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v= (L) di/dt assistance
|
|
|
| I've asked inductor questions before and never fully understood the
answers.
First of all, I took calculus and learned derviatives as: X^3 d/dx = 3X^2.
The voltage across an inductor is given by: v= (L) di/dt
I'm unsure how to take the derivative of that formula because there is a di
in the numerator.
I understand it can be said as delta i / delta t. But this would be in the
case of t approaching 0 or a linear change in current; i.e. a linear ramp.
What if I had a poteniometer that was changing at an acceleration rate
instead of a linear rate? Then my i would not be linear, it would be
curved.
What if I wanted to know the current at exactly X seconds not using a delta
assumption?
Any calculus assistance will be appreciated.
p.s. I am aware this formula can be used in Diff EQ and a formula will show
the voltage at every point in time.
| |
| Salmon Egg 2008-02-28, 3:25 am |
| In article <Xns9A51DCDD0A45Bnobodynobodycom@216.196.97.136>,
Steve <nobody@nobody.com> wrote:
> I've asked inductor questions before and never fully understood the
> answers.
>
> First of all, I took calculus and learned derviatives as: X^3 d/dx = 3X^2.
Certainly a better way of writing this is: d/dX (X^3) = 3x^2.
>
> The voltage across an inductor is given by: v= (L) di/dt
>
>
> I'm unsure how to take the derivative of that formula because there is a di
> in the numerator.
>
> I understand it can be said as delta i / delta t. But this would be in the
> case of t approaching 0 or a linear change in current; i.e. a linear ramp.
>
>
> What if I had a poteniometer that was changing at an acceleration rate
> instead of a linear rate? Then my i would not be linear, it would be
> curved.
>
>
> What if I wanted to know the current at exactly X seconds not using a delta
> assumption?
>
>
> Any calculus assistance will be appreciated.
>
> p.s. I am aware this formula can be used in Diff EQ and a formula will show
> the voltage at every point in time.
Your misunderstanding is mostly of the physics. The mathematical aspect
is not so important. In uyour context, I presume, you need to have a
specific functional form for i. Not to be original, suppose i(t) = t^3
and L = 2 henries, what would v be as a function of time?
Bill
| |
| daestrom 2008-02-28, 5:26 pm |
|
"Steve" <nobody@nobody.com> wrote in message
news:Xns9A51DCDD0A45Bnobodynobodycom@216.196.97.136...
> I've asked inductor questions before and never fully understood the
> answers.
>
> First of all, I took calculus and learned derviatives as: X^3 d/dx =
> 3X^2.
>
> The voltage across an inductor is given by: v= (L) di/dt
>
>
> I'm unsure how to take the derivative of that formula because there is a
> di
> in the numerator.
>
Not sure why you want the derivative of it in the first place. That would
give you dv/dt on left and d^2i/dt (second derivative of i) on the right.
To understand the voltage/current relationship, suppose the applied voltage
is v = ksin(wt). Substitution would give you...
ksin(wt) = (L) di/dt
k/L sin(wt) = di/dt
Now, *integrate* both sides....
k/L ((1/w)-cos(wt)) = i
(ignoring the integration constant for now...)
Note how i is a negative cosine function while we started with v as a sin
function. This shows the current is 90 degrees out of phase with the
voltage (current lagging voltage).
If the applied voltage is something more complex than a simple sinusoid
function, the integration can get a bit more difficult, but the principle is
the same.
Or, if you want you can use a current source driving current through the
inductor. Let us say the current source is a sin function such that i =
ksin(wt). We find the derivative of i...
di/dt = d(ksin(wt) / dt = kw cos(wt)
Substituting in the formula for the inductor....
v=(L) di/dt = (L) kw cos(wt)
Again, the shift from sin to cosine shows the voltage out of phase by 90
degrees from the current (voltage leading current). And again, if the
injected current is something other than a sinusoid, the principle is the
same, but you may have more work to do to arrange so that i can be reduced
to di/dt.
Hope this helps...
daestrom
| |
|
| "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
news:47c73d94$0$6509$4c368faf@roadrunner.com:
>
> "Steve" <nobody@nobody.com> wrote in message
> news:Xns9A51DCDD0A45Bnobodynobodycom@216.196.97.136...
>
> Not sure why you want the derivative of it in the first place. That
> would give you dv/dt on left and d^2i/dt (second derivative of i) on
> the right.
>
> To understand the voltage/current relationship, suppose the applied
> voltage is v = ksin(wt). Substitution would give you...
>
> ksin(wt) = (L) di/dt
> k/L sin(wt) = di/dt
>
> Now, *integrate* both sides....
>
> k/L ((1/w)-cos(wt)) = i
> (ignoring the integration constant for now...)
>
> Note how i is a negative cosine function while we started with v as a
> sin function. This shows the current is 90 degrees out of phase with
> the voltage (current lagging voltage).
>
> If the applied voltage is something more complex than a simple
> sinusoid function, the integration can get a bit more difficult, but
> the principle is the same.
>
> Or, if you want you can use a current source driving current through
> the inductor. Let us say the current source is a sin function such
> that i = ksin(wt). We find the derivative of i...
>
> di/dt = d(ksin(wt) / dt = kw cos(wt)
>
> Substituting in the formula for the inductor....
>
> v=(L) di/dt = (L) kw cos(wt)
>
> Again, the shift from sin to cosine shows the voltage out of phase by
> 90 degrees from the current (voltage leading current). And again, if
> the injected current is something other than a sinusoid, the principle
> is the same, but you may have more work to do to arrange so that i can
> be reduced to di/dt.
>
> Hope this helps...
>
> daestrom
>
>
Question about your answer. When you took the integral of both sides,
does the i come out as a constant because d/dt disappeared.
That example is fine if I'm applying a sine wave, but what if I have a
DC source? Using deltas will be a good approximation, however, it's not
exact because the current is di/dt and not delta i / delta t.
I believe I'm confusing this too much. 
| |
| daestrom 2008-03-02, 5:25 pm |
|
"Steve" <nobody@nobody.com> wrote in message
news:Xns9A55686E761FEnobodynobodycom@216.196.97.136...
> "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
> news:47c73d94$0$6509$4c368faf@roadrunner.com:
>
>
> Question about your answer. When you took the integral of both sides,
> does the i come out as a constant because d/dt disappeared.
>
> That example is fine if I'm applying a sine wave, but what if I have a
> DC source? Using deltas will be a good approximation, however, it's not
> exact because the current is di/dt and not delta i / delta t.
>
> I believe I'm confusing this too much.
Your questions may require more math than I can muster. But you can only
use di/dt and its integrals on smooth continuous functions, not step
functions. So for DC with a step change on/off, I don't think you can use
smooth integrals.
Where....
k/L sin(wt) = di/dt
integrated to
k/L ((1/w)-cos(wt)) = i
No, 'i' is not a constant . The right hand side is the instantaneous
current, which happens to be a function of time (say, f(t) = i ). Reversing
things and taking the derivative...
k/L ((1/w)-cos(wt)) = f(t) = i
k/l sin(wt) = f'(t) = di/dt
The sine function works out nicely in part because the function is
continuous and we can ignore any transient parts by only looking at the
'steady-state'. If you replace the applied voltage with something like a
step change at time 0, where at all times less than zero the voltage and
current are zero and then voltage steps to some fixed positive value, then
you have....
v=0, i=0 (t<0)
v=k, di/dt = v/L (t>0)
Notice that there is no 'upper limit' to current, it just keeps growing
forever (at the rate di/dt = v/L). This is because there is no resistance
in this 'theoretical' circuit. Of course a real circuit has a resistance,
which means we apply Kirchoff's law to get....
0=v - Ri - Ldi/dt (t>0)
Solving this R-L circuit can be found in many calculus textbooks. If you
ignore the transient part, obviously i=v/R for the steady-state where di/dt
= 0 (kind of the definition of 'steady-state' in this case). But the
transient part can be 'interesting' and will show the current rise from zero
up to the steady-state is a function of e^(-R/Lt). Hence the 'time
constant' for such a circuit is TC=R/L
Does that help at all??
daestrom
| |
|
| "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
news:47cb26a1$0$6138$4c368faf@roadrunner.com:
>
> "Steve" <nobody@nobody.com> wrote in message
> news:Xns9A55686E761FEnobodynobodycom@216.196.97.136...
>
> Your questions may require more math than I can muster. But you can
> only use di/dt and its integrals on smooth continuous functions, not
> step functions. So for DC with a step change on/off, I don't think
> you can use smooth integrals.
>
> Where....
> k/L sin(wt) = di/dt
>
> integrated to
>
> k/L ((1/w)-cos(wt)) = i
>
> No, 'i' is not a constant . The right hand side is the instantaneous
> current, which happens to be a function of time (say, f(t) = i ).
> Reversing things and taking the derivative...
>
> k/L ((1/w)-cos(wt)) = f(t) = i
>
> k/l sin(wt) = f'(t) = di/dt
>
> The sine function works out nicely in part because the function is
> continuous and we can ignore any transient parts by only looking at
> the 'steady-state'. If you replace the applied voltage with something
> like a step change at time 0, where at all times less than zero the
> voltage and current are zero and then voltage steps to some fixed
> positive value, then you have....
>
> v=0, i=0 (t<0)
>
> v=k, di/dt = v/L (t>0)
>
> Notice that there is no 'upper limit' to current, it just keeps
> growing forever (at the rate di/dt = v/L). This is because there is
> no resistance in this 'theoretical' circuit. Of course a real circuit
> has a resistance, which means we apply Kirchoff's law to get....
>
> 0=v - Ri - Ldi/dt (t>0)
>
> Solving this R-L circuit can be found in many calculus textbooks. If
> you ignore the transient part, obviously i=v/R for the steady-state
> where di/dt = 0 (kind of the definition of 'steady-state' in this
> case). But the transient part can be 'interesting' and will show the
> current rise from zero up to the steady-state is a function of
> e^(-R/Lt). Hence the 'time constant' for such a circuit is TC=R/L
>
> Does that help at all??
>
> daestrom
>
>
>
Well yes, a switch would be consider linear. I'm thinking more of say a
poteniometer where I'm changing the rate of current from 1 amp / second,
2 amps / second, then 4 amps / second, then 8 amps / second.
Of course this would be difficult to achieve but in that example a
linear ramp would clearly not be accurate. So using delta i / delta t
would be incorrect.
Now using that example, the current would be changing on a curve.
I think I was pampered with d/dx (x^3) = 3x^2.
Now I'm having problems trying to solve (basic) electronic equations.
| |
| daestrom 2008-03-04, 8:25 pm |
|
"Steve" <nobody@nobody.com> wrote in message
news:Xns9A56CE84BA4B1nobodynobodycom@216.196.97.136...
> "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
> news:47cb26a1$0$6138$4c368faf@roadrunner.com:
>
<snip>
>
> Well yes, a switch would be consider linear. I'm thinking more of say a
> poteniometer where I'm changing the rate of current from 1 amp / second,
> 2 amps / second, then 4 amps / second, then 8 amps / second.
>
> Of course this would be difficult to achieve but in that example a
> linear ramp would clearly not be accurate. So using delta i / delta t
> would be incorrect.
>
> Now using that example, the current would be changing on a curve.
>
Principles are the same, just the math can get ugly.
Suppose you have....
di/dt = t^2
Integrating (assuming i = 0 at t=0)...
i = 1/3 t^3 + 0
Obviously....
v = L di/dt = L t^2
Instead, using your example....
t=0 di/dt = 1A/s
t=1 di/dt = 2 A/s
t=2 di/dt = 4 A/s
t=3 di/dt = 8 A/s
This is a 'doubling' function so it's exponential. This one I can see from
inspection....
di/dt = 2^(t)
So...
v = L (2^(t))
Integrating this one is a bit harder...
di/dt = 2^(t)
di/dt = e^(t*ln2)
i = int(e^(t*ln2)) dt
Now, I *think* that you can integrate this term by setting u=(t*ln2) and
du=ln2dt. Then knowing int(e^u du) = e^u
i=1/ln2*int(e^u du)
i=1/ln2*e^u
i= 1/ln2 *e^(t*ln2)
-or-
i = 1/ln2 * 2^(t)
But I could be mistaken here.
> Now I'm having problems trying to solve (basic) electronic equations.
Well, if it was *easy*, you wouldn't have to go to school to learn it now,
would ya? :-)
The 'trick' is if you can find a function that describes di/dt in the first
place. The more complicated that function is, the harder it is to integrate
it to find 'i'.
daestrom
| |
| Paul Hovnanian P.E. 2008-03-05, 9:25 pm |
| Steve wrote:
>
> I've asked inductor questions before and never fully understood the
> answers.
>
> First of all, I took calculus and learned derviatives as: X^3 d/dx = 3X^2.
>
> The voltage across an inductor is given by: v= (L) di/dt
>
> I'm unsure how to take the derivative of that formula because there is a di
> in the numerator.
You take the derivative of i with respect to t because i is actually
some function of t.
In other words, i changes as t changes (or di/dt = 0).
In order to make use of the above relationship, you have to know what
i(t) is or know what v(t) is and work backwards to find i(t)
--
Paul Hovnanian mailto:Paul@Hovnanian.com
------------------------------------------------------------------
"Grant me the strength to change what I can, the ability to accept
what I can't, and the incapacity to tell the difference."
-- Calvin (of Calvin and Hobbes)
| |
| Bob Penoyer 2008-03-06, 3:25 am |
| On Wed, 27 Feb 2008 20:42:23 -0600, Steve <nobody@nobody.com> wrote:
>I've asked inductor questions before and never fully understood the
>answers.
>
>First of all, I took calculus and learned derviatives as: X^3 d/dx = 3X^2.
>
>The voltage across an inductor is given by: v= (L) di/dt
>
>
>I'm unsure how to take the derivative of that formula because there is a di
>in the numerator.
>
>I understand it can be said as delta i / delta t. But this would be in the
>case of t approaching 0 or a linear change in current; i.e. a linear ramp.
>
>
>What if I had a poteniometer that was changing at an acceleration rate
>instead of a linear rate? Then my i would not be linear, it would be
>curved.
>
>
>What if I wanted to know the current at exactly X seconds not using a delta
>assumption?
>
>
>Any calculus assistance will be appreciated.
>
>p.s. I am aware this formula can be used in Diff EQ and a formula will show
>the voltage at every point in time.
First, instead of X^3 d/dx = 3X^2, the correct form is
d/dX X^3 = 3X^2. That is, you take the derivative with respect to X
'of' X^3. This dictates the order of the operator in the equation.
Second, the function v = L di/dt can simply be interpretted as saying
that the voltage across an inductor is proportional to the rate of
change of current through the inductor. (The constant of
proportionality is L.) So, if the current changes faster the voltage
developed across the inductor will be larger. If a current that drives
an inductor is suddenly switched off, for example, then di/dt will be
very large, so v will be very large. This accounts for the term
"inductive kickback."
Let's take the case you describe that causes the variation of the
current to accelerate. For example, let i(t) = K t^2 where K is a
constant.
Since v = L di/dt, then
v = L d(K t^2)/dt = 2 L K t
What does the result mean? It means that the voltage across the
inductor continues to increase linearly with time. The "scale factor"
is 2LK. So at t = 0, v = 0. At t = 1, v = 2LK. At t = 2, v = 4LK. And
so on.
| |
|
| "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
news:47cde496$0$22823$4c368faf@roadrunner.com:
>
> "Steve" <nobody@nobody.com> wrote in message
> news:Xns9A56CE84BA4B1nobodynobodycom@216.196.97.136...
> <snip>
>
> Principles are the same, just the math can get ugly.
>
> Suppose you have....
>
> di/dt = t^2
>
> Integrating (assuming i = 0 at t=0)...
>
> i = 1/3 t^3 + 0
>
> Obviously....
>
> v = L di/dt = L t^2
>
> Instead, using your example....
>
> t=0 di/dt = 1A/s
> t=1 di/dt = 2 A/s
> t=2 di/dt = 4 A/s
> t=3 di/dt = 8 A/s
>
> This is a 'doubling' function so it's exponential. This one I can see
> from inspection....
>
> di/dt = 2^(t)
>
> So...
>
> v = L (2^(t))
>
> Integrating this one is a bit harder...
>
> di/dt = 2^(t)
> di/dt = e^(t*ln2)
>
> i = int(e^(t*ln2)) dt
>
> Now, I *think* that you can integrate this term by setting u=(t*ln2)
> and du=ln2dt. Then knowing int(e^u du) = e^u
>
> i=1/ln2*int(e^u du)
> i=1/ln2*e^u
> i= 1/ln2 *e^(t*ln2)
> -or-
> i = 1/ln2 * 2^(t)
>
> But I could be mistaken here.
>
>
> Well, if it was *easy*, you wouldn't have to go to school to learn it
> now, would ya? :-)
>
> The 'trick' is if you can find a function that describes di/dt in the
> first place. The more complicated that function is, the harder it is
> to integrate it to find 'i'.
>
> daestrom
>
>
After reading your explainations and playing with some circuits in
MultiSim, I believe I have a better understanding.
I was trying to understand the formula (v = L di/dt) without understanding
what I was doing or inputting a function.
If I understand everything correctly:
* Any linear change such as a continous increase (or decrease) in current
over time will be linear and thus making the formula delta i / delta t.
* If I'm applying some function to a circuit, then I need to solve for it
such as what you did.
I did have some issues when simulating. I was unable to obtain a higher
voltage than my input. Adjusting the input frequency only caused the
voltage across the inductor to equal my input voltage but only for a
different time period (obviously equal to the inductor formula).
It would have been nice to generate 100 volts with 5 volts input. Most
online HV circuits I found online contained step-up transformers and any
inductor sites beats the inductor formula to death without explaining it.
Another problem I saw: I had a 1H in series with a 100 ohm resistor and
opened the circuit (via a switch) and the voltage across the inductor was a
triangle wave (above and below ground) that appeared to never hault.
| |
| daestrom 2008-03-13, 8:25 pm |
|
"Steve" <nobody@nobody.com> wrote in message
news:Xns9A6040C77061nobodynobodycom@216.196.97.136...
> "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
> news:47cde496$0$22823$4c368faf@roadrunner.com:
>
<snip for brevity>[color=darkred]
> After reading your explainations and playing with some circuits in
> MultiSim, I believe I have a better understanding.
>
Glad I could help...
> I was trying to understand the formula (v = L di/dt) without understanding
> what I was doing or inputting a function.
>
> If I understand everything correctly:
>
> * Any linear change such as a continous increase (or decrease) in current
> over time will be linear and thus making the formula delta i / delta t.
>
> * If I'm applying some function to a circuit, then I need to solve for it
> such as what you did.
>
> I did have some issues when simulating. I was unable to obtain a higher
> voltage than my input. Adjusting the input frequency only caused the
> voltage across the inductor to equal my input voltage but only for a
> different time period (obviously equal to the inductor formula).
This may be a limit of multisim (never used it), or the way you set it up,
don't know.
>
> It would have been nice to generate 100 volts with 5 volts input. Most
> online HV circuits I found online contained step-up transformers and any
> inductor sites beats the inductor formula to death without explaining it.
>
Well 20:1 is asking a bit much, but it certainly is possible to use an
inductor to 'pump' to a higher voltage. One simple circuit is to arrange
the DC supply to feed through the inductor and return via a FET transistor.
Between the inductor and FET, tap off with a diode to feed a capacitor. Now
arrange a 'suitable oscillator' to turn on the FET so the inductor basically
shorts the DC supply. When current builds up, suddenly switch off the FET.
The inductor magnetic field will collapse and create a voltage spike that
will momentarily forward bias the diode and push some charge into the
capacitor. Once inductor current drops near zero, turn on the FET and start
over again. This uses the strong magnetic field built up in the inductor
when it's 'shorting' the power supply to 'pump' charge the capacitor. But
as with all things, there are limits to how much load you can draw off the
high-voltage capacitor because if you draw more charge off than the 'pump'
circuit can put in, voltage on the capacitor drops.
(I'm sure someone here could devise a better circuit or tell me where I may
have made a mistake with this, but the principle is used in a lot of DC-DC
converters that are used for step-up)
> Another problem I saw: I had a 1H in series with a 100 ohm resistor and
> opened the circuit (via a switch) and the voltage across the inductor was
> a
> triangle wave (above and below ground) that appeared to never hault.
>
That certainly is a multisim issue. What should happen is you get a very
high voltage 'spike', but it has a very short duration. A 'real' switch
would end up arcing momentarily from the high voltage spike and the energy
in the inductor would be dissipated in the arc.
daestrom
| |
| Don Kelly 2008-03-14, 3:25 am |
| ----------------------------
"Steve" <nobody@nobody.com> wrote in message
news:Xns9A6040C77061nobodynobodycom@216.196.97.136...
> "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
> news:47cde496$0$22823$4c368faf@roadrunner.com:
>
>
> After reading your explainations and playing with some circuits in
> MultiSim, I believe I have a better understanding.
>
> I was trying to understand the formula (v = L di/dt) without understanding
> what I was doing or inputting a function.
>
> If I understand everything correctly:
>
> * Any linear change such as a continous increase (or decrease) in current
> over time will be linear and thus making the formula delta i / delta t.
>
> * If I'm applying some function to a circuit, then I need to solve for it
> such as what you did.
>
> I did have some issues when simulating. I was unable to obtain a higher
> voltage than my input. Adjusting the input frequency only caused the
> voltage across the inductor to equal my input voltage but only for a
> different time period (obviously equal to the inductor formula).
>
> It would have been nice to generate 100 volts with 5 volts input. Most
> online HV circuits I found online contained step-up transformers and any
> inductor sites beats the inductor formula to death without explaining it.
>
> Another problem I saw: I had a 1H in series with a 100 ohm resistor and
> opened the circuit (via a switch) and the voltage across the inductor was
> a
> triangle wave (above and below ground) that appeared to never hault.
>
---------
Part of the problem is just what circuit you are trying to simulate and the
source involved (voltage or current) and it may be that the simulation
software simply doesn't like the initial conditions given or that the model
is not realistic (e.g. switches in series with an ideal current source would
be a serious problem- a non-ideal source can be converted to a voltage
source behind the switch). The circuits involved don't have the complexity
to need a simulator so you should be able to get a good handle on it without
one.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
| |
|
| "Don Kelly" <dhky@shaw.ca> wrote in
news:EAmCj.86204$w94.33585@pd7urf2no:
> ----------------------------
> "Steve" <nobody@nobody.com> wrote in message
> news:Xns9A6040C77061nobodynobodycom@216.196.97.136...
> ---------
> Part of the problem is just what circuit you are trying to simulate
> and the source involved (voltage or current) and it may be that the
> simulation software simply doesn't like the initial conditions given
> or that the model is not realistic (e.g. switches in series with an
> ideal current source would be a serious problem- a non-ideal source
> can be converted to a voltage source behind the switch). The circuits
> involved don't have the complexity to need a simulator so you should
> be able to get a good handle on it without one.
MultiSim allows you to set the initial conditions. Typically I leave it
at the default (let the simulator determine initial conditons), however,
this time I have it set to 'zero'.
I have a 1H inductor directly to ground; the 1H makes the math easier.
The switch allows for a linear ramp instead of the FET; it was causing
non-linearity most likely because of the internal capacitance.
I do, however, get double the voltage across the inductor than I
calculate. As an example, my switch takes 9.9us to open (as measured in
the program) and it's telling me the current is 7.4mA. That should be
[7.4mA / 9.9us] * 1H = 747v. but my measured voltage is 1492v. I'm not
sure if this is a fault of the program since I have an inductor directly
to ground. The circuit turns into a Diff EQ if I add a resistor and
makes the math tricky to deal with while "learning" the basics.
| |
| Don Kelly 2008-03-18, 3:25 am |
| ----------------------------
"Steve" <nobody@nobody.com> wrote in message
news:Xns9A63F14FBE8Anobodynobodycom@216.196.97.136...
> "Don Kelly" <dhky@shaw.ca> wrote in
> news:EAmCj.86204$w94.33585@pd7urf2no:
>
>
>
>
> MultiSim allows you to set the initial conditions. Typically I leave it
> at the default (let the simulator determine initial conditons), however,
> this time I have it set to 'zero'.
>
> I have a 1H inductor directly to ground; the 1H makes the math easier.
> The switch allows for a linear ramp instead of the FET; it was causing
> non-linearity most likely because of the internal capacitance.
>
> I do, however, get double the voltage across the inductor than I
> calculate. As an example, my switch takes 9.9us to open (as measured in
> the program) and it's telling me the current is 7.4mA. That should be
> [7.4mA / 9.9us] * 1H = 747v. but my measured voltage is 1492v. I'm not
> sure if this is a fault of the program since I have an inductor directly
> to ground. The circuit turns into a Diff EQ if I add a resistor and
> makes the math tricky to deal with while "learning" the basics.
>
I have no idea of what multisim is doing but it appears that you are
calculating an average voltage over the switch opening. I do have a problem
with the "measured" value -if it is the multisim result, it isn't actually
measured -however, it does agree with the maximum value assuming such a ramp
down.
I don't know enough about multisim (having never used it) to say what it is
actually doing. Is it trying to represent an ideal current source with a
switch in series? If so, problems occur. If you are representing an ideal
current source that ramps down from 7.4 to 0 in 9.9 microseconds- that is
another animal and your value of 750V is good. Otherwise???
Again, exactly what circuit is it that you are trying to represent? A great
deal of the problem appears to be that you are trying to model an
unrealistic system. Resistance always exists so there is no such thing as a
pure inductance.
I would suggest that for true learning purposes, you ignore multisim which
is a tool for those that have a firm hand on basics. It is not a substitute
for the work involved in learning the basics (as you have found) , including
the dread DEQ (1st and second order are sufficient at this stage and the
solutions of such are very straightforward and simple - by the book .
Don Kelly dhky@shawcross.ca
remove the X to answer
| |
| D. Ismay 2008-03-22, 1:25 pm |
| yes, i know this is late, nevertheless...
Steve wrote:
[...]
>
> I was trying to understand the formula (v = L di/dt) without understanding
> what I was doing or inputting a function.
>
> If I understand everything correctly:
>
> * Any linear change such as a continous increase (or decrease) in current
> over time will be linear and thus making the formula delta i / delta t.
that is accurate, and also true for non-linear change in current. the
issue is, you cannot develop a voltage across a pure inductance without
a change in current through the inductance.
<snip>
> I did have some issues when simulating. I was unable to obtain a higher
> voltage than my input. Adjusting the input frequency only caused the
> voltage across the inductor to equal my input voltage but only for a
> different time period (obviously equal to the inductor formula).
use a current source, not a voltage source.
<snip>
> Another problem I saw: I had a 1H in series with a 100 ohm resistor and
> opened the circuit (via a switch) and the voltage across the inductor was a
> triangle wave (above and below ground) that appeared to never hault.
the time constant for L-R circuits is, naturally, L/R. your description
suggests that you simulated an unrealistic condition -- you open
switch, current in your inductor must drop to zero, but if you don't add
another conductive path somewhere the delta-i has no place to "go".
your simulator then does whatever it can do, and the result isn't
necessarily realistic.
hth, hand.
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