Hello everyone,

I built very simple boost converter. It works pretty well. I thought I understood the working of the circuit until I checked the circuit with scope.

The blue curve corresponds to measurement across collector and emitter. The yellow between base and emitter. I thought whenn one signal turns on the other one would turn off. Both the grounds of scopes were connected to negative terminal or emitter. But, here both seems to turn on at the same time but in opposite direction. I am bit confused. Can someone explain how this works.

What is this circuit called boost circuit, flyback converter?

 

You do realize the the collector votage is high when the transistor is off and low when it is on, right?

Bob

 

You do realize the the collector votage is high when the transistor is off and low when it is on, right?

Bob

That's what I expected. But,collector voltage when transistor is on.

 

That's what I expected. But,collector voltage when transistor is on.

At what point on the waveform are you referring to?

 

The center tap of the primary is at AC ground, making the two ends of the primary have opposite phase.

How it works: Same circuit but driving a white LED.


The circuit could not be simpler than this. The transistor, 1K resistor and the tapped inductor form a blocking oscillator. When the power button is pressed, the transistor is biased on through the 1 k resistor. Voltage that appears from the tap on the inductor to the collector causes the voltage on the 1K resistor to be even higher than the battery voltage, thereby providing positive feedback. Also because there is voltage across the inductance between the tap and the collector of the transistor, collector current increases with time (this is in addition to a starting value that relates to the current supplied to the base, but this part of the collector current is rather small. Because of the positive feedback the transistor stays saturated until something happens to change its base current.

At some point the IR drop across the inductor from the tap to the collector approaches the battery voltage (actually battery voltage - VCEsat). As this happens voltage is no longer induced in the winding from the tap to the 1k resistor and the base voltage starts to drop, and this forces the base voltage to go negative, thereby accelerating the switching off of the transistor. Now, the transistor is off, but the inductor continues to source current and the collector voltage rises.

Quickly, the collector voltage gets high enough for the LED to conduct current, and it does for a little while, until the inductor runs out of current, then as the collector voltage starts to ring toward ground base voltage swings positive again, turning the transistor on again for another cycle.

 

At what point on the waveform are you referring to?

Sorry, typo there. I expected yellow curve to be off when blue was on and vice versa. But, what I see is opposite polarity.

 

The center tap of the primary is at AC ground, making the two ends of the primary have opposite phase.

How it works: Same circuit but driving a white LED.
View attachment 226515

The circuit could not be simpler than this. The transistor, 1K resistor and the tapped inductor form a blocking oscillator. When the power button is pressed, the transistor is biased on through the 1 k resistor. Voltage that appears from the tap on the inductor to the collector causes the voltage on the 1K resistor to be even higher than the battery voltage, thereby providing positive feedback. Also because there is voltage across the inductance between the tap and the collector of the transistor, collector current increases with time (this is in addition to a starting value that relates to the current supplied to the base, but this part of the collector current is rather small. Because of the positive feedback the transistor stays saturated until something happens to change its base current.

At some point the IR drop across the inductor from the tap to the collector approaches the battery voltage (actually battery voltage - VCEsat). As this happens voltage is no longer induced in the winding from the tap to the 1k resistor and the base voltage starts to drop, and this forces the base voltage to go negative, thereby accelerating the switching off of the transistor. Now, the transistor is off, but the inductor continues to source current and the collector voltage rises.

Quickly, the collector voltage gets high enough for the LED to conduct current, and it does for a little while, until the inductor runs out of current, then as the collector voltage starts to ring toward ground base voltage swings positive again, turning the transistor on again for another cycle.

Thanks for the explanations

These measurements were made under DC bias.

 

In oscillator circuit either you can have a mechanism to "kick" the circuit into operation or the transistors has to be biased so that noise can do the kicking.

 

The circuit is usually called a "Joule Thief" because it lights a 1.8V red LED when the battery is almost dead at 0.8V.
Here is the same schematic with the coil drawn a little differently:

 

Thanks everyone for the explanations. Additionally I measured voltage across a low ohm resistor between emiiter and the negative terminal. It's made the picture bit more clearer.