Joule Thief 101

[MileHigh]

Brad:

Are you unaware that the base current is additive to the collector current?,and if we increase the current to the base,then we increase the current flow through the inductor in the case of the common JT circuit.

A typical standard Joule Thief has a 1K-ohm base resistance, correct?  A standard Joule Thief is designed to have the transistor act as an ON-OFF switch.

That means that with a 1K base resistor that the transistor will be fully ON, and the collector-emitter voltage will be at a minimum.  If you reduce the base resistance to 700 ohms, then you will increase the base current, but the transistor as a switching device will still be fully ON, with the same minimum collector-emitter voltage.

When the transistor is acting like an ON-OFF switch, when it is ON and the battery voltage is constant, in the case of a 1K-ohm base resistor, or in the case of a 700-ohm base resistor, there will be no increase in current flow through the inductor, it will be the same. The factor that is limiting the current flow is the resistance of the inductor, it has nothing to do with the amount of base current flowing into the transistor.

In addition, if the battery voltage is 1.2 volts, then the maximum current flowing through the coil will be proportional to (1.2/coil_resistance).  If the battery voltage decreases to 1.0 volts, then the maximum current flowing through the coil will be proportional to (1.0/coil_resistance).  Therefore, there will be a decrease in the maximum current flow through the coil when the battery voltage is lower, resulting in a decrease in the initial current flow through the LED and therefore a dimmer LED for a lower battery voltage.

MileHigh

[MileHigh]

Brad:

1-we can increase the current flowing to the base of the transistor, as we decrease the base resistance along with the supply voltage drop.
2-which switches the transistor on harder and longer, resulting in a pulse width increase,
3-thus maintaining a magnetic field of the same value as the battery voltage drops

1.  You can increase the current flowing into the base of the transistor, but it will just be "wasted current" that does nothing because the transistor is already fully ON.

2.  The transistor will not switch on "harder" because it is already fully ON.   Why should it switch on longer?  Why?  What is the mechanism that determines the length of time that the transistor stays ON?  It is certainly not "switching harder" that makes the transistor supposedly stay on longer.

3.  The current flow though the inductor (a.k.a. the "magnetic field") will principally be controlled by (battery_voltage/inductor_resistance) because when the transistor is switched fully ON, there is a very low and constant voltage drop across the collector-emitter junction and that collector-emitter voltage drop will be the same if you switch the base resistance from 1K-ohm to 700 ohms.

MileHigh

[tinman]

Brad:

A typical standard Joule Thief has a 1K-ohm base resistance, correct?  A standard Joule Thief is designed to have the transistor act as an ON-OFF switch.

That means that with a 1K base resistor that the transistor will be fully ON, and the collector-emitter voltage will be at a minimum.  If you reduce the base resistance to 700 ohms, then you will increase the base current, but the transistor as a switching device will still be fully ON, with the same minimum collector-emitter voltage.

When the transistor is acting like an ON-OFF switch, when it is ON and the battery voltage is constant, in the case of a 1K-ohm base resistor, or in the case of a 700-ohm base resistor, there will be no increase in current flow through the inductor, it will be the same. The factor that is limiting the current flow is the resistance of the inductor, it has nothing to do with the amount of base current flowing into the transistor.

In addition, if the battery voltage is 1.2 volts, then the maximum current flowing through the coil will be proportional to (1.2/coil_resistance).  If the battery voltage decreases to 1.0 volts, then the maximum current flowing through the coil will be proportional to (1.0/coil_resistance).  Therefore, there will be a decrease in the maximum current flow through the coil when the battery voltage is lower, resulting in a decrease in the initial current flow through the LED and therefore a dimmer LED for a lower battery voltage.

MileHigh

MH
You need to have a good look at the basic JT circuit--the one you like,or call the JT circuit.
You have two coil's that provide current to build the magnetic field--not one.
As you decrease the resistance to the base,you increase the current flowing to the base,and this current is additive to the collector current due to the way the two coils are linked together,and thus the magnetic field strength can be maintained by reducing the base resistance value, so as to provide the same amount of power flowing through the LED from the kickback as the battery voltage drop's.

I will draw up a quick schematic of my test setup,and post it here. When you see the test setup,you will know that the measurements taken,and the statements i provided are correct.


Brad

[tinman]

@MH

Below is my test setup schematic.
I have added a diode and cap to the JT circuit,so as the LED see's only a stable DC current at all time's.
The LED is a 2.6 volt LED,and i have used a 2.5k VR instead of the 1k resistor. This is needed so as i can get the voltage across the cap down to the rated 2.6 volts of the LED,when the circuit is supplied with 1.5 volt's. If i use a 1k pot,the cap will charge up to 3.1 volt's,and the LED passes out. At 1.2 volt's,we are very close to the 1k ohm resistance value,and so very close to your standard JT circuit.

I have a 10 ohm CVR,as a 1 ohm is far to noisy at these low power levels,and the 10 ohm CVR allows for very clear scope traces.

I have the LED across a 4700uf cap,and this gives us a very clear/clean voltage value across the LED.

With this setup,i can clearly show a maintained voltage across the LED as the supply voltage is reduced. This can only mean that the magnetic field strength is being maintained as the input voltage is dropped.

With the 10ohm CVR,i can also clearly show the pulse width increase-and there for the current increase, as the base resistance is reduced along with the input voltage--all while maintaining 2.6 volts across the LED, with a smooth DC current flow through the LED.


Brad

[tinman]

Brad:



   



MileHigh

1.  You can increase the current flowing into the base of the transistor, but it will just be "wasted current" that does nothing because the transistor is already fully ON.

MH
1.2v/1kohms is 1.2mA. This is not enough current to fully switch on the transistor,as the transistor is a current device,unlike the FET which requires very little current,but a higher voltage.

2.  The transistor will not switch on "harder" because it is already fully ON.

It will not be fully on at a low base current of 1.2mA.

Why should it switch on longer?  Why?  What is the mechanism that determines the length of time that the transistor stays ON?  It is certainly not "switching harder" that makes the transistor supposedly stay on longer.

The reason it stays on longer, is because the transistor is switching on harder,in that the required switching current to the base has now increased.

3.  The current flow though the inductor (a.k.a. the "magnetic field") will principally be controlled by (battery_voltage/inductor_resistance) because when the transistor is switched fully ON, there is a very low and constant voltage drop across the collector-emitter junction and that collector-emitter voltage drop will be the same if you switch the base resistance from 1K-ohm to 700 ohms.

As i stated,the lower the supply voltage get's,the less base current there is available to switch the transistor fully on. There is also the fact that the base current is additive to the collector current in this situation. So the least resistance to that current flowing to the base,the more current flows through the inductor as a whole.

Brad