Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Rosemary Ainslie]

vi dt is not power, it is energy. power is vi.

Bubba - why are you going on about this?  Energy and power are generic terms.  Applied to electric energy then Power is still volts times amps x time.  And when time is factored in then its represented as Joules.  I just don't see your point.

Rosemary

[alexandre]

Hello,
I was wondering how much would the batteries last, if they weren´t getting any charge. This is a very crude aproximation:

If you use 6 12v batteries to power a DC load dissipating 50 watts, the current is around 50/72 which is 0.7 amps .  Dividing the AH rating of the batteries, lets say 100 AH, by the current, we get the run time of 142.8 hours.

If the load is pulsed, it seems there will be extra run time, even more so with lead acid batteries.

What is needed is replication and continuous operation of the heater, this is better than the the measurements route. Any takers?

A simple PIC microcontroller + drive transistor could set the gate voltage and provide a safety circuit breaker function.

Best,
Alex

[cHeeseburger]

For anyone dubious about the idea that a shunt containing an inductance whose reactance is actually larger than the shunt resistance at the frequencies of interest could actually distort the direction of current flow and change the areas under the positive and negative portions of the trace, here is another even more astounding and revealing demonstration:

Here we have a current generator set up to ramp up and down linearly from zero to two Amperes.  The true current flow is always positive as shown by the trace taken across the pure shunt resistance.  It never once goes below zero.

Yet the voltage as taken across the R+L shows huge amounts of reverse (negative) “current flow”.  If we believed that, that is.

You see, the whole idea of a shunt is based on using a pure resistance to obtain the analog of the current flowing through it by looking at the voltage drop and applying Ohm’s law E=IR.  A resistor’s voltage drop across its terminals is a pure function of the instantaneous current flowing through the resistor.

The relationship between voltage drop and current is entirely different in inductors and capacitors.  In an inductor, the voltage measured across its terminals depends ONLY on how fast the current is changing and whether it is rising or falling and not at all on the actual amount or the polarity of current.  Inversely, in a capacitor, the current flow through it depends ONLY on how fast the voltage across its terminals is changing and not at all on the value of that voltage.

So, a pure inductor will have no voltage drop (zero) across its terminals no matter what the current flowing as long as that current is not changing.  If the current is changing at a steady rate (di/dt is a constant slope), the voltage across the inductor will be a non-zero value and will also remain steady.  If the current is rising, the voltage will be positive.  If the current is declining, the voltage will be negative.

If the rate of change of current (di/dt) suddenly changes from one slope to another, there will be a “spike” of voltage produced.  This is (di/dt)/dt, the rate of change of the rate of change.  That is why we see the spikes each time the slope of the current instantly changes on our trapezoid waveform, shown here.  If there were no resistance in series to damp these spikes, they would be infinite in amplitude if the change in slope was instantaneous NO MATTER HOW LARGE OR SMALL THE ACTUAL LEVEL OF CURRENT WAS.

So using a shunt that has internal inductance with a reactance far larger than its resistance at the frequencies of interest will primarily show the rate of change of the current and not the current’s actual value.

The scope traces here and in my sim of Rose’s circuit shown earlier prove the point that an inductive shunt can and does show negative voltages any time the true current is declining EVEN WHEN THE CURRENT ITSELF IS ALWAYS A POSITIVE NUMBER.

Humbugger

[Bubba1]

Bubba - why are you going on about this?  Energy and power are generic terms.  Applied to electric energy then Power is still volts times amps x time.  And when time is factored in then its represented as Joules.  I just don't see your point.

Rosemary

Yeah, I see that.  Maybe it doesn't matter.

[twinbeard]

Hi Rosemary,

So back to applications.  In your opinion, would it be possible to drive the existing resistive heating element in an off the shelf home water heater appliance from this circuit?  If not, what is necessary to retrofit an existing heater with a replacement element suitable for use with the circuit?  I think it is a easier path to mass implementation to provide a modular kit suitable to upgrade the existing devices in use, as opposed to the larger manufacturing requirements and hence consumer investment required
to replace existing systems altogether.

Cheers,
Twinbeard