If there is a capacitor between two diodes, will this circuit burn the capacitor?

sky999

If there is a capacitor between two diodes, will this circuit burn the capacitor? [Copy link]

 

The +7V on the left is a common power supply. The current is relatively small, but it is always on. The +5V on the right is normally suspended, but sometimes it is inserted and the current is relatively large.

The +5V on the right will disconnect the +7V on the left through another MOS tube control, which means that normally C1 is 7V. When 5V is inserted, 7V will be disconnected.

If the on-off frequency of +5V is too high, will the capacitor C1 burn out? Behind N1 is the TP4065 charging circuit

maychang

"If the on-off frequency of +5V is too high, will the capacitor C1 burn out?"

The on-off frequency will reach hundreds of kHz before it will have a significant impact on C1.

littleshrimp

The diode C1 will only be charged by 7V or 5V and discharged by the load. When the capacitor is charged to 7V, 5V will be turned on and will not affect the capacitor.

sky999

maychang posted on 2022-11-16 21:41 "If the on-off frequency of +5V is too high, will the capacitor C1 burn out? 】 The on-off frequency will reach hundreds of kHz before it will have a significant impact on C1.

Normally it won't have any impact, right?

tagetage

maychang posted on 2022-11-16 21:41 "If the on-off frequency of +5V is too high, will the capacitor C1 burn out? 】 The on-off frequency will reach hundreds of kHz before it will have a significant impact on C1.

I don't understand what the impact will be, can you explain it in detail?

maychang

tagetage posted on 2022-11-16 22:23 I don't understand what the impact will be, can you explain it in detail? ?

Capacitors are not ideal and always have certain losses, especially electrolytic capacitors. The loss of capacitors is manifested as heat, so capacitors cannot pass too much AC current, otherwise they may be damaged due to excessive temperature. The capacitor C1 in this circuit is an electrolytic capacitor, so it is also subject to this rule.

damiaa

The power supply is switched manually. I don't think it will happen that often. Don't worry too much about C1.

fxyc87

TP4065 cannot use 7V voltage, right?

tagetage

tagetage posted on 2022-11-16 22:23 I don't understand what the impact will be, can you explain it in detail? ?

Thanks. Got it.

mingplus

If there is no soft-start when switching between power sources, you need to consider whether C1, D2 & D1 can withstand the inrush current at the moment of switching. If there is soft-start, there will be no problem.

sky999

fxyc87 posted on 2022-11-17 09:00 TP4065 cannot use 7V voltage, right?

Up to 8V is fine

sky999

maychang published on 2022-11-17 07:59 Capacitors are not ideal and always have certain losses. Especially electrolytic capacitors. The loss of capacitors is manifested as heat, so capacitors cannot pass too large AC...

If in this circuit, the voltage on both sides is 5V, with a fluctuation of ±0.1V, that is, it is possible that they are both 5.0V at the same time, what will happen?

maychang

sky999 posted on 2022-11-17 14:32 If in this circuit, the voltages on both sides are 5V, with a fluctuation of ±0.1V, that is, it is possible that both are 5.0V at the same time, then there will be...

If both sides are 5V, then the load (N1) can only get 4.3V~4.6V voltage, because 1N4001 has a considerable forward voltage drop. As for "what will happen then?", most of the time nothing will happen.

sky999

maychang posted on 2022-11-17 15:31 If both sides are 5V, then the load (at N1) can only get 4.3V~4.6V voltage, because 1N4001 has a considerable forward voltage drop. As for "what will happen then...

I mean, if the cathode voltage of a diode is greater than the anode voltage, the diode will be cut off, right? What if the voltages at both ends of the diode are equal? Let's assume that the cathode of the diode is 4.6V, the anode of the diode is 5V, and the voltage drop of the diode is exactly 0.4V. Here, 5V and 4.6V are two power supplies. Will these two power supplies become parallel?

maychang

sky999 posted on 2022-11-17 16:58 I mean, if the cathode voltage of the diode is greater than the anode voltage, then the diode will be cut off, right? What if the voltages at both ends of the diode are equal? Then...

"If the cathode voltage of a diode is greater than the anode voltage, the diode will be cut off, right? What happens if the voltages at both ends of the diode are equal?"

The voltage across the diode is equal, so there must be no current flowing through it.

maychang

sky999 posted on 2022-11-17 16:58 I mean, if the cathode voltage of the diode is greater than the anode voltage, then the diode will be cut off, right? What if the voltages at both ends of the diode are equal? Then...

"Assuming the cathode of the diode is 4.6V, the anode of the diode is 5V, the voltage drop of the diode is exactly 0.4V"

This 0.4V is the forward voltage drop of the diode, which is caused by a certain current passing through the diode.

Moreover, the cathode of the diode is 4.6V and the anode is 5V, which does not mean that the voltages at both ends of the diode are equal.

maychang

sky999 posted on 2022-11-17 16:58 I mean, if the cathode voltage of the diode is greater than the anode voltage, then the diode will be cut off, right? What if the voltages at both ends of the diode are equal? Then...

"The 5V and 4.6V here are two power supplies. Will these two power supplies be connected in parallel?"

5V and 4.6V are two power supplies. If there is no load or other circuit, the diode will pass current (quite small), and the 5V power supply will be discharged and the 4.6V power supply will be charged. At this time, the two power supplies do not form a parallel relationship.

If there is a load, then both power supplies may be in a discharge state, and a small current flows through the diode. At this time, the two power supplies can be said to be in a parallel relationship because both discharge to the load.

sky999

maychang posted on 2022-11-17 17:12 "Assuming that the cathode of the diode is 4.6V, the anode of the diode is 5V, and the voltage drop of the diode is exactly 0.4V" This 0.4V is the forward voltage of the diode...

Oh, I see. That is to say, even if both ends are 5.00V, no current will flow to the positive electrode of the diode.

吾妻思萌

『If the switching frequency of +5V is too high, will it burn the capacitor C1? 』 The switching frequency will reach hundreds of kHz before it will have a significant impact on C1.

Teacher, I remember someone said that capacitors are passive components. They are hard to be damaged except for overvoltage and overload.

maychang

Azuma Simeng posted on 2023-1-19 06:50 Teacher, I remember someone said that capacitors are passive components. Except for overvoltage and overload, they seem to be difficult to damage

Passive components are nothing more than resistors, capacitors, and inductors. Among these three components, capacitors may be the most easily damaged. However, this has a lot to do with the manufacturing process. When the process level is improved, the damage rate is often reduced a lot.