take that doesn't require such a high V. For example, when I simulate in LTspice, I often get large spikes at switch off events. This would mean that dI/dt were large. I'm not sure how you can have such a large change in current if the inductor opposes it? I feel like I'm going in circles lol
--- End quote ---
There are multiple inductances in a real circuit, though it's not apparent if your model contained them.
The dI/dt indeed can be quite high, but not in the main inductor.
Consider at the instant of commutation, you are asking the switch's current to drop to zero, and the diode's to step up from zero.
The inductance of those components, specifically the loop between them, experiences that dI/dt, and if it has inductance (in a physical circuit, it necessarily must -- this happens to be equivalent to saying the circuit has nonzero length and the speed of light is finite), that loop will generate a proportional voltage.
--- Quote ---I get confused and this is not as heavy on my mind but figured you could elucidate. I've read that current in an inductor wants to remain that current, even after energizing voltage had been removed,
--- End quote ---
When a voltage is "removed", it goes to zero. In other words, the inductor has been shorted.
--- Quote ---Also, attached is an example of the equation previously posted. How do they know that the current drops to 0 in the time they've decided to switch off? I see that the time off is 10ms, but what if it was 5ms or 100ms? How do they know it drops to 0 in 10ms? They integrated V/L(dt)?
--- End quote ---
Really poorly phrased prompt, the discharge voltage is undefined. They seem to be making the assumption that it is constant, in which case that will be true.
They specify only a switch in the circuit -- a highly nonlinear component. The impedance changes from finite(?) to infinite. The risetime of a switch can be fractional nanoseconds; the dI/dt will be extraordinary. Immediately, the switch breaks down, arcing.
In purely ideal terms, the switch goes to infinite impedance in zero time, the dI/dt is infinite, and the peak voltage is infinite for infinitesimal time. The voltage is a Dirac delta. This has infinite bandwidth ("DC to light" as they say), which is impossible.
The initial setup is questionable as well: if the source is a Thevenin equivalent, then we can charge the inductor from zero with enough source voltage and resistance that the current reaches the desired peak value, and switch it then; that's fine, but it's not quiescent. If we set V / R = Ipk, then the charging curve is exponential and it takes infinite time to reach Ipk. (The problem does not specify how long it's been charging for, but infinity is an acceptable duration in a theoretical question.) We could also craft a waveform which charges the inductor rapidly at first, then holds it at 0V to maintain the desired current. In both of these cases, the outcome does not depend on when we throw the switch, which seems appropriate. However, if the source is not a Thevenin equivalent, but a true current source, we run into the same problem, because if the inductor's initial current is at all mismatched to the source's current, a Dirac delta of flux will be developed as the currents equalize. Which again is impossible.
The usual textbook setup for this type of problem is to simply assume the inductor has a given initial current. No source is needed, we can simply invoke an initial condition. The switch is then shorting the inductor, holding its charge until opened.
A more physical circuit would represent the components as nonideal equivalents: Thevenin source; switch of finite resistance (on and off), capacitance and speed, and perhaps breakdown voltage as well; inductor with DCR and parallel impedances (core or radiation loss, capacitance, etc.). We can make meaningful predictions in this case: the realistic equivalent to a Dirac delta is a sharp spike with its parameters (rise time, peak voltage, resonant frequency, ringing, etc.) fully described by the RLC values in the circuit.
More generally, we are imposing limits on the bandwidth of the circuit. A Dirac delta has infinite bandwidth, but real circuits do not.
Incidentally, real signals have an exponential frequency cutoff at some point, which is why there is currently no such thing as a "DC to light" squarewave, for instance.
Tim
TheDood:
--- Quote from: T3sl4co1l on January 30, 2020, 03:29:17 am ---
--- Quote from: TheDood on January 30, 2020, 02:38:49 am ---I thought that the voltage across the inductor reversed polarity upon the event at which the energizing voltage was "instantaneously" removed? Im getting confused on how to clamp the voltage if it's reversing?
--- End quote ---
This phrasing suggests a circuit where the driving impedance rises suddenly. Like switching a relay coil. Where you don't care what happens when it turns off, you just want the current gone.
Of course we don't want unconstrained voltage, that damages transistors. So even in this simple case, we use a diode to clamp the voltage, which keeps the driving impedance low. The inductor voltage is always under control, always well defined.
In a switching converter, the inductor voltage is always well defined: either Vin when switched on, or Vin-Vout (for the boost case shown) when switched off (and discharging). There is a third state, if the inductor has finished discharging: voltage rings down to zero.
--- Quote ---For a cct with a 12V source, single inductor, and switch:
State 1:
Open cct
Energizing V = 12V
Voltage across inductor = 0
State 2:
Closed cct
Energizing V = 12V
V across inductor = 11.99 V
State 3:
Closed cct
Energizing V = 12V
V across inductor = 5V
--- End quote ---
Where's the voltage dropping? There's no resistor shown in the circuit.
--- Quote ---State 5:
Open cct
Energizing V = 12V
V across inductor = -12V
--- End quote ---
This is missing the output voltage. Is it 24V then?
--- Quote ---If the energizing V is 12V but I'm trying to push current onto a 13V cap, does discharge time drop in duration compared to the charge time? If I charged a coil for 1s at 12V and placed a 13V cap (cap charged to 13V) in its discharge path, the time of discharge is reduced from the time of charge up, correct? So that if 50% duty cycle were my max ON duration, and I was boosting V, that I shouldn't be worried about discharge time as it'll be less than the charge time?
--- End quote ---
Assuming you're still talking about the boost configuration, no, discharge time is inversely proportional to discharge voltage. For a 13V output, V_L(disch) = -1V. The equal flux condition is V_L(chg) * t_chg = V_L(disch) * t_disch. If flux is 12Vs then discharge is 1V and 12s.
If the switch turns on during the discharge, current begins at whatever it left off at. If the inductor is 12H, the peak current is 1A (12Vs / 12H = 1A; henries have units of flux per ampere). Say you turn on after 6 seconds: current starts at 0.5A, and an on-time of for example 0.5s is needed to raise it back to 1A. This is CCM.
--- Quote ---Also is my assumption that, the total energy stored in the mag field is being discharged completely or almost completely (depending on hysteresis) true?
--- End quote ---
If I_L = 0 at the end, yes.
--- Quote ---That if V = J/C, and I = C/s, that the energy will dissipate completely regardless the resistance/impedance in its path?
--- End quote ---
Energy only dissipates if given a path to dissipate through. Inductance is lossless. You can short an ideal inductor at 0V forever and it will hold its magnetic charge (current and flux and energy). Superconductors show this.
A regular conductor will eventually dissipate its energy through the voltage dropped across its resistance.
In switching converters, we choose DCR to be small compared to L*Fsw, so that converter efficiency is worthwhile.
--- Quote ---Like a transformer maintains power but the V,I of the primary is different that the V,I of the secondary but total power is conserved (ideally)?
--- End quote ---
An ideal transformer stores zero energy. A nonideal transformer is a transformer in parallel (and series) with inductance. The elements can be analyzed independently.
--- Quote ---The inductor charges up with whatever V, but it reaches a certain energy before cct is opened, and if a cap charged to a higher V is put in the inductor discharge path, that the energy discharged will still be equal to the energy stored,
--- End quote ---
It won't, actually. The inductor is in series with the DC source, which boosts its energy. You get some energy for free. Which is intuitive, as in the complete absence of switching, the output is still at least Vin.
A buck-boost configuration (in essence, the output cap is connected between IN and OUT, not GND and OUT) does satisfy this condition.
--- Quote ---Attached is what I think you were saying to me, but I'm not certain.
--- End quote ---
You've shorted the inductor in reverse with a diode, so it can never discharge into the capacitor.
--- Quote ---If it is what you were indicating, it looks like an additional buffer for the transient(?), but I'm still not certain on exactly why or how we get such transients if there's a route for the current to take that doesn't require such a high V. For example, when I simulate in LTspice, I often get large spikes at switch off events. This would mean that dI/dt were large. I'm not sure how you can have such a large change in current if the inductor opposes it? I feel like I'm going in circles lol
--- End quote ---
There are multiple inductances in a real circuit, though it's not apparent if your model contained them.
The dI/dt indeed can be quite high, but not in the main inductor.
Consider at the instant of commutation, you are asking the switch's current to drop to zero, and the diode's to step up from zero.
The inductance of those components, specifically the loop between them, experiences that dI/dt, and if it has inductance (in a physical circuit, it necessarily must -- this happens to be equivalent to saying the circuit has nonzero length and the speed of light is finite), that loop will generate a proportional voltage.
--- Quote ---I get confused and this is not as heavy on my mind but figured you could elucidate. I've read that current in an inductor wants to remain that current, even after energizing voltage had been removed,
--- End quote ---
When a voltage is "removed", it goes to zero. In other words, the inductor has been shorted.
--- Quote ---Also, attached is an example of the equation previously posted. How do they know that the current drops to 0 in the time they've decided to switch off? I see that the time off is 10ms, but what if it was 5ms or 100ms? How do they know it drops to 0 in 10ms? They integrated V/L(dt)?
--- End quote ---
Really poorly phrased prompt, the discharge voltage is undefined. They seem to be making the assumption that it is constant, in which case that will be true.
They specify only a switch in the circuit -- a highly nonlinear component. The impedance changes from finite(?) to infinite. The risetime of a switch can be fractional nanoseconds; the dI/dt will be extraordinary. Immediately, the switch breaks down, arcing.
In purely ideal terms, the switch goes to infinite impedance in zero time, the dI/dt is infinite, and the peak voltage is infinite for infinitesimal time. The voltage is a Dirac delta. This has infinite bandwidth ("DC to light" as they say), which is impossible.
The initial setup is questionable as well: if the source is a Thevenin equivalent, then we can charge the inductor from zero with enough source voltage and resistance that the current reaches the desired peak value, and switch it then; that's fine, but it's not quiescent. If we set V / R = Ipk, then the charging curve is exponential and it takes infinite time to reach Ipk. (The problem does not specify how long it's been charging for, but infinity is an acceptable duration in a theoretical question.) We could also craft a waveform which charges the inductor rapidly at first, then holds it at 0V to maintain the desired current. In both of these cases, the outcome does not depend on when we throw the switch, which seems appropriate. However, if the source is not a Thevenin equivalent, but a true current source, we run into the same problem, because if the inductor's initial current is at all mismatched to the source's current, a Dirac delta of flux will be developed as the currents equalize. Which again is impossible.
The usual textbook setup for this type of problem is to simply assume the inductor has a given initial current. No source is needed, we can simply invoke an initial condition. The switch is then shorting the inductor, holding its charge until opened.
A more physical circuit would represent the components as nonideal equivalents: Thevenin source; switch of finite resistance (on and off), capacitance and speed, and perhaps breakdown voltage as well; inductor with DCR and parallel impedances (core or radiation loss, capacitance, etc.). We can make meaningful predictions in this case: the realistic equivalent to a Dirac delta is a sharp spike with its parameters (rise time, peak voltage, resonant frequency, ringing, etc.) fully described by the RLC values in the circuit.
More generally, we are imposing limits on the bandwidth of the circuit. A Dirac delta has infinite bandwidth, but real circuits do not.
Incidentally, real signals have an exponential frequency cutoff at some point, which is why there is currently no such thing as a "DC to light" squarewave, for instance.
Tim
--- End quote ---
Oh man, thank you!
Yes I forgot an R. Oops!
I guess the fly back is not what you were indicating, I see what you're saying, at first I thought it would increase the cap V in parallel, but now I see it just shorts.
I think what you said about 13V output and 12V in, and the voltage across the inductor being -1V, is finally clicking, but I'm still not sure. In the State 1 ect scenario, at the end I've a -12V across inductor, and you quoted and said I'm missing the output V. The total cct V at that point would be 24V because 12Vin - -12V across inductor = 24V? Or why did you come up with 24V and were you saying that the Vout of inductor was 24V or entire cct? (Refer to attachment for better clarification, sorry its sloppy, but notice V's above components to see if I'm comprehending, 1 and 5 are open cct, 2-4 are closed cct). Also there'd only be -12V across inductor if the dI/dt satisfied that, correct?
As far as Dirac delta, and Thevenin, Ill have to look those terms up to grasp the last part of your post and I think Ill be picking at this for awhile. I think I caught a hint or 2, thanks.
I'm still not sure what you mean by clamping the V of the inductor. The Vin is fluctuating at 120Hz, so to clamp the V or keep the V constant I'd imagine you'd have to have another cap but before the inductor? Initially I thought that this type of config would produce poor PF, with only peak charging times, but after some thought maybe this is what you're referring too?
Ill work to quote and organize my questions better.
EDIT-
Imagine the R was before the switch in the 5 state drawing..
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