Oscillator Powering 6 Modified Led bulbs

[magnetman12003]

The voltage of any neon you get, will be dependent upon which transistor you are using, if you are using a 100 volt transistor, then a 100 volt neon would be good.
So, a higher voltage transistor would be best in these types of radiant spike circuits and in case you forget to connect the load for some reason, high voltage would be present across the transistor.
peace love light

Do you think a C4423 transistor could be used in place of the TIP3055 transistor in the 12 volt circuit.  It has a voltage rating of 400 volts and 10 amps.  How about a BUH1015  which has a voltage rating of 700 volts?   I have lots of each.

[SkyWatcher123]

Hi magnetman, that would be a good choice, it would be safety margin and would be more reliable.
If anyone is going to use these light circuits while they are away, i would also add a fuse of proper rating.
peace love light

[magnetman12003]

Hi magnetman, that would be a good choice, it would be safety margin and would be more reliable.
If anyone is going to use these light circuits while they are away, i would also add a fuse of proper rating.
peace love light
[/quote

Do you have an idea what base resistor value is needed-watts also. if I use a BUH1015 transistor instead of a TIP3055 in your 12 volt circuit?
The BUH1015 is used in the horizontal output section of a TV.  Very fast switching large transistor. Shown is my monster ferrite ring compared to a lemon. If I don't use it I will sell it. Its already been taped Coil winding comes next.

[gyulasun]

....
Do you have an idea what base resistor value is needed-watts also. if I use a BUH1015 transistor instead of a TIP3055 in your 12 volt circuit?
The BUH1015 is used in the horizontal output section of a TV.  Very fast switching large transistor. Shown is my monster ferrite ring compared to a lemon. If I don't use it I will sell it. Its already been taped Coil winding comes next.

Hi magnetman,

Here is a data sheet for your transistor: http://pdf.datasheetcatalog.com/datasheets/105/365081_DS.pdf 
and the parameter  h_FE  (DC current gain) is shown to be between 7 to 14 only (page 2, in the middle of tabelle Electrical Characteristics).   This means that to have a collector current of say 100 mA, you need to pump roughly 100mA/10mA=10mA base current via a resistor from the supply (I considered the typical  h_FE = 10 from data sheet, and h_FE = I_C/I_B 
Now to provide 10 mA base current, you would need to use a resistor which provides this from your input supply voltage. IF you use 5V DC input, then base resistor R could be  (5V-0.8V)/10mA = 420 Ohm or if you use 12V then R=(12V-0.8V)/10mA = 1120 Ohm. (The 0.8V I considered is the forward bias voltage drop between the base and the emitter.)
The power dissipation in such resistor may be evaluated too,  base current squared times the resistor, i.e.  .01A*.01A*1120 Ohm= 112 mW  or for a 420 Ohm at 5V DC supply it would be 42 mW.

Of course this is an approach in theory but can still serve to get an insight, the AC feedback voltage transformed back from the collector winding via the coupling coil to the base will modify the base current. 

A good approach in practice would be to use a 100 Ohm  3 W resistor in series with a 1 kOhm potmeter rated for also at least 2W or higher.  This is not an overkill but precaution because making the base current adjustable is always advisable to find a "sweet" operating point for the transistor and if you happen to bring the wiper of the potmeter to zero Ohm during the tests then the base current would increase to 112 mA via the series 100 Ohm from a 12V supply voltage and then the dissipation in the 100 Ohm would increase to 1.25 W.

Gyula

[magnetman12003]

Hi magnetman,

Here is a data sheet for your transistor: http://pdf.datasheetcatalog.com/datasheets/105/365081_DS.pdf 
and the parameter  h_FE  (DC current gain) is shown to be between 7 to 14 only (page 2, in the middle of tabelle Electrical Characteristics).   This means that to have a collector current of say 100 mA, you need to pump roughly 100mA/10mA=10mA base current via a resistor from the supply (I considered the typical  h_FE = 10 from data sheet, and h_FE = I_C/I_B 
Now to provide 10 mA base current, you would need to use a resistor which provides this from your input supply voltage. IF you use 5V DC input, then base resistor R could be  (5V-0.8V)/10mA = 420 Ohm or if you use 12V then R=(12V-0.8V)/10mA = 1120 Ohm. (The 0.8V I considered is the forward bias voltage drop between the base and the emitter.)
The power dissipation in such resistor may be evaluated too,  base current squared times the resistor, i.e.  .01A*.01A*1120 Ohm= 112 mW  or for a 420 Ohm at 5V DC supply it would be 42 mW.

Of course this is an approach in theory but can still serve to get an insight, the AC feedback voltage transformed back from the collector winding via the coupling coil to the base will modify the base current. 

A good approach in practice would be to use a 100 Ohm  3 W resistor in series with a 1 kOhm potmeter rated for also at least 2W or higher.  This is not an overkill but precaution because making the base current adjustable is always advisable to find a "sweet" operating point for the transistor and if you happen to bring the wiper of the potmeter to zero Ohm during the tests then the base current would increase to 112 mA via the series 100 Ohm from a 12V supply voltage and then the dissipation in the 100 Ohm would increase to 1.25 W.

Gyula

Thank you much for your input help on this Gyula.  I am not familiar how to use the transistor tables at all. I need now is the resistor since I already have the 1k pot, transistor,cap and diode