Joule Thief 101

[partzman]

webby1

You were correct regarding the instantaneous voltages across the resistors at T=5   For the question regarding the general shape of the voltages and currents when T>5 secs, I attached a sim showing the results which will speak more clearly than an explanation with words.

The Vin and Vout traces are superimposed until T=5 when Vin goes to zero and Vout shows the voltage across the 1.6666 ohm and 10 ohm resistors respectively. The current from L1 for each resistance is shown and marked as voltage/current pairs. To be clear, R1 has been stepped with 1.6666 ohms and 10 ohms to produce the overall plot with two pairs of voltage and current from T5 to the end.

There are also two cursors on the plot indicating the time and current magnitude for each pair at one time constant or T=L/R which indicates the currents have dropped ~63% from the starting value of 2.4 amps.

Several things to note and that is, the polarity of the voltage(s) across R1 and the shape of the voltage and current waveforms.

Also, we can conclude from this that there is no value of "fixed" resistance that can reproduce the same current waveform as MH's from T5 to T8.

partzman

[partzman]

webby1,

I've attached a sim that uses a non-linear resistor R1 from T5 to T8 that creates a complimentary negative going current ramp to the voltage generated positive going current ramp from T0 to T3. This is shown for comparison to the fixed resistor R1s in the previous sim.

Since dI = E*t/L and L is fixed at 5H and we wish to start with I = 2.4 amps at T5 and ramp to 0 amps at T8, we can solve for R using either t or E. I chose to solve using t as follows.  With R=E/I and dI = E*t/L, through substitution and simplification we arrive at R = L/t.

The denominator t is actually T-dt with T equal to the overall time from T5 to T8 to ramp from 2.4 amps to zero amps and dt is the incremental change in time during this period. However, R1 is calculated during the entire plot as is seen in the red "5/(8-time)" trace and at t5, the resistance is 1.66666 ohms which is the starting value needed to produce 4 volts at 2.4 amps. Also at T5, S1 is opened to disconnected Vin from L1 and S2 then connects R1 to L1.

To clarify, the label at the far left of the plot is the resistance of R1 so "100s-1" should read 100 ohms, etc.

As we approach T8, "T-dt" approaches zero and R1 approaches infinite resistance.

partzman

[MileHigh]

webby1,

I've attached a sim that uses a non-linear resistor R1 from T5 to T8 that creates a complimentary negative going current ramp to the voltage generated positive going current ramp from T0 to T3. This is shown for comparison to the fixed resistor R1s in the previous sim.

Since dI = E*t/L and L is fixed at 5H and we wish to start with I = 2.4 amps at T5 and ramp to 0 amps at T8, we can solve for R using either t or E. I chose to solve using t as follows.  With R=E/I and dI = E*t/L, through substitution and simplification we arrive at R = L/t.

The denominator t is actually T-dt with T equal to the overall time from T5 to T8 to ramp from 2.4 amps to zero amps and dt is the incremental change in time during this period. However, R1 is calculated during the entire plot as is seen in the red "5/(8-time)" trace and at t5, the resistance is 1.66666 ohms which is the starting value needed to produce 4 volts at 2.4 amps. Also at T5, S1 is opened to disconnected Vin from L1 and S2 then connects R1 to L1.

To clarify, the label at the far left of the plot is the resistance of R1 so "100s-1" should read 100 ohms, etc.

As we approach T8, "T-dt" approaches zero and R1 approaches infinite resistance.

partzman

Awesome work Partzman.  Getting the current curve to ramp down in a linear fashion with a variable resistor is a not-too-difficult thought experiment but it is really cool, even beyond cool, to see you work the variables to define the variable resistor as a function of time for the simulation and then run it and show the fruits of your labour.

I can imagine the usual grumbling, "That's not ever going to happen in real life," and "That's useless and I can't see any reason for doing that."

Well, what about a rocket?  When a rocket launches it obeys the usual f = ma to accelerate and get off the ground.  However, the mass of the rocket 'm' is actually 'm(t).'  In other words the mass of the rocket is a function of time t and always decreasing because it is burning fuel, i.e.; f = m(t)a.   But then there is another thing to consider, let's assume that the force from some rocket engines is proportional to the amount of fuel in the engine, and so we can say that the force is also a function of time.  So you get f(t) = m(t)a.   So rewriting it, you get a = f(t)/m(t).   Now your variable resistor that is a function of time is starting to look a lot more interesting.

Finally, when it comes to launching a rocket, you can't forget the air resistance that is slowing the rocket down.  Let's say that we say that the air resistance is a function of both the height h and the velocity v.  i.e; The force of the air resistance fair = fair(h,v)

So now you have a new formula for the acceleration of the rocket:  a = [f(t) - fair(h,v)]/m(t).

Now all of a sudden that resistance that's a function of time starts to look pretty sweet.

[partzman]

Awesome work Partzman.  Getting the current curve to ramp down in a linear fashion with a variable resistor is a not-too-difficult thought experiment but it is really cool, even beyond cool, to see you work the variables to define the variable resistor as a function of time for the simulation and then run it and show the fruits of your labour.

I can imagine the usual grumbling, "That's not ever going to happen in real life," and "That's useless and I can't see any reason for doing that."

Well, what about a rocket?  When a rocket launches it obeys the usual f = ma to accelerate and get off the ground.  However, the mass of the rocket 'm' is actually 'm(t).'  In other words the mass of the rocket is a function of time t and always decreasing because it is burning fuel, i.e.; f = m(t)a.   But then there is another thing to consider, let's assume that the force from some rocket engines is proportional to the amount of fuel in the engine, and so we can say that the force is also a function of time.  So you get f(t) = m(t)a.   So rewriting it, you get a = f(t)/m(t).   Now your variable resistor that is a function of time is starting to look a lot more interesting.

Finally, when it comes to launching a rocket, you can't forget the air resistance that is slowing the rocket down.  Let's say that we say that the air resistance is a function of both the height h and the velocity v.  i.e; The force of the air resistance fair = fair(h,v)

So now you have a new formula for the acceleration of the rocket:  a = [f(t) - fair(h,v)]/m(t).

Now all of a sudden that resistance that's a function of time starts to look pretty sweet.

Thanks MH,

partzman

[tinman]

Going back to the simple JT,which circuit would be more efficient?

We shall use MH's term for efficiency as stated on this thread in reply 166

For purposes of a fair discussion let's put aside the "Thief" part of the Joule Thief that can extract energy from nearly dead batteries.  In other words, let's just look at lumens per watt of supplied power.

So what is a JT circuit?
Well to also quote MH on post 166

When is a circuit a Joule Thief or not?  I think that there is a simple answer to that one.  If the circuit can power a LED with a battery whose output voltage is lower than the normal drive voltage for the LED, and the LED is driven using the technique of a discharging inductor acting as a current source, then you have a Joule Thief.

As some would remember,MH later on decided that the first circuit-circuit 1 is the JT circuit,although it would seem that this go's against his first post on what a JT circuit is.

But for the sake of the discussion,lets just stick to the two circuits below.
Which do you believe to be the more efficient circuit out of the two?


Brad