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It's 1 mH
So...you want to limit the current to some value (Isat) while allowing 5v across the coil?
Generally, a buck converter does this by varing the duty cycle while maintaining a constant frequency.
Maybe a current limiter is more appropriate.
Reactance of pulsed DC
- Thread starter eromlignod1
- Start date
Generally, a buck converter does this by varing the duty cycle while maintaining a constant frequency.
Maybe a current limiter is more appropriate.
What I want is to select an inductor with enough impedance that I don't need current/voltage control at all. Again, a formula for reactance under a DC square wave would let me easily do this.
Don
Don
In which case just select one with enough resistance.
With a DC drive, the overall current depends only on the resistance. (See post #7)
(by the way, it will get hot)
The problem with that is the current will not necessarily get to the DC current level in each pulse. At high enough frequency it will get nowhere near it.
To the TS: You have stated pulsed DC. What is the state of the driver when it is off? Is it high impedance (not connected) or is it connected to ground. If I know that, I will create a simulation for you. 15 minutes with LTSPICE will give you what you need to know.
Bob
To the TS: You have stated pulsed DC. What is the state of the driver when it is off? Is it high impedance (not connected) or is it connected to ground. If I know that, I will create a simulation for you. 15 minutes with LTSPICE will give you what you need to know.
Bob
At high frequency, it’s more likely not less likely to be in continuous current mode. As soon as it is In continuous current mode, the average current depends only on the resistance.
Each coil is dedicated to a specific frequency. By selecting a coil with the proper impedance I should be able to get whatever current I want.
The coil will be fired with a sinking MOSFET. There is also a snubber diode to freewheel the inductive kick.
Don
The coil will be fired with a sinking MOSFET. There is also a snubber diode to freewheel the inductive kick.
Don
What are you saying? The reactance doesn't affect the current through the coil?
If I select by resistance I'd be assuming that the frequency had no effect on the magnetic pull. This is untrue.
Don
If I select by resistance I'd be assuming that the frequency had no effect on the magnetic pull. This is untrue.
Don
Audioguru again
- Joined Oct 21, 2019
- 6,826
The first post says the frequency of the pulses is only a 60Hz squarewave then the current in the 10mH inductor (or maybe it is 1mH) is almost DC for half the time.
MaxHeadRoom
- Joined Jul 18, 2013
- 30,665
Reactance does not come into play with a inductance fed with DC or DC pulse.
Reactance occurs with the application of AC in a circuit containing capacitance or inductance
Exactly, if the frequency is high enough for the current to fail to reach zero when the transistor is off, then the current depends ONLY on the resistance.
If you split the driving waveform into its Fourier components, there will be a large component at f=0, namely its average DC value.
Z=√(R^2+(2πfL)^2),
where f=0, then it’s just R.
There will be some high frequency ripple superimposed on the average DC value due to the higher frequency components of the driving waveform, but the DC value will be the most important because I depends on 1/f for the higher frequencies.
Then why does the current display on my power supply decrease significantly as I increase the frequency?
Don
Don
But it soon gets changed (by post#4) to “a few kHz” and they are “relay-like” coils at the beginning.
Maybe there is something your are not telling us about the waveform.
maybe the rise and fall times are slow
maybe there is some ac coupling
maybe the duty cycle reduces due to delays.
because otherwise, the current should fall as the frequency is reduced (As it goes into discontinuous current mode)
I did an informal test last night. I had a 106 (10 mH) coil connected to +5Vdc. I measured the DC resistance of the coil to be 14 ohms. The other terminal I connected to the drain of an N-type power MOSFET. The source was connected to ground. I connected a snubber diode across the inductor. I drove the gate with a function generator set to produce a square wave (0 to 5V). I connected a scope to the drain connection. I started at 4000 Hz.
The scope displayed a clean, crisp square wave. The current display on the power supply said "73 mA". As I decreased the frequency, the the current pull increased. By the time I got down to 50 Hz, the current had increased to over double. At that point the square wave was still clean and the current was steady (not running away).
Don
The scope displayed a clean, crisp square wave. The current display on the power supply said "73 mA". As I decreased the frequency, the the current pull increased. By the time I got down to 50 Hz, the current had increased to over double. At that point the square wave was still clean and the current was steady (not running away).
Don
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Could it be due to saturation? At lower frequencies will it spend more time saturated.
I think the reactance play a part in the coil, but that it is actually a combined effect with the coil resistance.
To keep things simpler.
Impedance = reactance + coil resistance = \[ \sqrt{(2 \pi f L)^2 + R^2} \] like mentioned prior https://forum.allaboutcircuits.com/threads/reactance-of-pulsed-dc.187688/post-1743620 Ian0.
That square root is used as that reactance is out of phase from R, in a perfect case when current is max, voltage is zero, and when voltage is max current is zero. This would be true say for a sine wave AC.
square waves can be broken down into its sine waves components, e.g. fourier series
https://mathworld.wolfram.com/FourierSeriesSquareWave.html
hence for simplicity sake, we'd assume that the primary frequency is dominant and consider that as a primary frequency as a simplification.
(it isn't correct, but it is a simplification that approximates it)
Then that equation above would be true. i.e. if you increase frequency the reactance in the coil increase and as a result total impedance increase, and less current flows in the coil.
to be 'fussy' and to consider the complete maths would possibly involve considering differential equations such as
\[ V_{in} = L\frac{di}{dt} + V_R \]
solving that differential equation would probably derive the exact solution, but is likely difficult (and your Vin is actually the square wave, making exact solutions even more difficult to find, possibly it may have solutions only for initial value problems).
some substitutes are to consider I or V as some function of \[ e^{j\omega t} \] and work that as solutions
which is actually the sine AC signals with complex components.
I've been trying to work that out for a buck converter but stumbled on the math.
https://forum.allaboutcircuits.com/...ing-simulation-differential-equations.187476/
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Maybe it's not that tough. All periodic waveforms are made up of a sum of sine and cosine functions. The voltage wave is just an amplitude V multiplied by a unity wave function. When the V = L(di/dt) is integrated for a sine wave, we integrate
di = V/L cos(wt) dt
so
i(t) = V/wL sin(wt)
the current becomes V/wL (the amplitude of the current) multiplied by another unity wave function. Since all the components of a periodic wave are sine or cosine, all integrate as another sine or -cosine with the w coming out as 1/w and you still get V/wL and a unity wave function. So the reactance is always wL.
For example, a unity square wave is just
square(t) = 4/pi * [sin(wt) + 1/3*sin(3wt) + 1/5*sin(5wt) + ...]
This produces a square wave from -1 to +1, and the voltage formula would be just
v(t) = V * square(t).
Every term of the unity function would integrate to 1/w multiplied by a sine or -cosine and the w would come out and combine with the V/L as V/wL from the distributive property.
Actually, if all periodic waves are explained by a Fourier series, then even a DC-offset wave should have a Fourier series. Perhaps X=wt for any periodic wave.
Don
di = V/L cos(wt) dt
so
i(t) = V/wL sin(wt)
the current becomes V/wL (the amplitude of the current) multiplied by another unity wave function. Since all the components of a periodic wave are sine or cosine, all integrate as another sine or -cosine with the w coming out as 1/w and you still get V/wL and a unity wave function. So the reactance is always wL.
For example, a unity square wave is just
square(t) = 4/pi * [sin(wt) + 1/3*sin(3wt) + 1/5*sin(5wt) + ...]
This produces a square wave from -1 to +1, and the voltage formula would be just
v(t) = V * square(t).
Every term of the unity function would integrate to 1/w multiplied by a sine or -cosine and the w would come out and combine with the V/L as V/wL from the distributive property.
Actually, if all periodic waves are explained by a Fourier series, then even a DC-offset wave should have a Fourier series. Perhaps X=wt for any periodic wave.
Don
Thread seems to be going in circles. TS keeps posting math and reasoning yet is largely ignoring application notes by users asking for more information and how to apply the data. He seems convinced of his methods.
At this point, I would really like to see some examples otherwise this is another 'my word vs yours' thread.
