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That's a pretty crappy circuit. Note that the boost converter is running completely open loop. There is no feedback that shuts it down when its output gets high enough. You don't show what the voltages of the zener and the linear regulator are, but most likely the zener is there just to make sure the input doesn't get above what the cap and the linear regulator can handle. The linear regulator then produces the nice and steady output voltage.

The reason I say this is a crappy circuit is because it is quite wasteful. That's usually a bad thing when running from a battery. Instead of adding feedback to the boost switcher, the extra power just gets wasted in the zener and the linear regulator. It would only take one more transistor to turn on when the regulator has a bit more voltage than it really needs. This transistor would kill the oscillations of Q1, thereby shutting down the boost converter until the voltage dropped again. This is essentially adding some loose regulation to the switcher output.

Added:

I see from the comments that there is interest in discussing how to regulate the switcher so that it's not running open loop.

As Russell and I both mention, in this case a NPN transistor that pulls the base of Q1 low is one means to kill the oscillations. Now the problem becomes turning on this transistor when the switcher output gets high enough. In the context of this circuit, as Russell has already mentioned, the simplest way is to have the bottom of the zener go into the base of this second oscillation-killing transistor. I'd also put a resistor from that base to ground to make sure this transistor doesn't come on just due to leakage. When the switcher output gets high enough, the zener conducts, which turns on the new transistor, which kills the oscillations so that the switcher stops making high voltage until that voltage goes a little lower again.

A totally different way to get a "voltage is high enough" signal is what Russell alluded to in a comment. This is putting a PNP transistor around the regulator such that it turns on when the input of the regulator is the B-E drop of the transistor above the regulator output. That threshold-detecting transistor would then be used to turn on the oscillation-killling transistor. I go into more detail about this method of threshold detecting as feedback to a switcher at http://electronics.stackexchange.com/a/149990/4512.

Added 2:

I see you have now added a updated schematic. Yes, that's exactly what Russell and I are talking about.

I would only make a small refinement by adding a resistor from the base of Q2 to ground. This guarantees some minimum current thru D2 before the switcher is shut down. If you don't do this, the voltage across D2 could be significantly less than its zener rating. Look at the datasheet for D2. It's voltage will be guaranteed only above some minimum current. Without knowing anything about that zener, I'd aim for about 500 µA. Figure the Q2 base voltage will be 600 mV, so that make the resistor 1.2 kΩ.

That's a pretty crappy circuit. Note that the boost converter is running completely open loop. There is no feedback that shuts it down when its output gets high enough. You don't show what the voltages of the zener and the linear regulator are, but most likely the zener is there just to make sure the input doesn't get above what the cap and the linear regulator can handle. The linear regulator then produces the nice and steady output voltage.

The reason I say this is a crappy circuit is because it is quite wasteful. That's usually a bad thing when running from a battery. Instead of adding feedback to the boost switcher, the extra power just gets wasted in the zener and the linear regulator. It would only take one more transistor to turn on when the regulator has a bit more voltage than it really needs. This transistor would kill the oscillations of Q1, thereby shutting down the boost converter until the voltage dropped again. This is essentially adding some loose regulation to the switcher output.

Added:

I see from the comments that there is interest in discussing how to regulate the switcher so that it's not running open loop.

As Russell and I both mention, in this case a NPN transistor that pulls the base of Q1 low is one means to kill the oscillations. Now the problem becomes turning on this transistor when the switcher output gets high enough. In the context of this circuit, as Russell has already mentioned, the simplest way is to have the bottom of the zener go into the base of this second oscillation-killing transistor. I'd also put a resistor from that base to ground to make sure this transistor doesn't come on just due to leakage. When the switcher output gets high enough, the zener conducts, which turns on the new transistor, which kills the oscillations so that the switcher stops making high voltage until that voltage goes a little lower again.

A totally different way to get a "voltage is high enough" signal is what Russell alluded to in a comment. This is putting a PNP transistor around the regulator such that it turns on when the input of the regulator is the B-E drop of the transistor above the regulator output. That threshold-detecting transistor would then be used to turn on the oscillation-killling transistor. I go into more detail about this method of threshold detecting as feedback to a switcher at https://electronics.stackexchange.com/a/149990/4512.

Added 2:

I see you have now added a updated schematic. Yes, that's exactly what Russell and I are talking about.

I would only make a small refinement by adding a resistor from the base of Q2 to ground. This guarantees some minimum current thru D2 before the switcher is shut down. If you don't do this, the voltage across D2 could be significantly less than its zener rating. Look at the datasheet for D2. It's voltage will be guaranteed only above some minimum current. Without knowing anything about that zener, I'd aim for about 500 µA. Figure the Q2 base voltage will be 600 mV, so that make the resistor 1.2 kΩ.

That's a pretty crappy circuit. Note that the boost converter is running completely open loop. There is no feedback that shuts it down when its output gets high enough. You don't show what the voltages of the zener and the linear regulator are, but most likely the zener is there just to make sure the input doesn't get above what the cap and the linear regulator can handle. The linear regulator then produces the nice and steady output voltage.

The reason I say this is a crappy circuit is because it is quite wasteful. That's usually a bad thing when running from a battery. Instead of adding feedback to the boost switcher, the extra power just gets wasted in the zener and the linear regulator. It would only take one more transistor to turn on when the regulator has a bit more voltage than it really needs. This transistor would kill the oscillations of Q1, thereby shutting down the boost converter until the voltage dropped again. This is essentially adding some loose regulation to the switcher output.

Added:

I see from the comments that there is interest in discussing how to regulate the switcher so that it's not running open loop.

As Russell and I both mention, in this case a NPN transistor that pulls the base of Q1 low is one means to kill the oscillations. Now the problem becomes turning on this transistor when the switcher output gets high enough. In the context of this circuit, as Russell has already mentioned, the simplest way is to have the bottom of the zener go into the base of this second oscillation-killing transistor. I'd also put a resistor from that base to ground to make sure this transistor doesn't come on just due to leakage. When the switcher output gets high enough, the zener conducts, which turns on the new transistor, which kills the oscillations so that the switcher stops making high voltage until that voltage goes a little lower again.

A totally different way to get a "voltage is high enough" signal is what Russell alluded to in a comment. This is putting a PNP transistor around the regulator such that it turns on when the input of the regulator is the B-E drop of the transistor above the regulator output. That threshold-detecting transistor would then be used to turn on the oscillation-killling transistor. I go into more detail about this method of threshold detecting as feedback to a switcher at http://electronics.stackexchange.com/a/149990/4512.

That's a pretty crappy circuit. Note that the boost converter is running completely open loop. There is no feedback that shuts it down when its output gets high enough. You don't show what the voltages of the zener and the linear regulator are, but most likely the zener is there just to make sure the input doesn't get above what the cap and the linear regulator can handle. The linear regulator then produces the nice and steady output voltage.

The reason I say this is a crappy circuit is because it is quite wasteful. That's usually a bad thing when running from a battery. Instead of adding feedback to the boost switcher, the extra power just gets wasted in the zener and the linear regulator. It would only take one more transistor to turn on when the regulator has a bit more voltage than it really needs. This transistor would kill the oscillations of Q1, thereby shutting down the boost converter until the voltage dropped again. This is essentially adding some loose regulation to the switcher output.

Added:

I see from the comments that there is interest in discussing how to regulate the switcher so that it's not running open loop.

As Russell and I both mention, in this case a NPN transistor that pulls the base of Q1 low is one means to kill the oscillations. Now the problem becomes turning on this transistor when the switcher output gets high enough. In the context of this circuit, as Russell has already mentioned, the simplest way is to have the bottom of the zener go into the base of this second oscillation-killing transistor. I'd also put a resistor from that base to ground to make sure this transistor doesn't come on just due to leakage. When the switcher output gets high enough, the zener conducts, which turns on the new transistor, which kills the oscillations so that the switcher stops making high voltage until that voltage goes a little lower again.

A totally different way to get a "voltage is high enough" signal is what Russell alluded to in a comment. This is putting a PNP transistor around the regulator such that it turns on when the input of the regulator is the B-E drop of the transistor above the regulator output. That threshold-detecting transistor would then be used to turn on the oscillation-killling transistor. I go into more detail about this method of threshold detecting as feedback to a switcher at http://electronics.stackexchange.com/a/149990/4512.

Added 2:

I see you have now added a updated schematic. Yes, that's exactly what Russell and I are talking about.

I would only make a small refinement by adding a resistor from the base of Q2 to ground. This guarantees some minimum current thru D2 before the switcher is shut down. If you don't do this, the voltage across D2 could be significantly less than its zener rating. Look at the datasheet for D2. It's voltage will be guaranteed only above some minimum current. Without knowing anything about that zener, I'd aim for about 500 µA. Figure the Q2 base voltage will be 600 mV, so that make the resistor 1.2 kΩ.

That's a pretty crappy circuit. Note that the boost converter is running completely open loop. There is no feedback that shuts it down when its output gets high enough. You don't show what the voltages of the zener and the linear regulator are, but most likely the zener is there just to make sure the input doesn't get above what the cap and the linear regulator can handle. The linear regulator then produces the nice and steady output voltage.

The reason I say this is a crappy circuit is because it is quite wasteful. That's usually a bad thing when running from a battery. Instead of adding feedback to the boost switcher, the extra power just gets wasted in the zener and the linear regulator. It would only take one more transistor to turn on when the regulator has a bit more voltage than it really needs. This transistor would kill the oscillations of Q1, thereby shutting down the boost converter until the voltage dropped again. This is essentially adding some loose regulation to the switcher output.

That's a pretty crappy circuit. Note that the boost converter is running completely open loop. There is no feedback that shuts it down when its output gets high enough. You don't show what the voltages of the zener and the linear regulator are, but most likely the zener is there just to make sure the input doesn't get above what the cap and the linear regulator can handle. The linear regulator then produces the nice and steady output voltage.

The reason I say this is a crappy circuit is because it is quite wasteful. That's usually a bad thing when running from a battery. Instead of adding feedback to the boost switcher, the extra power just gets wasted in the zener and the linear regulator. It would only take one more transistor to turn on when the regulator has a bit more voltage than it really needs. This transistor would kill the oscillations of Q1, thereby shutting down the boost converter until the voltage dropped again. This is essentially adding some loose regulation to the switcher output.

Added:

I see from the comments that there is interest in discussing how to regulate the switcher so that it's not running open loop.

As Russell and I both mention, in this case a NPN transistor that pulls the base of Q1 low is one means to kill the oscillations. Now the problem becomes turning on this transistor when the switcher output gets high enough. In the context of this circuit, as Russell has already mentioned, the simplest way is to have the bottom of the zener go into the base of this second oscillation-killing transistor. I'd also put a resistor from that base to ground to make sure this transistor doesn't come on just due to leakage. When the switcher output gets high enough, the zener conducts, which turns on the new transistor, which kills the oscillations so that the switcher stops making high voltage until that voltage goes a little lower again.

A totally different way to get a "voltage is high enough" signal is what Russell alluded to in a comment. This is putting a PNP transistor around the regulator such that it turns on when the input of the regulator is the B-E drop of the transistor above the regulator output. That threshold-detecting transistor would then be used to turn on the oscillation-killling transistor. I go into more detail about this method of threshold detecting as feedback to a switcher at http://electronics.stackexchange.com/a/149990/4512.