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Old 05-02-2010, 11:22 AM
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Harvey Harvey is offline
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Back To Basics Part 2 : The First Half Of The Transaction

I'll bet you thought I was going to leave you all hanging with my questions in this post:

Mosfet Heating Circuits

Nah, I wouldn't do that to you guys so here come the answers:

Here are some questions to answer regarding Test #13:

What is the frequency?
That depends on what part of the circuit we look at. But generally we use the gate pulse frequency:

426.0 kHz
Rise 768.4ns
Fall 513.0ns
High 13.61V
Low -4.800V

What is the impedance of the load resistor?
The classical formula for impedance of an RL circuit is Z = √ R + X

R = 9.73
X = 2πfL
2π = 6.28
f = 426000

Now for L we need to determine the inductance in Henries:

The formula for inductance of a single coil winding in microhenries is (rN/9r+10l) (For reference, one inch = 25.4mm)
r = (32mm / 2) / 25.4 = 0.625"
N = 48, the number of windings
l = length of all the windings and spacing. 20 AWG = 0.032" = 0.8128mm thick, so the length = (48 * 0.8128mm + 47* 1mm)/25.4 = 3.39"

So, L in Henries = ((0.625 * 48) / ((9 * .625) + (10 * 3.39))) * 10^-6 = ((0.390625 * 2304) / (5.625) + (33.9))*10^-6 = (900/39.525)*10^-6 = 0.00002277H

So, X = 6.28 * 426000 * 0.00002277 = 60.92Ω (That is the inductive reactance of Glen's resistor at that frequency)

So, Z = √ (9.73 + 60.92) = 61.69Ω

With this information we can also determine the Phase Angle of the current in this inductive resistor using the formula: θ = arctan(X/R). For Glen's resistor, we expect the current to lag the voltage by 80.92

What is the real power being dissipated? P = IR | I = E/Z | E = (24.77 / (Z + 2 + 0.25))* Z = 23.90V
So, I = 23.90 / 61.69 = 387mA. Thus the real power dissipated is 1.46W

What is the apparent power involved, yet returned? P = IX
So, 387mA * 60.92Ω = 9.14 volt-amps. ( I added this question here)

How does a square wave affect the impedance of an inductor? This was a trick question - Inductors do not have impedance unless coupled with capacitance or resistance. However, inductors do have impulse response and reactance, both of which can be affected by the slope of the waveform. This was introduced here to get people to understand that there is more happening here than even a basic classical AC approach can address. The risetime and falltime of the waveforms associated with the inductive-resistor play an important role in the actual instantaneous reactance.

If the resistor was non-inductive, would the frequency make any difference as long as the duty cycle remained constant? No, the power relationships remain the same regardless of frequency as long as the duty cycle remains constant in purely resistive applications.

================================================

So when is power delivered from the battery to the load?
This is when the gate pulse is at a positve voltage relative to the MOSFET source pin. Note that there is a low voltage reading of negative 4.8V on the gate pin. We don't have the MOSFET Source pin value recorded in the screen shot to go with that, but we do have the data dumps and we could look up what the source reading was when the gate was at that level. This may be an interesting exercise for you to do as it can enlighten you as to how a 555 timer can allow a negative voltage to occur on the output pin 3 and where that negative energy comes from.

We find then, that when we apply the classical AC (or pulsed DC) treatment to Glen's Test #13, we expect to see a power dissipation of 1.46W and an overall power swing around 10.6W which is mostly apparent power. This is much closer to what the actual data showed, with a true average of 1.3W when we include the high resolution 2S/div data, with the lower resolution 40S/div data. Each of these samples were averaged independently, and then all of those averages were averaged together to reach 1.3W. That is about as true as we could get considering the entire test covered a full hour and we only have 0.00462 seconds (4.62ms)of data for the entire test. A lot can happen in the other 3599.99538 seconds we have no data for. Sure would be nice to have an RSA6000 to get some seamless data with.

Now this puts us back to asking the question: If we are only showing, both by projections and by data, ~1.5W of dissipation, how do we account for the 5.5W of thermal output?

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Last edited by Harvey; 05-04-2010 at 06:45 AM.
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