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Blocking oscillators

:: EPE Chat Zone ­:: ­Radio Bygones Message Board :: » EPE Forum Archives 2010 - » Archive through 15 January, 2010 » Blocking oscillators « Previous Next »

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bruce
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Username: bruce

Post Number: 235
Registered: 04-2008

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Posted on Friday, 15 January, 2010 - 11:52 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Hi,
I have a theoretical question about blocking oscillators. Some time ago, IU had a circuit which used a blocker to illuminate a white LED using a single 1.5V cell, a bit like this:


image/bmpBlocker A
BlockA.bmp (16.1 k)


Now, I used this circuit to generate a 5V rail at a few mA by charging a capacitor via a diode thus:

image/bmpBlocker B
BlockB.bmp (25.3 k)


Now for the question: when the transistor saturates ( if thats what happens ) there is a voltage spike on the collector. I cant conceptualise whats happening. Is the spike positve or negative wrt to ground? I cant tell whats going on with my scope. If the spike is -ve, then I need to reverse the diode.
The reason I ask is that the flash circuit in disposable cameras nearly always generates a negative 300V rail wrt ground and I always wondered why.

Bruce
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gajjer
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Username: gajjer

Post Number: 280
Registered: 05-2007

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Posted on Friday, 15 January, 2010 - 03:41 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Hi Bruce
here is my guess at what is happening.
When you turn on, the base resistor provides drive to turn on the transistor. A voltage develops across the collector coil as the current ramps up. The coil in the base circuit also has a voltage develop across it that adds to the base drive. At some point the collector current will stop increasing and the base drive coil will no longer be assisting the drive. Collector current reduces and now the base coil drives the opposite way and starts turning off the transistor. When it gets back to zero the process repeats.
Now the voltage on the collector will be near zero when it is on and a high +ve when it turns off. So your diode is the right way round as you have drawn it - albeit that it looks like a zener rather than an ordinary diode! Its the wobbly hands I know! Just about every faculty you have starts to let you down just when you need them most!!


cheers mate
gaj
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ianfield
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Username: ianfield

Post Number: 23
Registered: 08-2007

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Posted on Friday, 15 January, 2010 - 05:14 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

The question of why a camera flash BO inverter produces a negative voltage is answered by the fact that the capacitor charging circuit is in series with the base drive circuit.

This is done so the large charging current to charge the capacitor from nothing, also produces large current pulses through the base, as the capacitor nears fully charged the current through the base "throttles back" - saving energy and only making up loss due to leakage.

Obviously the pulses of current driving the base must be positive in polarity so the inevitable rectifier diode will have it's cathode pointing toward the base - and conversely it's anode toward the HV winding's output, making the output negative.

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