When the circuit is first turned on the voltage on pin 2/6 is low, this triggers the 555 which turns on pin 3. The 10n cap then charges through the 33k resistor. When the voltage on pin 2/6 reaches 2/3 Vcc the output turns off and the 10n capacitor starts to discharge through the 33k resistor into pin 3. When the voltage drops to below 1/3 Vcc, the process repeats.
When the 555 is triggered the NPN is turned on, the exact current allowed through the coil depends on the 1k resistor and which transistor was chosen. You can vary this by change the values of these two components or changing the timing values presented by the 10n cap or the 33k resistor.
Since the current through a coil cannot change immediately it slowly ramps up when the 555 is triggered. If the timing interval is too long the current can ramp up to an extremely high value. You thus have to make sure the current is limited either by component values or the timing.
Something else to take note of is that the voltage at the collector of the transistor can be extremely high (couple of hundred volts easy), it is pretty likely that you are blowing the NPN transistor after only a short operating time. I would recommend putting in some feedback at the output. Use a voltage divider on the output of the transformer, into an NPN transistor that pulls the reset line low (it should have a pullup). This will stop over-voltage killing your circuit.
So to summarize, there are three problems with this design:
- Timing values probably aren't calculated correctly.
- Current isn't limited, either by timing or components.
- No protection against over-voltage
As for the voltage decreasing over time, pulling 1A from a 9V will degrade its ability to supply power pretty quickly, this will drop the output voltage/current which affects the circuit. To test this you can monitor the voltage of the 9V battery while the circuit is operating, you should see it drop drastically.
It takes almost 2 mA just to charge and discharge the gate of your MOSFET. You're also wasting about 5 mA in R1, since it is grounded through pin 7 about half the time. Your voltage feedback divider is drawing about 1 mA from the high-voltage rail, which translates to more than 20 mA at the input.
There's a problem with using a 555 to drive a large MOSFET: The limited output current of the 555 means that the MOSFET can't switch quickly from full-off to full-on and back again. It spends a lot of time (relatively speaking) in a transition region, in which it dissipates a significant amount of your input power instead of delivering that power to the output. The MOSFET has a total gate charge of 63 nC, and the 555 has a maximum output current of about 200 mA, which means it takes a minimum of 63 nC / 200 mA = 315 ns to charge or discharge the gate. If you're using a CMOS 555, the output current is much less and the switching time is correspondingly longer.
If you add a gate driver chip between the 555 and the MOSFET (one that's capable of peak currents of 1-2A), you'll see a marked increase in overall efficiency. A real boost controller chip will often have such drivers built in.
If you're serious about developing switchmode power converters, you definitely need to get an oscilloscope so that you can see these effects for yourself.
That regulator design is also rather crappy for another reason. The power through a boost mode converter is regulated by varying the duty cycle of the switching element. In this circuit, the feedback is created by using a transistor to pull down on the control voltage node of the 555, which reduces the upper switching threshold. However, because of the way the 555 is constructed, this also reduces the lower switching threshold by a proportional amount. This means that the change in duty cycle as the ouptut voltage rises is much less than you might otherwise think. It has a bigger effect on the frequency of the output pulses, but this isn't relevant. Again, switching to a proper boost controller chip would solve this problem.
By the way, the "regulator" part of the circuit is NOT using the input voltage as its reference, it's using the forward voltage of Q1's B-E junction as its reference.
As Spehro points out, a 100 µH inductor at a switching frequency of 30 kHz — nominal on time = 16 µs — with a 9V source is going to reach a peak current of 1.44 A. This is really abusing the hell out of a 9V battery, not to mention the I2R losses in both the inductor and the MOSFET. This is also uncomfortably close to the saturation current of the inductor, which only exacerbates the losses.
Best Answer
The discharge slope, or output low time, only depends on R3 when R3 is too low. It's also worth noting that R1 and R3 in your schematic form an equivalent resistance. Is there a reason they are separate? Is R1 fixed and R3 a pot in practice? If not, you can just use one resistor.
From the wikipedia article, "Particularly with bipolar 555s, low values of \$R_1\$ must be avoided so that the output stays saturated near zero volts during discharge, as assumed by the above equation. Otherwise the output low time will be greater than calculated above."
In normal operation, the voltage drop inside the 555 timer's discharge pin is fixed while the capacitor is discharging. The bulk of the voltage drop is across the R1||R3 resistance. Thus, the discharge pin is at nearly 0V and the capacitor discharges at the normal rate. However, when the R1||R3 resistance is too low, there is less voltage drop across these resistors and more internal to the timer. Then, the discharge pin is no longer near 0V and the capacitor takes longer to discharge. When R3 = 0, the same thing is happening, just to an extreme. There is theoretically no voltage drop across R3||R1. Instead all the voltage drop from Vcc to GND is internal to the 555 timer. Thus, the capacitor is simply held at about Vcc.
As the other suggested, it's good to look at the internals of the 555 timer to understand what is really going on. I've had to do the same in the past. These java applets are a good way to play around with circuit and see what is really going on.
http://falstad.com/circuit/e-555int.html - 555 Internals
http://falstad.com/circuit/e-555square.html - Astable Multivibrator