Thread: Mosfet Heating Circuits View Single Post

05-04-2010, 06:43 AM
 Harvey Gold Member Join Date: Jul 2009 Posts: 1,137
Back To Basics - Part 3 : The Other Half Of The Transaction

Hi Guys,

I have the utmost confidence that those following my last two post regarding questions for Test #13 will have realized that so far I have only covered half of the AC transaction - the positive half of the cycle. So far we have discussed the dissipated power in the load resistor that we can attribute to battery power being delivered to it. However, we all know that there is more to the story. We have 9.14 volt-amps of apparent power involved that we really haven't given a fair hearing.

We know, that after a period of time that the MOSFET is in the ON condition, that conventional DC current flows through the load resistor and if allowed to continue long enough, that resistor will charge magnetically in the inductive winding it possesses. This means that it becomes a DC solenoid and a magnetic field surrounds it with a B vector running straight through its center lengthwise.

How long does it take to build the field and how much energy is stored in the field?

There is a relationship between the electrical current and the building of the magnetic field. At time zero (0t) when we first turn on the MOSFET, 100% of the current is not yet flowing. At time one (1t), 63.2% will be flowing, leaving 36.8% not yet flowing. At 2t, an added 63.2% of the remaining 36.8% will be flowing leaving 13.54% of the overall current not yet flowing. At 3t, an added 63.2% of the remaining 13.54% will be flowing, leaving 4.98% not yet flowing. At 4t an added 63.2% of the remaining 4.98% will be flowing, leaving 1.83% not yet flowing. And finally at 5t an added 63.2% of the remaining 1.83% will be flowing, leaving 0.675% not yet flowing. So, after 5t we find that our current is flowing at 99.33% of its total value - therefore we generally consider an inductor to be fully charged after 5 time constants. You can go to infinity and it will never reach 100% because each addition of current flow is always 63.2% of the remaining amount. The formula for the forgoing is It = E/R (1 - e ^ (tR / L) where e is the natural log constant 2.71828. This resolves down to each time constant, t = L / R seconds. This is good for us, because we know the R = 9.73 ohms from Glen's Baseline profile test and we know the L in Henries from Back To Basics Part 2 as 0.00002277H. Therefore, we can easily calculate how long it will take for our inductor to reach maximum current flow.

Time in seconds to full charge = 5 * (L / R) = 5t
5t = 5 * 0.00002277 / 9.73
5t = 2.34µs

So that answers our first question - Glen's resistor is fully charged after 2.34µs from the MOSFET coming into full conduction. I looked up the stats on the IRFPG50 just for grins here and the turn on delay for this device is 19ns with a 35ns rise time under the test conditions of max current and Vdd = 500V. But we can probably safely say that from the time the gate goes high, the inductor is fully charged 2.394µs later.

Now for the second question as to how much energy is stored in that inductor if we fully charge it.

Energy Stored in an Inductor

Energy Stored in an Inductor

From this we learn that the energy stored is 1/2 LI². So in Glen's case it is 0.5 * 0.00002277 * 0.387 * 0.387 = 0.000001705120065 J or 1.7µJ.

So if we are operating at 426kHz, how long do we have DC current building the field?
That one is easy. Our full cycle time is the reciprocal of our frequency, therefore a complete cycle is 1/426000 seconds long, or 2.34µs long and our ON period is 57.32% of that or 1.35µs. In test #13, the inductor does not fully charge. If a full charge is 2.34µs, then 1t is 0.468µs. 1.35 / 0.468 = 2.9t or 94% Charged. We might add here that this means our stored energy is only 1.6µJ then.

So this brings us to the other half of the transaction, the half where we turn off the MOSFET. And this is also where things change and must be analyzed differently. At this point we have 1.6µJ stored in the inductor as a magnetic field and we take the gate terminal low. The IRFP50 has a turn off delay of 130ns and a fall time of 36ns. So we can expect at 166ns delay from the time we signal it to turn off to the time the inductor begins to collapse it's field. This can be seen in the screen shots and the actual delay can vary depending on residual gate charge that is not scavenged away by the 555. We find that typically, the field is fully collapsed about 234ns later. So it charges in 1350ns and discharges in 234ns. At this point, the field is fully collapsed and has been converted to a voltage potential commonly referred to as a BEMF spike, at the drain of the MOSFET. There is a difference between BEMF and CEMF. BEMF is a voltage which when allowed to act on a circuit will cause a back flow of current after the magnetic field is fully collapsed. CEMF on the other hand, is a voltage which acts against, or counter, to an incoming voltage of opposite polarity in an AC system and causes a clash of currents. Simply put, the BEMF occurs when a switch stops current flow, CEMF occurs when the power supply polarity is reversed. So now, we are at the time where our BEMF spike is at it's maximum potential. The MOSFET is OFF. The voltage can only flow through one path, it must turn around and flow back into the load resistor. This point is very important: The BEMF seeks to distribute itself evenly throughout all the conductors connected to it. In this case, the conductors become charged evenly (given the time to fully distribute).

So how much voltage would we expect in the BEMF pulse?
Interestingly, this depends on how fast the field is allowed to collapse and how much energy is stored in the field. We can use this formula: E = -NdΦB / dt where N is the number of turns (48) and ΦB is the flux density in Webers and E is the electromotive force. Now, the thing to keep in mind here is that the electromotive force is across the coil, not across the battery. But how do we determine what our Webers are? One simple way is to relate it to the Joules we have already determined. A Weber is the same as Joules divided by Amps. Now doing it this way can introduce a compound error if we have made any mistakes in determining the Joules and good practice would demand that we use another approach so that we can compare the results. Perhaps someone would like to double check me on this by using another means like m^2∙kg∙s^-2∙A^-1. For now, let’s use the easy way. So we have 1.7µJ and 387mA: This gives us 0.000001705120065 / 0.387 = 0.000004405995 Webers, or 4.405995 µW. So now we can get back to our equation: - N ∙ dΦB / dt where N = 48 Turns, dΦB = 0.000004405995 Webers to Zero (that is our change in flux) and dt = 0.000000234 Seconds (that is our change in time). So what is our result? -903.79 Volts.

Now to explain the negative in that equation: Remember my comment above that I said was important to remember? The -903.79V is across the coil and tells us something according to Lenz’s Law and that is why there is a negative on there. When the field collapses, the voltage is negative with reference to the voltage that created the field. Recall that we had a positive voltage of 23V across the coil when we created the field, and now, leaving our measurement equipment in the same place, we will see a negative voltage across the coil. How does this relate to the Battery? Well, for one thing, our Battery is our point of reference in Glen’s data gathering. And we have the interesting condition where our point of reference is isolated from the coil by the OFF condition of the MOSFET. Additionally, the very place we expect a negative nine hundred and three volts we leave that connected right to the Battery Positive. Now, we are either going to drive that B+ down 903Vor the other end of the coil is going to go up 903V but one way or the other we are going to see -903V across that coil. Well, as reality would have it, the Battery holds its own and clamps that end of the coil at our 24.77V relative to B- and this means the floating end (Drain connection) of the coil will go up to a +903.79V + 24.77V for a grand total of +928.56V relative to Glen’s ground reference.

How much real dissipated power can we expect from the BEMF Spike?

This is where things get interesting, because now we are 400ns into our MOSFET OFF period and at the peak of our BEMF spike. What will happen next? We have 1.7µJ of energy sitting on a small section of wire between the coil and MOSFET Drain just chomping at the bit to go somewhere. Recall that I said earlier that the BEMF spike is looking to distribute itself equally over the conductors attached to it. When it does, current flows back from the small piece of wire there, through the coil, to the battery plates. But this transaction is not happening at the lazy frequency of 426 kHz. No, this is where the slope of that falling voltage changes all of the dynamics. The fall is so fast, that to our coil it looks like a 2.5MHz signal (400ns wavelength) Now we have to go back to square one and recalculate our inductive reactance using that frequency to know what our impedance will be during the return trip, so we can determine our current flow and thus our power dissipation in the resistor. X = 2πfL = 6.28 ∙ 2500000 ∙ 0.00002277 = 357.67Ω.
Z = √ R² + X² = √9.73² + 357.67² = 357.80 Ω. To determine the amperage, we now take our 903.79V / 357.80Ω and we get 2.53A. Now we can determine how much of that is dissipated from I²R, 2.53² ∙ 9.73 = 62.08W. Is that real power? Yes, but it is peak power of a sinus waveform and is reduced to zero watts over a 100ns period. So the RMS value would be 0.707 ∙ 62.08 for that 100ns period, or 43.89W. Now, it would seem reasonable to double this time, because we have to apply this same value while we are building the BEMF spike - in that case, it would be for 200ns for both the rise and fall of the BEMF spike.

So how does this work out for one full cycle? We have 1.46W during the entire ON time and we have 43.89W during 200ns of OFF time and relatively zero watts during the other 0.974µs of OFF time. This comes out to an average of 4.47W of real power dissipation for each full cycle of the MOSFET. So we are still about 1W short of the thermal output recorded during the test – where does that extra energy come from?

Where does the extra watt come from?

One thing we know about this circuit, is that it is very aperiodic when operating in its preferred mode of oscillation. This means that the frequency is always changing. The calculations I have done in this Back To Basics three part post are on one single cycle, 2.34µs based on a snapshot of the real events. As mentioned in part two, there is a huge area of time that we have no data on and cannot say what is really happening there. We found other spikes in Test #22 that are too narrow to be picked up with the 10,000 sample resolution even at 2µs. So, it could be that the extra watt is buried somewhere in those unrecorded time intervals. The opposite could be argued as well. The unrecorded time intervals could demonstrate that when 100% of the data is collected together uninterrupted that we really may have no classical explanation for where the extra energy is coming from.

I look forward to new tests giving us uninterrupted data capable of proving the case either way. One such test would be an endurance run that exceeds the capacity of the supply. Another test would be to run the device from a monitored power source with a BATCAP filter and a calorimeter cabinet (the entire circuit contained so all thermal energy is counted) – this would be easy to see over an extended period of several hours exactly how the input power relates to the output power.

Cheers!

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