Joule Thief 101

[tinman]

Sure, sorry, it's something that I often complain about too, but 1200 pixels wide is perfect for my monitor. I forget that other people may not have widescreen monitors. So I'll keep them to 800 pixels wide in the future. Thanks....

Meanwhile here is some raw data taken with the ultracap as power source. Every 30 seconds I recorded the Lux reading and the voltage in mV. These data can be used to generate curves and get total light output in Lux-seconds as I described earlier. It will take me some time to graph the data as I have to take the dog to the dogpark and do some other things. But if anyone else wants to do it..... welcome to it.

Good test TK.
There is one problem with it,but we can eliminate a couple of myths from that test.

The first one being that suggested by MH. It was his thought that by including the battery in series with the inductor and LED during the off time of the transistor,that the battery would be drained to a lower voltage,and so more of the energy left in the battery would be used. As we can see,that is not the case,as the cap was left with very close to the same voltage in it with both circuit's--circuit 2 slightly lower,which would indicate a slightly deeper drain of the battery.

Second.
We also see that circuit 2 had a longer run time,and drew less current over that time.
This means that circuit 2 would offer a deeper drain of the battery,and the recovery voltage over that battery would be less,once disconnected from the circuit.
As there is less current flowing through the capacitor during the run,that also means less waste heat dissipated by the capacitor,as the capacitor dose have internal resistance--which brings us to the problem with using a capacitor,in stead of a battery.

When using a battery(as the JT is designed for),the internal resistance of the battery will increase as the voltage drops across it. As circuit 1 has more current flowing through the battery at all times to that of circuit 2,and due to the internal resistance rising in the battery as the voltage across it drops,then the power dissipated as I/R losses(waste heat) in that battery increases per energy volume/per pulse ratio. In other words,the lower the battery voltage gets,the higher the amount of energy total per pulse,is lost to waste heat in the battery,and less is delivered to the LED.

There is also the fact that during the off time of the transistor in circuit 1,energy delivered to the LED is from both the inductor and the battery,and this is confirmed by both the shorter run time,and brighter LED. This also tells us that the battery in circuit 1 will run hotter ,and that it will also dissipate more energy as waste heat.
A battery is a resistor as well,and the resistance value increases as the battery voltage drops.
In regards to a capacitor,how dose the value of internal resistance change as the voltage drops ?.
Something to think about there TK  .

Brad

[tinman]

OK... Brad says that transistor type and coil turns don't matter as long as they are the same for both circuits. I don't fully agree with this but it certainly makes things easier for me. So for all tests, until further notice, I will be using the circuit I built and described in the other thread: MPSA18 transistor, 20+20 turns on the pot-core inductor, low-side current sensing using 0.1 ohm non-inductive CSR, and one of the LumiLed ultra-efficient LEDs for the load. I connect the LED, which is fixed to the inside end of my lightbox, to the circuit with a jumper cable about 18 inches long. All I have to do to switch between circuits is to change the pins where the jumper connects to the circuit board, so the LED isn't touched and remains in the exact same position during testing. These circuits runs at between 9 to 14 kHz or so, depending on voltage.

Lately though I have decided to use an ultracapacitor for the power source. I'll be using a Nesscap 10F 2.7V rated capacitor with 30mOhm (0.030 ohm) equivalent series resistance. See the data sheet attached below (which by the way describes an easy method to determine the actual capacitance of a test capacitor.)

Using the capacitor as power supply has several advantages. It eliminates noise caused by the power supply, it has a much lower impedance than the battery, and it allows one to track changes in light output over time as the voltage drops while running the circuit. Plotting the Lux value against seconds of runtime will allow one to generate a "lux vs. seconds" curve, the area of which will correspond to the total light output in Lux-seconds. Selecting a certain end-point voltage, say 0.450 V, and timing the time taken to reach that voltage from a given starting voltage, say 1.500 V, will give a consistent set of boundaries for measurement. Input power vs. Lux output readings can also be made along the way.

I've already done a couple of sample test runs using the ultracapacitor, charged to 1.5-1.6 V, and it works quite well. It's hard to get valid power readings from the scope at the lowest voltages due to the small current, but the input voltage and Lux values can be read quite precisely. It takes about 11-12 minutes for the voltage to go from 1.500 V to 0.450 V using Circuit 2 (LED across coil). The LED still produces measurable light output down to below 0.420 V after over 17 minutes.

Yes,we could run the test this way-using super caps. But as i stated in my last post,the cap will not include the impedance/internal resistance rise that you would get with a battery.

To carry out the test correctly using a cap in place of the battery,you would have to include a series variable  resister,and increase the resistance value to mimic that of the internal resistance value of the battery as the voltage drops. You would have to get some sort of graph plot from tests carried out on an actual battery,to give you some idea as to how much the internal value increases as the voltage drops. Once you have some estimates-say at 100mV intervals,you could then adjust you VR to mimic that internal resistance when using the cap,at those 100mV intervals.

This will also allow you to calculate the energy dissipated over that internal resistance,which is now our VR,for both circuits.

We can then go onto raising the light output of circuit 2,to match that of circuit 1,and see if there is any difference between P/in for both circuits,and run time duration.
Perhaps a VR on the base would allow us to raise the light output for circuit 2,to match that of circuit 1.

I will have my test setup completed tonight.


Brad

[tinman]

Quote TK:
I couldn't get my power supply to set precisely at 1.5 volts; the voltage monitor showed 1.62 volts for the tests I have run. I checked input voltage and current both with DMMs and with oscilloscope and got essentially the same results. The output ran one LumiLed super-efficent LED in my lightbox, with the ExTech LT300 lightmeter, with sensor 18 inches away from the LED. As you can see from the image of the test circuit below, all I had to do to change between the circuits was to flip the LED connector over and attach it to the other output pins. The actual position of the LED in the lightbox is exactly the same in both cases, there is absolutely no difference in the two setups except how the LED is connected to the board.

TK
Is there a reason that you have your light meter 18 inches away from the LED?--is this some sort of specified measurement distance for the light meters?.

The reason i ask is as follows.
I am using the Digitech QM1587 light meter. I have it in a sealed box,where the distance between the meter and LED is 8 inches--maybe to close?. For the LED,i am using the array from one of those cheap LED torches from the $2.oo store,that has 9 small LEDs in the array,and runs on 3 AAA batteries.
Anyway,i was testing various inductors- wound as per the JT way,and was averaging 1995 LUX per watt--as per the calculations you used on your testing.

So, Circuit 1 ran at an average input power of 90 mW and produced 63.9 lux at the sensor, for an efficiency of 710 lux per Watt.
Circuit 2 ran at an average input power of 40 mW and produced 30.0 lux at the sensor, for an efficiency of 750 lux per Watt.

As you can see,my value per watt is much higher than yours,so im guessing that is because of the close proximity of the LEDs to the light meter sensor,and also because i am using an LED array of 9 LEDs,and not just one as you are,although mine would be just cheapies,and not super efficient one's like you are using.

Anyway,regardless of that,here is what happened.
I decided to dig out my hybrid toroid coil,and give that a whirl in the JT configuration.
Here are the measurements from that test.
P/in was 1.48 volts @ 20mA-measured with both scope and DMMs,and within 2% of each other.
So P/in= 29.6mW
Light out as per light meter was 177LUX.
177/29.6= 5.97 LUX per mW--X 1000=5970 LUX per watt.
But it gets even more interesting.
To achieve the same 177 LUX value,using straight DC from my PSU,it takes 2.98v @ 10.3mA
That is 30.69mW 

There is more good news yet.
The circuit ATM has long wires and clip lead all over the show,and have not as yet started fine tuning by way of different value base resistances. So time to clean up the circuit by shortening all the lead's,use heavy gauge wire,and replace the 1k ohm base resistor with a VR.

I have the distance of your light meter from the LED,but could you give me the measurements for your light box,and the inside finish --E.G,is it a gloss finish that can reflect the light of the inner walls of the box,or is it a flat finish,where little reflection would take place of the inner walls of the light box.

Cheers

Brad

[MileHigh]

Brad:

Well, this is turning out to be a big showcase for your powers of reasoning.  Let's have a look.

Well you screwed that up MH.
Nice try,but far from an even playing field.
Both inductors are supplied with the same(thats right MH--the same) amount of energy,and they are the same inductor. How you ever decided to make the input energies different--well who knows where your head is at some times.

The inductors now have the same  !SAME! amount of stored energy.
So now tell us,which of the two circuits will deliver the most of there stored energy to the LED.

That is how simple it is MH.
God knows how you keep screwing these things up

Lord knows indeed, he certainly works in mysterious ways with you.  I post a simple example to look at how two different load types with different duty cycles will affect the losses in a seven ohm internal resistance for the power source, and you talk about something totally different and unrelated in your response.  It's like you are on the Bizzaro World debating team, and you are in higher esteem the more illogical and/or non-responsive your debating points are.

The first one being that suggested by MH. It was his thought that by including the battery in series with the inductor and LED during the off time of the transistor,that the battery would be drained to a lower voltage,and so more of the energy left in the battery would be used. As we can see,that is not the case,as the cap was left with very close to the same voltage in it with both circuit's--circuit 2 slightly lower,which would indicate a slightly deeper drain of the battery.

But I was talking about a battery as the source and this test is being done with a capacitor so you can't draw any conclusions.  In addition, it makes no sense whatsoever to mention "circuit 2 slightly lower,which would indicate a slightly deeper drain of the battery" because the difference between the two final recorded capacitor voltages is insignificant.  Anybody that understands science would ignore the voltage difference in this case.  The debate attendees from Bizarro World are giving you a round of applause.

As there is less current flowing through the capacitor during the run,that also means less waste heat dissipated by the capacitor,as the capacitor dose have internal resistance--which brings us to the problem with using a capacitor,in stead of a battery.

But TK said this. "I'll be using a Nesscap 10F 2.7V rated capacitor with 30mOhm (0.030 ohm) equivalent series resistance."  Therefore the equivalent series resistance of the capacitor can be considered negligible and can be ignored.  The debate attendees from Bizarro World are giving you a another round of applause.

In regards to a capacitor,how dose the value of internal resistance change as the voltage drops?

Here is where you get a thunderous round of applause from your Bizarro fans for all of your "bench smarts."

To carry out the test correctly using a cap in place of the battery,you would have to include a series variable  resister,and increase the resistance value to mimic that of the internal resistance value of the battery as the voltage drops. You would have to get some sort of graph plot from tests carried out on an actual battery,to give you some idea as to how much the internal value increases as the voltage drops. Once you have some estimates-say at 100mV intervals,you could then adjust you VR to mimic that internal resistance when using the cap,at those 100mV intervals.

You get a smattering of applause from the Bizarro Word fans.  It's not really correct to say that the internal resistance of the battery increases as the "voltage drops."  You don't necessarily see a voltage drop on a high-internal-resistance battery with a voltmeter, do you?  The internal resistance of the battery increases as the state of charge of the battery decreases.  Six years on the bench and your t-shirt looks great.

Perhaps a VR on the base would allow us to raise the light output for circuit 2,to match that of circuit 1.

Now you've got your Bizarro World fans really excited and they are waving their frying pans.

MileHigh

[ramset]

MH
Quote
Lets showcase your reasoning /Snip/ from bizzaro world"
end quote

I could not agree more ...you have an opportunity to get "infront" of the investigation and lead by example
and yet you chose to "lead from behind"

I have to say ,yes it is always much safer to lead from the back ....

But Quite Bizarre indeed.. if one truly expects to be taken seriously!

Our world is filled with "Monday morning quarterbacks"!

Not just one mans opinion ....

Chet K