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From: "Fritz Schlunder"
Subject: Re: Flash Inverter Analysis - Help!
Date: Mon, 13 Jan 2003 08:50:56 -0700
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"Guy" wrote in message
> I took apart a disposable camera to study the flash circuit, after 1.5
> i still can't convince myself exactly how this circuit functions. I pretty
> peeved as i have a degree in electronics and feel i should be able to do
> So could you guys take a look at http://homepage.ntlworld.com/guy.stanley
> and tell me whats going on?
> I realise that C1 is part of an oscillator used to generate AC and thus
> allow TR1 to generate 300V from the 1.5V battery. Also once the voltage
> across C2 reaches approx -300V (with respect to GND) the MOV begins to
> conduct and switches on Q1 thus disabling the inverter.
Yeah from the looks of it to me this is indeed the function of Q1 and the
> The part i'm struggling with is exactly how the oscillator functions. I
> that C1 can only charge to a max of 0.6V due to Q2 turning on and clamping
> the voltage. Q2 on makes Q3 switch on and current will flow through TR1's
> primary. Due to the high turns ratio a large voltage will appear across
> secondary, since node P1 is 0.6V node P3 must go negative to maintain the
> voltage across TR1's secondary.
> This is where i get stuck. D1 is now forward biased and this suggests
> current being drawn from C2, But where does it go then? I can't see the
> current path back to GND and thus the other plate of C2. The only route is
> through the secondary coil but how can this be as the current would be
> flowing from negative to positive then.
This is a most interesting circuit that someone probably put allot of love
into designing despite its apparant simplicity. I've been looking at the
circuit for quite some time now and I think I may understand how it
functions (though I can't be 100% sure without seeing and playing with the
circuit myself, including measuring all the currents in the circuit during
First lets make the schematic slightly clearer by placing the polarity dots
on the transformer. Both dots should appear on the topside of each coil (or
alternatively both on the bottomsides of the coils).
During the transformer on time both Q2 and Q3 are conducting. Intitially Q2
conducts due to the current through R1, but then as Q3 conducts the
transformer secondary provides positive feedback to the base drive of Q2 in
turn providing good base drive to Q3. During this on time period current is
flowing out the top of the secondary of the transformer, through the forward
biased base-emmitter junction of Q2, in through the positive terminal of C2,
out the negative terminal of C2, and then through diode D1 back to the other
side of the transformer winding. There is nothing too particularly
incredible about this part of the cycle of operation, except to realize that
the current is not limited by any conventional means, but rather the
parasitic resistances of the battery, primary and secondary resistances,
primary and secondary coupling, and possibly the betas of transistors Q2 and
Q3. Thus this circuit is quite stressful on the components (and not very
efficient) but the stress is tolerabe since it deals with such low power
If I understand the circuit operation correctly (and the schematic was drawn
correctly, which I believe it was), then it seems to me the funky stuff
really begins when it comes time for transformer reset.
Eventually the core of TR1 will saturate. When it does, the core will want
to reset as with any normal saturable transformer blocking oscillator.
Thus, the voltage accross the secondary begins to fall prior to that of its
peak value when the core was not yet in saturation. As the voltage falls,
the voltage at P3 becomes less negative than the voltage on capacitor C2,
thus diode D1 becomes reverse biased. Here is where the mysterious stuff
starts happening. It appears diode D1 has a quite slow reverse recovery
time (my guess is it might be a 1500V standard recovery rectifier). So when
diode D1 becomes reversed biased, it doesn't full block like it is supposed
to, instead a fairly substantial reverse recovery current begins to flow
through it. So, current starts flowing out the anode of diode D1, down into
the negative terminal of C2, out the positive terminal of C2, up through the
grounded side of C1 (and up through the negative terminal of BAT1, out
BAT1+, through SW1, through R1), in through the top of the secondary of TR1,
out the bottom of the secondary of TR1, and finally back into the cathode of
Thus potential P1 falls below ground potential, but not hundreds of volts
below ground since diode D1 is still blocking most of the transformer reset
voltage (though not all of it). Capacitor C1 was sized such that the
reverse recovery currents through D1 were not sufficiently high to cause the
base emitter junction of Q2 to avalanche (thus reducing the beta of Q2 and
eventually destroying Q2).
Thus since potential P1 becomes negative, Q2 turns off, which in turn turns
off Q3 thus the transformer primary is no longer forced to ~1V and
transformer reset can commence. Interestingly the transformer is only
allowed to reset for as long as the reverse recovery currents through diode
D1 last. When diode D1 starts fully blocking resistor R1 will recharge C1
up to about ~0.6V and then transistor Q2 and Q3 turns back on and the cycle
Since the time allowed for transformer reset is dictated by the reverse
recovery time of D1, this means that the core of TR1 may not be fully reset
after the first cycle of operation (IE immediately after SW1 is depressed).
Since some residual B may exist in TR1 this means that the time before it
saturates again for the second cycle of operation will probably be shorter
than the time for the first cycle. Eventually a steady state is found where
the reset time is just long enough. Thus to some degree the frequency of
operation of this circuit is probably determined by the reverse recovery
characteristics of diode D1. A faster reverse recovery time will result in
a higher frequency, but if too fast a recovery rectifier is used the circuit
will stop working altogether and smoke might start comming out of Q3.
It is interesting to note that since the reverse recovery currents of diode
D1 flow through C2, C2 is somewhat discharged during each reset cycle, but
since the forward currents during the on time are much larger than the
reverse recovery currents the net result is the capacitor will eventually
This is a rather devious yet fascinating circuit which I believe the
original designer deserves kudos for. On the other hand the circuit is
still more complicated than a basic saturating transformer blocking
oscillator, so just how many kudos is uncertain. Since it relies on so many
parasitic elements it probably isn't very suitable for scaling up to higher
power levels either...
> I have other queries as well but if someone can explain the circuit
> operation, especially with regard to the above then i might be able to
> those answers out for myself.
> Thanks in advance
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