Hello, Could I ask a question?
- Thread starter opalstear
- Start date
Ahhh, nope - it won't work. Sorry.
You're trying to use N-ch MOSFETs, but no way to get the gate 10v higher than the source terminals.
You've drawn the body diodes backwards.
You've put a label next to the three-input OR gate indicating that it's supposed to be a "discharge mode" - but not what it's supposed to discharge?
You have a six-input AND gate with five inputs disconnected or "floating". This means the output will always be a logical "0", so the output from the inverter will always be a logic "1". However, since the gate of the MOSFET won't go any higher than the supply voltage, the MOSFET will make a nice room heater right until it burns up.
Over on the right, you have a blob (maybe a resistor?) labeled "Load or charger". Ahh, which is it? Are you trying to make a general purpose supply, or is that going to be an input from a battery charger, or do you want to charge batteries with it? Unfortunately though, the diode to the left of that will act as a short across the "load or charger", and will either cause the supply on the left to burn up, or it will burn up itself.
You're trying to use N-ch MOSFETs, but no way to get the gate 10v higher than the source terminals.
You've drawn the body diodes backwards.
You've put a label next to the three-input OR gate indicating that it's supposed to be a "discharge mode" - but not what it's supposed to discharge?
You have a six-input AND gate with five inputs disconnected or "floating". This means the output will always be a logical "0", so the output from the inverter will always be a logic "1". However, since the gate of the MOSFET won't go any higher than the supply voltage, the MOSFET will make a nice room heater right until it burns up.
Over on the right, you have a blob (maybe a resistor?) labeled "Load or charger". Ahh, which is it? Are you trying to make a general purpose supply, or is that going to be an input from a battery charger, or do you want to charge batteries with it? Unfortunately though, the diode to the left of that will act as a short across the "load or charger", and will either cause the supply on the left to burn up, or it will burn up itself.
