Unfortunately, transformers can deliver only AC signals since the core flux must be reset each half cycle.
If you let it relax to remanence (or more likely, a little bit below that, because if the node is left "floating", circuit capacitance will hold up the voltage a little while longer), then you're only using half the flux capacity of the core.
If that's enough, there's no problem.
Tim
The important part is you must wait until enough negative flux has been delivered, before starting another cycle. This causes the positive duty cycle to be limited by the negative peak voltage limit.
Failure to do so will cause the flux to ratchet upwards, until the core saturates, and you'll run out of drive voltage (and draw short-circuit current from the driver) at an inconvenient moment.
Would'nt the regular cores be optimized for as low remanance as possible, I mean I don't know the exact values but would the Brem be like under 10% of Bsat... and I am still trying understand your point...
what you are trying to say is that the entire range of flux in the core is from -Bsat to +Bsat so if I'm using from 0 to +Bsat I'm using only 50% of the flux capacity of the core and lower than 50% if lets say Brem is 10% then I would actually using 40% of the total flux capacity, is this right?
On another note if a core has a saturating flux density Bmax of 1T does it mean it can go from +0.5T to -0.5T on the BH curve or is it actually +1T to -1T?, because if its the latter then I would be using the fully flux density of the core from 0 to 1T right?
I fail to understand why the flux would continue to build up towards saturation when the current is cycling from 0 to max should'nt the flux follow the current every cycle and also move between 0 to +Bmax or +Brem to +Bmax ?
Not current -- voltage. You're driving it with a constant voltage driver (usually).
If the source impedance is changing (i.e., a switch that turns off, and no diode or switch turns on to 'catch' it), then current will decay to zero, and flux with it (well, towards remanence of course).
For this reason, even with a balanced signal source (consider TL598 for instance), it's a good idea to include a coupling capacitor in series with the gate drive transformer; or for a single-ended case, to use a switch that turns off between cycles, so the "reset" flux can take care of itself.
Tim
The induction voltage is equal to number of turns time time derivative of the flux. So the flux is proportional to integrated voltage. As the flux is limited, the long time average of the voltage has to be zero, or there will be a DC current superimposed, that reduces the useful flux even more.
A free wheeling diode slow's down the speed of field decay. A suitable resistor (or zener diode) in series to the diode can speed up the decay and still limit the voltage. With 50% PWM ratio and a free wheeling diode the current will usually not decay all the way to zero, unless the coil resistance is rather high. The negative voltage needs to be on average as high as the the positive when driving the FET on - so negative peak voltage should be even higher (e.g. 1.5 times).
It is Faradays induction law, or one of the Maxwell equations that gives the voltage for one turn as the derivative of the flux through that turn.
Looking at transformers it is the area under the voltage curve that sets the flux. This is coming just from integrating the law of induction. There is no need to calculated the current and magnetic curve of the core. It is only winding resistance that adds a little voltage - so not something for an near ideal transformer.
The way an inductor / transformer works it that the induction voltage compensated most of the external voltage and only a small fraction divided by coil resistance drives the magnetizing current. In the ideal transformer you neglect the magnetizing current and the voltage drop it causes.
The shown curve is an ideal case, ignoring coil and switch / diode resistance. In real world (and no superconductors), the flux would slowly decrease. Depending on the coil this could be in the ms to 10 µs time scale - so for the shown example one might hardly see the flux creeping down. With 50% PWM ration the flux would not at all reach zero before the next puls comes.
A simple experiment would be using just inductor, resistor and free-weeling diode and look at the current when sending pulse groups (e.g. 10 pulses with 5 or 50% PWM, and than a longer break). Once the current does not go to zero the current will add up over the pulses.
It might be even enough to look at this in a simulation.
Or with more analytical correctness:
The plot of V and Phi is true, when V == EMF. That is, the electromotive force (the part that comes from Faraday's Law) in the core itself.
What you measure at the pins is "dirtied" by stuff like DC resistance and stray/leakage inductance.
If the core acts more or less like a regular inductor (current proportional to flux), then nonzero flux will also have nonzero current, and that current will drop voltage across DCR. Thus, over time, flux decays to zero, following the L/R time constant of the source resistance + coil DCR, and winding inductance.
Regardless of resistance, this much is true: to make flux move quickly, you must apply a large voltage. Doesn't matter whether it's a positive or negative voltage (with respect to the two terminals of the winding), for an increase or decrease in flux.
Tim
Thanks again Tim!
I'm still trying to understand what you are trying to say and while searching on the topic on the web I found this image
It show's that when the voltage goes to zero the flux tends to maintain to constant value, i thought this was wrong wouldn't the flux start going to zero when the voltage drops to 0...
* Saturable Core Model, copied from:
* _SPICE Models For Power Electronics_, Meares and Hymowitz.
*
.SUBCKT INDSAT 1 2 PARAMS: VSEC=1e-4 LMAG=1e-5 LSAT=1e-7 FEDDY=1e6
F1 1 2 VM1 1.0
G2 2 3 1 2 1.0
E1 4 2 3 2 1.0
VM1 4 5 0.0
RX 3 2 1E9
CB 3 2 {VSEC/500} IC=0
RB 5 2 {LMAG*500/VSEC}
RS 5 6 {LSAT*500/VSEC}
VP 7 2 250
VN 2 8 250
D1 6 7 DCLAMP
D2 8 6 DCLAMP
.MODEL DCLAMP D ( CJO={3*VSEC/(6.28*FEDDY*500*LMAG)} VJ=25 RS={LSAT/VSEC} )
.ENDS
I use this model:Code: [Select]* Saturable Core Model, copied from:
* _SPICE Models For Power Electronics_, Meares and Hymowitz.
*
.SUBCKT INDSAT 1 2 PARAMS: VSEC=1e-4 LMAG=1e-5 LSAT=1e-7 FEDDY=1e6
F1 1 2 VM1 1.0
G2 2 3 1 2 1.0
E1 4 2 3 2 1.0
VM1 4 5 0.0
RX 3 2 1E9
CB 3 2 {VSEC/500} IC=0
RB 5 2 {LMAG*500/VSEC}
RS 5 6 {LSAT*500/VSEC}
VP 7 2 250
VN 2 8 250
D1 6 7 DCLAMP
D2 8 6 DCLAMP
.MODEL DCLAMP D ( CJO={3*VSEC/(6.28*FEDDY*500*LMAG)} VJ=25 RS={LSAT/VSEC} )
.ENDS
You'll need to paste this into a subcircuit file, and load that file as a model or library. RTFM
It works by transforming terminal voltage into current in a capacitor; as the capacitor charges, flux increases. A diode clamps the flux, providing an exponential cutoff: flux cannot practically increase beyond this point. A second capacitor (the diode capacitance) even provides a lowpass filter characteristic, which has the effect of reducing flux at high frequencies: the permeability of the core drops. (The core also becomes lossy, i.e., the permeability has an imaginary component. Inductive reactance is already imaginary, so imaginary * imaginary = resistive loss! ) A voltage source measures the current flow into the clamp diode section, relaying this back to the terminals; thus terminal current is controlled by flux, and attempting to push excessive flux into the core will draw a huge current (just as should be the case!).
Which explains what the variables control:
VSEC: saturation flux in volt-seconds
LMAG: magnetizing inductance in henry (default, un-saturated inductance)
LSAT: saturated inductance in henry (this should be mu_r times smaller than LMAG)
FEDDY: cutoff frequency where permeability drops (as if due to eddy currents in a conductive core).
This is equivalent to a ferrite core with cross-sectional area A_e, turns N, permeability mu_r, inductivity A_L, and:
VSEC = A_e * N
LMAG = A_L * N^2
LSAT = A_L * N^2 / mu_r
A typical #77 ferrite might be mu_r = 2000 and FEDDY = 1e6 or so. High mu (up to 10-15k) ferrites have lower FEDDY (down to about 0.3e6), and lower loss, lower permeability types have higher FEDDY. Laminated iron will have FEDDY inversely proportional with lamination thickness, with several kHz being typical. (More complicated core loss curves might simply be modeled with an RLC network in parallel with the core, rather than at the flux level. For instance, the diffusion characteristic that nanocrystalline steel has.)
Tim
Stepper motors are usually driving the coils in a + 0 - 0 + sequence. So the flux will reverse.
For unipolar drive, there are usually two coupled coils like a center taped transformer. With unipolar drive it is a bad idea to have direct freewheeling diodes, as there will be a kind of transformer function from the other half coil.
The same is true for many latching relays with 2 separate coils: with free-wheeling diodes more current / a longer pulse is needed. So at least add a series resistor to the diode.