Where can I find this component? -Small transformer for charging flash

ian field

Joined Oct 27, 2012
6,536
I saw that, but I don't understand how you can have "2.7 – 5.5 V input; 350 V output" if the ratio is only 1:15.
Can someone explain?
Its a flyback converter.

The transistor switches on to pass current through the primary to build up magnetic flux. When the transistor lets go, the current flow stops and the flux collapses abruptly - that produces a back emf several times the magnitude of the original applied voltage - that back emf is what's multiplied by the turns ratio.
 

Threeneurons

Joined Jul 12, 2016
30
What are you using it for ?

Easiest route is just to gut another disposable camera. In this age of digital cameras, are they even still selling disposable film cameras ? Most "magnetics", such as those transformers, are custom wound, and not sold by any electronics distributor, off the shelf.

If you can tell us, your intent, then we make give you an alternative.
 

ian field

Joined Oct 27, 2012
6,536
What are you using it for ?

Easiest route is just to gut another disposable camera. In this age of digital cameras, are they even still selling disposable film cameras ? Most "magnetics", such as those transformers, are custom wound, and not sold by any electronics distributor, off the shelf.

If you can tell us, your intent, then we make give you an alternative.
There's a toy manufacturer who's name escapes me that does high tech toys - like pretend laptops.

One of their products is a kiddies digital camera, its very robust and most of the components are probably off the shelf items.

The photoflash inverter runs off 6V, unlike the 1.5V ones in disposables.
 

joeyd999

Joined Jun 6, 2011
5,237
I saw that, but I don't understand how you can have "2.7 – 5.5 V input; 350 V output" if the ratio is only 1:15.
Can someone explain?
IIRC, for flyback, the peak primary voltage is (D)/(1-D) times the excitation voltage where D is the duty cycle. The secondary peak voltage is N (turns ratio) times the peak primary voltage. So, the peak Vo = Vin * D/(1-D) * N.

For D close to one, the output voltages can get very high, the limitations being the insulation breakdown voltages between turns and windings, and the magnetic saturation limit of the core. The cores are gapped for the latter reason.
 

joeyd999

Joined Jun 6, 2011
5,237
I should point out that a flyback transformer is not really a "traditional" transformer. It's better to consider it a "two-port" choke or inductor. The first winding (magnetically) charges the core, the second winding discharges it. The two windings operate alternately and independently, unlike a traditional transformer.

Finally, the formula I gave above ignores losses which can be considerable in flyback designs. A major source of energy loss is due to parasitic inductance which tends to be quite high on flybacks. Other losses occur due to winding resistance and the tendency to operate near magnetic saturation.
 
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