Actually, it's the _holding current_ that is the value below which the anode current has to drop in order for the SCR (thyristor) to turn off. You haven't explicitly shown the SCR you're using, but if it's a BT149 whose datasheet you showed, then it has a typical latching current of 2mA and also a typical holding current of 2mA. That makes designing your circuit difficult. For the SCR to turn on, it needs about 0.6V on its gate and an inital current greater than the latching current. When that happens, the capacitor is charged up to about 2.6V, meaning that there is 1.4V across the 860R resistor, giving a current of 1.63mA. The rest of the current needed to exceed the latching current comes from the capacitor discharging into the SCR and we don't know how much that is. For the SCR to turn off, it needs to be passing a current less than the holding current. However, as the capacitor rapidly discharges the voltage across the SCR rapidly drops, which means that the gate becomes isolated while the current through the 860R resistor rises towards 4V/860R = 4.6mA. That means that either the SCR used had a holding current greater than 4mA (not typical, but possible), or the lack of gate current allows the SCR to turn off at a lower value than the holding current. In all of this, you can see we really don't have enough information to confidently design an oscillator. You just have to "suck-it-and-see" to find a working value for R1 depending on the SCR you're using. If R1 is too large, the SCR won't turn on; if R1 is too small the SCR will turn on, but not off again.
You can increase voltage and use TRIGGER DIODES ( DB3 for example) . Your schema has an error. You have to discharge C1 in series with resistor to limit high current flow between anode and cathode.
When the thyristor opens it creates a path of no resistance so the energy stored on C1 will start to empty to go down this path. Which will eventually shut off the Zener and Thyristor so C1 begins to fill again causing this process to loop. C1 will discharge if there is a path for it to discharge even when connected to the Voltage Source. You’d have to do the math, the resistors in the circuit are assisting to keep the current and voltages close to the desired levels for this loop to occur.
Nice one!! 👍Thanks so much for your excellent examples. great channel!!
Thank you so much for those information. I'm amateur and i love your videos
Actually, it's the _holding current_ that is the value below which the anode current has to drop in order for the SCR (thyristor) to turn off. You haven't explicitly shown the SCR you're using, but if it's a BT149 whose datasheet you showed, then it has a typical latching current of 2mA and also a typical holding current of 2mA. That makes designing your circuit difficult.
For the SCR to turn on, it needs about 0.6V on its gate and an inital current greater than the latching current. When that happens, the capacitor is charged up to about 2.6V, meaning that there is 1.4V across the 860R resistor, giving a current of 1.63mA. The rest of the current needed to exceed the latching current comes from the capacitor discharging into the SCR and we don't know how much that is.
For the SCR to turn off, it needs to be passing a current less than the holding current. However, as the capacitor rapidly discharges the voltage across the SCR rapidly drops, which means that the gate becomes isolated while the current through the 860R resistor rises towards 4V/860R = 4.6mA. That means that either the SCR used had a holding current greater than 4mA (not typical, but possible), or the lack of gate current allows the SCR to turn off at a lower value than the holding current.
In all of this, you can see we really don't have enough information to confidently design an oscillator. You just have to "suck-it-and-see" to find a working value for R1 depending on the SCR you're using. If R1 is too large, the SCR won't turn on; if R1 is too small the SCR will turn on, but not off again.
Done that many years ago. It is not difficult, with a single npn transistor and with some passive parts
You can increase voltage and use TRIGGER DIODES ( DB3 for example) . Your schema has an error. You have to discharge C1 in series with resistor to limit high current flow between anode and cathode.
I showed in my video that this oscillator works as is.
It is not impossible. You can create a Phase Shift Oscillator using only single BJT without inductor.
I don't understand why that capacitor c1 discharges while connected to vcc
When the thyristor opens it creates a path of no resistance so the energy stored on C1 will start to empty to go down this path. Which will eventually shut off the Zener and Thyristor so C1 begins to fill again causing this process to loop.
C1 will discharge if there is a path for it to discharge even when connected to the Voltage Source.
You’d have to do the math, the resistors in the circuit are assisting to keep the current and voltages close to the desired levels for this loop to occur.
@@burgercheezes thanks for the reply 🤩
This is a bit difficult to understand... 🫤
Do you have a more simple video about thyristors?
ruclips.net/user/shortsDmVQGEhwGx4?si=dNhvyKtxFwqy5FIT
Please bro some examples