Hi! Thank you very much for your video. Very helpful! At about 9:05 you said that the output of the regulator needs a load. But what if the LED burns over time due to aging and it would wouldn't be replaced for a certain period of time, can the L7805 get damaged? Thank you in advance.
A component can fail open or closed (short). If it fails open then no current will be drawn. A constant current driver will then increase the output voltage in an attempt to drive the current. On the other hand, if the LED fails shorted, the constant current driver will reduce the output voltage to limit and therefore maintain the constant current.
@@PizzeyTechnology Thank you very much for the answer. So if the component fails open, the regulator will slightly increase the voltage but since no current is flowing, there is no danger for the regulator to get destroyed, even not after a long period of time. But if the component fails short, the current flow from the regulator will be very high. But since the regulator has a built-in short circuit protection, I suppose that the regulator will not overheat or even get damaged, even if after a long period of time. In that case, the regulator could even act as a short circuit protection to the circuitry as well as for the power supply prior to the regulator. Are my thoughts correct? Thank you very much ones again for your help 👍 Btw, subscribed to your channel.
@@sesvid Nearly there, but not quite right. If the load fails "open" the constant current driver will not "slightly increase the voltage" -- it will increase the voltage as much as it can. This will be limited by the supply voltage (Vs or Vin) and the regulator drop (Vd). However, with no load current there will be no significant current through the voltage regulator. Power dissipated (P = VI) will therefore be minimal. No power = no heating. Thanks for the subscription.
Hi! Thanks for the video! I was looking for a temporary solution to drive a LED and your solution is quite nice!
Glad I could help
Can you make a video on this but for LM317 for constant current source? Will be used for laser TTL.
Hi! Thank you very much for your video. Very helpful!
At about 9:05 you said that the output of the regulator needs a load. But what if the LED burns over time due to aging and it would wouldn't be replaced for a certain period of time, can the L7805 get damaged?
Thank you in advance.
A component can fail open or closed (short). If it fails open then no current will be drawn. A constant current driver will then increase the output voltage in an attempt to drive the current. On the other hand, if the LED fails shorted, the constant current driver will reduce the output voltage to limit and therefore maintain the constant current.
@@PizzeyTechnology Thank you very much for the answer. So if the component fails open, the regulator will slightly increase the voltage but since no current is flowing, there is no danger for the regulator to get destroyed, even not after a long period of time.
But if the component fails short, the current flow from the regulator will be very high. But since the regulator has a built-in short circuit protection, I suppose that the regulator will not overheat or even get damaged, even if after a long period of time. In that case, the regulator could even act as a short circuit protection to the circuitry as well as for the power supply prior to the regulator. Are my thoughts correct?
Thank you very much ones again for your help 👍
Btw, subscribed to your channel.
@@sesvid Nearly there, but not quite right. If the load fails "open" the constant current driver will not "slightly increase the voltage" -- it will increase the voltage as much as it can. This will be limited by the supply voltage (Vs or Vin) and the regulator drop (Vd). However, with no load current there will be no significant current through the voltage regulator. Power dissipated (P = VI) will therefore be minimal. No power = no heating. Thanks for the subscription.
would LDO voltage regulators work in this instance too?
Hyy, I want to design 100v 40 khz supply using half bridge topology can you for me? I wish you will help me for the same... greetings from india..
Best of luck...