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Registered Member #3343
Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Hi Gamma,
>>Assuming that the ferrite inner core has a diameter of 20mm, so the inner ferrite area is 0.000314 square meter,
>>Assuming that the coil has 300 turns,
>>Assuming that the core is a ferrite of commercial quality,
>>Assuming that the discharge of C2 (input cap) may reach 1.00A peak,
The maximun induction may be calculated as:
Bmax= L * Ipk / (N *A)
Bmax = 0.012*1/(300 * 0.000314) = 0.127T
Conclusion: Considering that commercial ferrites start to saturare at 0.2~~0.3tesla, your coil is ok. The core is not in the saturation....... ASSUMING that the input data is 100% correct.
If you do not tell us the value of C2 we can not proceed with the calculations .....
Registered Member #5323
Joined: Fri Jun 15 2012, 02:14PM
Location:
Posts: 104
Newton, C2 = 1000 pf at the present time, but I vary C2 value from time to time in order to change the spark gap frequency, still undecided what value is best for C2. Electra, you are right, diagram shows for a postive ion source, reverse D1 and C1 for negative operation.
Registered Member #5323
Joined: Fri Jun 15 2012, 02:14PM
Location:
Posts: 104
I'm using a 10mm blue led (rated at 3.2v typical, 20 ma, p/n 276-0006) with a 1k resistor across C1 22,000uf. With this large smoothing cap, no visible pulse is seen in the led, however, if C1 is reduced significantly, the led will pulse in sync with whatever frequency the spark gap is set to.
Registered Member #3343
Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Gamma:
Now the situation has changed, AASSSUMING THAT THE ANTENA PROVIDE 10MICROAMPERES..... Time of a 10uA to charge a 0.001uF cap with 1000V: I=C*dv/dt >> dt=C*dv/I = 0.001*0.000001*1000/0.000010 = 0.1second
Energy in a 0.001uF cap charged to 1000V: E= 0.5*C*V*V = 0.5*0.001*0.000001*1000*1000 = 0.0005joules or 0.5 milijoules. This 0.5milijoules are tranfered to the coil during the discharge. or The capacitor discharge 0.5 milijoules 10 times per second.
The discharging current: Charecteristic imppedance C2 with L1: Zo = (L/C)^1/2 Zo = (0.012 /(0.001 x 0.000001))^0.5 = 3464 ohms Peak Current = Ec/Zo Peak Current = 1000/3464 Peak Current = 0.289A.
The next step is calculate the best C1 capacitance, but before we need to comfirm some assumptions
USING THE 0.001 CAPACITOR CAN you tell how many ticks per second (or per minute )you have when the cap is discharging ? You may use a oscilicope or you paper tube to count the ticks.
Registered Member #5323
Joined: Fri Jun 15 2012, 02:14PM
Location:
Posts: 104
Newton:
Used a meter to measured the frequency at the junction of L1 and D1, it's 0.5 Hz, plus I can hear the emp from the sidacs on an AM radio each time they discharge C2. If I look very closely at the led, I can indeed barely see a slight brightening (then dimming) in perfect sync with the sidac discharges. (The junction of L1 and D1 is the only spot in the circuit that my meter will measure the Hz, I suppose after L1 the signal is too smooth to register on the meter).
Registered Member #3343
Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Recalculating... AGAIN !
The current that charge the C2 capacitor (0.001uF) from 0 to 1000 volts in half second:
I = C* dv/dt = 0.001*0.000001 *1000/0.5 = 0.2microampere
SOoo The capacitor C2 charging time is 0.5 second. The antena charging current is 0.2microampere. The C2 discharging current through L1 is 0.289ampere. ( as before calculated ) The energy tranfered from C2 to L1 in each discharge is the energy in a 0.001uF cap charged to 1000V: E= 0.5*C*V*V = 0.5*0.001*0.000001*1000*1000 = 0.0005joules or 0.5 milijoules.
And the discharging time (not considering the LED and C1) t= (L*C)^0.5 = 0.012 *0.001*0.000001)^0.5 = 3.46microseconds ( the frequency is 46 kilohertxz) (In your broadcast radio you may have a click every 46 kilohertz)
Now: You said that "If I look very closely at the led, I can indeed barely see a slight brightening ...in perfect sync with the sidac discharges" This info let me assume that the led is feed with 2volts and 2miliampere during 0.1second...
If your LED blinks with 2volts and 2miliampere during 0.1 second, the LED energy may be; Eled= 2*0.002*0.1 = 0.4 milijoule
Conclusion: The energy of the capacitor is reasonable tranfered to the led.
Please inform the capacitor C1 capacitance, The number of sidacs in serie, and disrupting voltage in each sidac, The C2 voltage rating i ( specifyif is Vac or Vdc or WVdc )
Registered Member #3700
Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107
As Steve have posted, " the key would be the extremely low leakage current of the gas tube switch" that for me means you shall find a extreme low leaksage for you spark gap circuit .
Typical leakage current for a sidacs could be 10microampere That means the most of the current collected by the antena is leaking by the sidac, And only 0.2microamperes ( as calculated by Newton ) is going to charge C2... Resulting in low input efficience.
However, The typical leakage current for a reed glass contact could be less than 0.001microamperes! So,if you use the reed glass contact (as a vacuum spark gap) all current collected by antena will be used in your converter!
Typical disrupt voltage of a 15mm reed contact is 300V, so use 3 or 4 reed contacts in series.
Im still wating for the COMPLETE circuit, including the input capacitor(full data) and the LED load ? And the data espec/ of all components (gas dischargind device) ? (only the circuit that works with sucess.)....
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