Hi, I'm learning about this circuit below and it is a bit confusing. Please reference the images below related to the question.

The circuit is using a PUT with a Capacitor. According to the book, the PUT starts out with the Anode being blocked until it reaches at least as much voltage as is applied at the Gate, which given the 15K/27K voltage divider below, appears to be 3.85V. That means the Anode should open and allow current to flow through to the Cathode of the PUT once the Anode voltage reaches 3.85V.

What's confusing me about Capacitors in general is that from my perspective here, the 6V power supply and the Capacitor are both directly connected to the Anode.

From reading materials, I've uncovered 2 important points:
- The 470K resistor being in series with a capacitor makes the resistor have 0V drop across it, since the capacitor has infinite resistance (as is shown in the second screenshot in Figure 2-80 at the bottom).
- The capacitor is "directly connected" to Ground, and the power supply apparently is not.

My confusion is this:

It seems to me that once the capacitor builds up enough charge (I'm guessing at least 6V since it should not stop charging until it reaches near 6V, which means the potential matches that of the Supply), it will suddenly discharge the 6V through the Anode. But why would the Anode only accept the 6V from the capacitor and not the power supply? It seems to me that the power supply and capacitor are directly connected so I would think the power supply's 6V would break the barrier at the gate.

If I understand partly, it's because the power supply is not directly connected to the ground, thus there is no potential difference to channel current through.

First Question:
Why would the Anode only accept the 6V from the capacitor and not the power supply when the Anode is blocked? Simply because it's directly connected to Ground?

Second Question:
If the Gate is set to 3.85V as I believe, does the capacitor start discharging itself at 3.85V or does it wait to build up to 6V before discharging?

Third Question (select the condition that actually applies):
- If the capacitor does start discharging itself at 3.85V, wouldn't the PUT only stay open for a fraction of a second, just long enough for the capacitor charge to drop down to 3.84V and then close immediately? Or does it stay open long enough to let the full 3.85V discharge?

- If the capacitor actually waits until near the 6V to start discharging itself, does that mean that (6V - 3.85V = 2.15V) 2.15V is what flows down through the PUT until the level reaches below 3.85V, at which point the PUT closes the Anode?

Fourth Question:
If I removed the capacitor and shorted that line, the circuit would function normally, just at a very fast pace, correct? But if I made the capacitor and Open circuit, then the circuit would not work since the supply is not connected to Ground in any way except the closed Anode path?


Fifth Question:
If I were to add a resistor before or after the LED, would it delay the discharge (otherwise it would discharge instantly)? This author seemed to have no problem omitting a protective resistor in this example.

Thanks in advance for any help! I'm trying to wrap my head around when and why I would use a capacitor.

 

First Question:
Why would the Anode only accept the 6V from the capacitor and not the power supply when the Anode is blocked? Simply because it's directly connected to Ground?

The capacitor never reaches 6V because the PUT fires when the anode voltage reaches Vg+0.7V = ~4.6V (see http://www.circuitstoday.com/programmable-ujt https://www.allaboutcircuits.com/textbook/semiconductors/chpt-7/unijunction-transistor-ujt/)

Second Question:
If the Gate is set to 3.85V as I believe, does the capacitor start discharging itself at 3.85V or does it wait to build up to 6V before discharging?

The capacitor begins discharging through the PUT & LED when the PUT anode voltage reaches ~4.6V and the PUT enters its negative resistance region.

Third Question (select the condition that actually applies):
- If the capacitor does start discharging itself at 3.85V, wouldn't the PUT only stay open for a fraction of a second, just long enough for the capacitor charge to drop down to 3.84V and then close immediately? Or does it stay open long enough to let the full 3.85V discharge?
- If the capacitor actually waits until near the 6V to start discharging itself, does that mean that (6V - 3.85V = 2.15V) 2.15V is what flows down through the PUT until the level reaches below 3.85V, at which point the PUT closes the Anode?

I think what happens is that the capacitor discharges until the anode voltage reaches the valley point, which is referenced on the forward voltage of the LED in this circuit.

Fourth Question:
If I removed the capacitor and shorted that line, the circuit would function normally, just at a very fast pace, correct? But if I made the capacitor and Open circuit, then the circuit would not work since the supply is not connected to Ground in any way except the closed Anode path?

If you short the line, it is almost equivalent to making the capacitor massive, so the pace would be extremely slow. Making an open circuit is equivalent to using a very small capacitor, so the pace would be extremely fast - possibly faster than the PUT is able to respond.

Fifth Question:
If I were to add a resistor before or after the LED, would it delay the discharge (otherwise it would discharge instantly)? This author seemed to have no problem omitting a protective resistor in this example.

Yes, it would delay the full discharge. A protective resistor is not necessary when the energy stored in the capacitor is not sufficient to burn out the LED.

 

Hi RBR1317, I appreciate your response. Would you mind clarifying the response to the first question as well? Let me rewrite it based on your response so the correct questions is being asked:

Why would the Anode only accept the voltage from the capacitor and not the power supply when the Anode is in a state of blocking the path to the Cathode? Simply because it's directly connected to Ground?

I'm having trouble understanding this phenomenon because it almost seems like the Anode is "favoring" the capacitor's voltage over the supply's voltage, when there is a direct line to the Anode from both the supply and the capacitor. I would think that the 6V at the supply would be enough to break the potential difference at the Gate (~4.6V). I know I'm obviously wrong about this, just don't understand why.

Thanks again!

 

Why would the Anode only accept the voltage from the capacitor and not the power supply when the Anode is in a state of blocking the path to the Cathode?

The PUT anode is connected directly to the capacitor so it must "accept" the voltage from the capacitor directly. The PUT anode is not connected directly to the power supply but rather through the 470K resistor, so the current from the power supply that charges the capacitor causes a voltage drop across the 470K resistor which is the difference between the power supply voltage and the capacitor voltage. As the capacitor charges, the charging current falls and the voltage drop across the resistor also falls while the capacitor voltage increases.