- #1
Jay_
- 183
- 0
I was trying to study the inductor role in Flyback converters and all, they seem to rely on the principle that the inductor is first energized while the switch is closed and then, it acts as a voltage source itself when the switch is opened.
All that is good, but I am confused with the POLARITY.
http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE5803
The simulations here seem to show all the current directions from negative to positive. Are they talking about the flow of electrons?
Either ways, I don't see when the inductor gets 65volts across it to power the neon lamp.
This is what I understood: First the inductor has a voltage of 10 when the switch is closed with A, then it goes reducing to zero. After the switch is closed with B, the voltage is formed across the inductor to power the neon-lamp.
But all the currents shown are negative to positive?
Can someone clear me on this? I am aware of the equations V = L(di/dt). So when the constant current becomes zero, voltage is there across the inductor, right? But the polarity?
All that is good, but I am confused with the POLARITY.
http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE5803
The simulations here seem to show all the current directions from negative to positive. Are they talking about the flow of electrons?
Either ways, I don't see when the inductor gets 65volts across it to power the neon lamp.
This is what I understood: First the inductor has a voltage of 10 when the switch is closed with A, then it goes reducing to zero. After the switch is closed with B, the voltage is formed across the inductor to power the neon-lamp.
But all the currents shown are negative to positive?
Can someone clear me on this? I am aware of the equations V = L(di/dt). So when the constant current becomes zero, voltage is there across the inductor, right? But the polarity?