Boost converter unable to support any load

[Los Frijoles]

I have built a boost converter as shown in the image below. Normally the pwm runs at a 70% duty cycle @ 15khz to give an output voltage of around 12V. However, when I connect any sort of load to the 12V out pin, the voltage immediately drops to 7-ish volts. The load is less than 100mA and actually may be less than 50mA (I do not have a very accurate ammeter on my 12V supply when I am running this thing without the boost converter).

Is there anything blatantly wrong with this circuit? According to my dad (who designs these for a living), a boost converter should not be dropping down that low on that small of a load. When I activate the PI loop that I have running on the pwm based on the analog feedback, it is unable to compensate for the voltage drop and hits the saturation point very quickly and the circuit has to be shut off.

 

[Prototyp_V1.0]

Mybe the inductor is too small.

Maybe there is something wrong with the logic.

Try a fixed 70% duty cycle an see what happens when you apply a load resistor.

Is the duty cycle at 70% when there is no load at all?

 

[Los Frijoles]

The duty cycle is at 70% no load. When I take off the load resistor, the voltage skyrockets as could be expected. I usually run the load with the load resistor still there. Could this be a problem?

The inductor is 3mH (not uH)...certainly that is enough?

 

[FvM]

May be it's a combination of non-optimal part values. The inductor series resistance and saturation current isn't known, the transistor is "huge" with a high capacitance, the diode is too small with a series resistance that causes a considerable voltage drop. The conroller should have at least a duty cycle limitation or better a current limit respectively use current mode.

70 % duyty cycle should give more than 15 V output, apparently a lot of losses exist in the circuit. They can be indentified by analyzing the waveforms.

 

[Prototyp_V1.0]

... The conroller should have at least a duty cycle limitation or better a current limit respectively use current mode.

Thats sure. If the duty cycle goes to 100% then you'll probably short circuit the 5v source, and there will be no voltage out.
So you have to check the duty cycle (scope) when a load resistor is connected.

 

[Los Frijoles]

Right now the controller limits the duty cycle to 95%. I think I should be using a different diode than a 1N4148, but I really don't have any others. The MOSFET was one that I pulled out of an old VCR some 10-odd years ago. I got the inductor out of one of those radioshack variety packs.

@FvM: What would be the optimum places to measure these waveforms for losses?

 

[Prototyp_V1.0]

Maybe Qi is totaly broken, so that the current that the cmos would normaly drawn, now comes from the output (open colector ttl?) of the controller itself.

 

[haff99]

I think the inductor is way too big. Just thinking about this for a moment, at no load the inductor current is at a steady state value. As soon as you apply a load the output voltage drop, the FB voltage drop, the NMOS switch stays on for a longer and longer time. However, the inductor current is going to ramp up slow since the slew rate of the current will be 5V/3mH(ignoring inductor resistance and Ron). It is going to take the inductor many cycles to build up the current to support any type of load at 12V. For instance you are going to get .001666A/uS slew rate with that inductor. If you dropped down to lets say a 100uH inductor that value goes up to 50mA/uS. Even at 15KHz, I think a 100uH should be plenty of inductance(you can probably go even smaller). So now you will see your output voltage dip and it should recover much faster, due to the inductor current ramping faster. I would bet if you let your simulation run out much longer with that large of an inductor you would see the output voltage recover eventually.