COP 17 Heater | Rosemary Ainslie

[Aaron]

note in first post

I have posted a red note at the top of my first post in this thread:

http://www.energeticforum.com/55875-post1.html

[MileHigh]

A different take on it

Two resistors and a single NPN transistor could be put between the 555 and the MOSFET gate to create a simple invertor. That's easier than redesigning your 555 circuit.

I concur that .99's analysis paper is correct.

Here is a way to view the energy being transferred into the coil: When the MOSFET switches on, high voltage is put across the resistor-inductor (Rosemary's case) or across the coil (Luc's case). At the beginning of the "on" cycle, the instantaneous resistance is very high, because the coil acts like an open circuit at the switch-on time. As time goes on the the instantaneous resistance starts to decrease, and approaches the the resistor-inductor (or 44-ohm) resistance of the resistive load. So this means that the current does NOT start flowing right away, like you had a pure resistive load. Ohm's Law does not apply. The current starts at zero and then increases.

It's that "extra" resistance that is "burning up" more power than the load resistor because we know that when two resistors are in series, the higher-value resistor burns off more of the power. For a few milliseconds the "extra resistance" is much larger than the load resistance.

So what is this mysterious "extra" resistor? It's the COIL "looking like" a dynamic resistor that changes value through time. However, this "extra" voltage drop across the inductor plus resistive load is not BURNING power, it's STORING power supplied by the voltage source as energy (power x time).

The "extra" resistor "disappears" after a certain amount of time, and does not generate any more voltage drop. At this point, the resistive load is responsible for all of the voltage drop. The coil has now "finished resisting" and is storing a certain number of Joules of energy. Joules of energy that came from the battery because the coil was "mimicking" a resistor.

Then the MOSFET switches off and the "extra resistor" in the coil now dumps it's stored energy into the load.

Two final points:

No energy discharged from the coil goes back to the souce battery at all. In Rosemary's case, all of the current + voltage discharge from the coil goes through the resistive part of the coil and the diode. In Luc's case, all of the coil energy goes through the diode pair and the 44-ohm resistor.

This can be easily tested with a shunt resistor and scope at the positive battery terminal. This can be "quick and dirty" tested by soldering two vanilla 0.6V red LEDs back to back, and putting them in series with the positive battery terminal and watching or scoping for any "back flashes."

There are no high voltage back-spikes in either setup when the coil discharges. In Rosemary's case, the coil will generate the diode's forward voltage drop, about 0.6-0.7 volts? In Luc's case, it will be somewhat higher. Incuctor's only generate the required voltage to keep the current flowing in the same direction. The "obstacle" in Rosemary's case is a diode. The higher the resistance of the "obstacle" the higher the voltage oputput and shorther the pulse generated by the coil. (In Rosemary's first circuit, the "resistance" is the MOSFET when it is off, and the coil *smacks* the MOSFET with a very nasty and short high voltage spike. The "body diode" inside the MOSFET is there to try to protect the rest of the MOSFET from that unfortunate occurance)

MileHigh

[poynt99]

Q2) If the Quantum article circuit has a 3.7% duty cycle at the Mosfet DRAIN - doesn't that mean that the Mosfet is allowing current to pass through from the battery at 3.7% of the time per pulse?

But my question 2 is not in reference to a scope shot? Just asking about an authentic 3.7% conducting time.

Aaron,

If the 3.7% duty cycle is being calculated using the voltage wave form seen on the MOSFET Drain, then the answer to question #2 is still NO.

If the duty cycle being referenced is in terms of MOSFET ON-TIME, then yes current will conduct for 3.7% of the total PERIOD.

So it depends on how the term "duty cycle" is being used. Technical people almost always use the term to mean the percentage of total PERIOD time that power is being delivered to a load, OR how long a digital pulse is "HI" in comparison to how long it is "LO", over one complete cycle.

To make things easy, in the case of semiconductor switching applications such as with Luc's and Rosemary's circuit, duty cycle simply means "how long is the switch ON compared to how long it is OFF, in one complete cycle.

An easy to remember rule, with N-Channel MOSFET's and NPN transistors, the switch of choice is "ON" when it's input is "ON".

Simple as that.

.99

[RAMSET]

New circuit

FuzzyTomCat

Google Translate


That circuit is 5 or so years old any history available?
Chet

[poynt99]

MileHigh

.99

[FuzzyTomCat]

FuzzyTomCat

Google Translate


That circuit is 5 or so years old any history available?
Chet

Hey Chet,

I'm looking as we speak, the first I saw it was a posting ( French version ) from "NerzhDishual" a OU member in TK's thread, anything I find will be posted including any response from the author ( e-mailed him in French )

Glen

[Inquorate]

Solving problems

Seems to me with my limited understanding of electronics that a lot of issues regarding on time / duty cycles etc could be resolved by implementing a suitably engineered mechanical switch.

What do you guys need exactly? Perhaps I cm contribute by making such a switch.

[Aaron]

duty cycle

My question was:

If the DRAIN has a 3.7% duty cycle, IS IT or IS IT NOT conducting current from a power source for 3.7% of the time per pulse?

The MOSFET switch needs to be set to exhibit a 3.7% ON duty cycle

NO.
When the drain signal is at battery voltage (high) the mosfet is OFF and the load is non-conducting.

PS - If you want to call the drain signal a 3.7 percent duty cycle, that's OK as long as you know the MOSFET is OFF at that time. Not ON.

Right, but I never mentioned anything about determining 3.7% on by doing the voltage check as explained. Just that if there is a 3.7% duty cycle then the switch is on 3.7% of the time and if the drain is on that long, the whole switch has 3.7% duty cycle.

[poynt99]

Yes

Aaron,

In short...

YES!

and if the drain is on that long

As long as in the quote you mean that the Drain is conducting current.

.99

[Aaron]

duty cycle

If the duty cycle being referenced is in terms of MOSFET ON-TIME, then yes current will conduct for 3.7% of the total PERIOD.

Technical people almost always use the term to mean the percentage of total PERIOD time that power is being delivered to a load

Exactly. And your definition of the duty cycle is the only definition I have ever seen and believe that just about anyone here recognizes that the intent of the circuit is for the battery to deliver current 3.7% of the time that the mosfet switch is on.

[Aaron]

led test?

An easy way to verify duty cycle is to put an led in series with the load isn't it? Led going to a photosensor and scope that or put a tiny solar cell from a calculator at the led and scope the output. That will give you exactly the duty cycle and frequency.

That would eliminate any confusion of anyone scoping the switch if that concept is confusing.

Then when the desired duty cycle and frequency is met, remove the led.

[Aaron]

coil discharge

No energy discharged from the coil goes back to the souce battery at all. In Rosemary's case, all of the current + voltage discharge from the coil goes through the resistive part of the coil and the diode.

Is this what is "supposed" to happen or did you actually verify this with this circuit?

[Aaron]

yes

As long as in the quote you mean that the Drain is conducting current.

Yes. That is exactly what I mean.

[poynt99]

Exactly. And your definition of the duty cycle is the only definition I have ever seen and believe that just about anyone here recognizes that the intent of the circuit is for the battery to deliver current 3.7% of the time that the mosfet switch is on.

_|-|________________________________|-|_______
..<------------------------------------->...<------
.................PERIOD(T) (100%)..............3.7%

Not quite Aaron. It's 3.7% of the PERIOD (T), not 3.7% of the time the MOSFET is ON.

.99

[poynt99]

A tentative confirmation

I ran a quick test of Rosemary's circuit in PSpice, and it does appear that current (and power) spikes do run backwards into the battery.

These appear to be rather narrow, about 20ns in width and depending on the presence or not of the flyback diode, are: -120mA and -3W (with diode), and -1.6A and -40W (without diode).

This is with using the proper 3.7% duty switching, and without any strange high frequency oscillation. Just simple switching.

Don't jump to conclusions please. Let me play with this a bit and I will report my findings.

.99