Editing MPPT Solar and Wind Power Boost Charge Controller

Jump to navigation Jump to search

Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.

The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then save the changes below to finish undoing the edit.

Latest revision Your text
Line 1: Line 1:
 
ALL MY PROJECTS ARE HERE: https://opencircuits.com/index.php?title=User:Definitionofis
 
ALL MY PROJECTS ARE HERE: https://opencircuits.com/index.php?title=User:Definitionofis
  
This circuit is an analogue computer when run without a microprocessor. As revolutions per minute voltage rises, current is allowed to rise proportionally. The variable loading math is P = I(V-offset) and (V-offset)/I = R. It behaves like a fixed resistor load, R, although it is feeding a battery which has a hugely variable resistance. Also I x R = V. The op amp pegs I=V. Read about source to load "impedance matching" elsewhere to understand the point of this.
+
This circuit is an analogue computer when run without a microprocessor. As revolutions per minute voltage rises, current is allowed to rise proportionally. The variable loading math is P = I(V-offset) and (V-offset)/I = R. It behaves like a fixed resistor load, R, although it is feeding a battery which has a hugely variable resistance) and I x R = V. The op amp pegs I equal to V. Read about source to load "impedance matching" elsewhere to understand the point of this.
  
Here is a statement on the circuit diagram that explains the concept that the I and V input values are doing something to the output voltage to accomplish the above described load relationship. You understand that you can put more or less load on a generator or solar panel. Here is how: [The IN voltage is raised by lowering the OUT volts. The IN current is raised by raising the OUT volts. Thus the OP AMP feeds an influence on FB which is confusing until you grasp this. The FB pin only needs +-50mV to move the output radically.] The input current and input volts you notice go in opposite directions. So you can move them around until they are equal, thus doubling current and doubling voltage is four times the power; 2 squared=4x power, 3 squared = 9x power, 4 squared = 16x power. If I and V are not made equal then power is indeterminate, unknown, not a squared relation, and the resistance V/I is changing and not maintaining impedance matching to the source of power.  
+
Here is a statement on the circuit diagram that explains the concept that the I and V input values are doing something to the output voltage to accomplish the above described load relationship. You understand that you can put more or less load on a generator. Here is how: [The IN voltage is raised by lowering the OUT volts. The IN current is raised by raising the OUT volts. Thus the OP AMP feeds an influence on FB which is confusing until you grasp this. The FB pin only needs +-50mV to move the output radically.] The input current and input volts you notice go in opposite directions. So you can move them around until they are equal, thus doubling current and doubling voltage is four times the power; 2 squared=4x power, 3 squared = 9x power, 4 squared = 16x power. If I and V are not made equal then power is indeterminate, unknown, not a squared relation, and the resistance V/I is changing and not maintaining impedance matching to the source of power.  
  
 
How do you set the offset? A variable resistor in the circuit does that and you must select a value which is a manually selected setting of the maximum power point. Thought experiment: You could put a high wattage fixed resistor as the load and move that resistance up and down to find the maximum power transfer point and then remove that resistor load and attach this circuit in its place and move the offset variable resistor to cause that same favourable voltage across the input terminals and then you know this circuit is acting like the same fixed resistor. If you attached directly to the battery without this circuit, you do not get a fixed resistor effect.
 
How do you set the offset? A variable resistor in the circuit does that and you must select a value which is a manually selected setting of the maximum power point. Thought experiment: You could put a high wattage fixed resistor as the load and move that resistance up and down to find the maximum power transfer point and then remove that resistor load and attach this circuit in its place and move the offset variable resistor to cause that same favourable voltage across the input terminals and then you know this circuit is acting like the same fixed resistor. If you attached directly to the battery without this circuit, you do not get a fixed resistor effect.

Please note that all contributions to OpenCircuits may be edited, altered, or removed by other contributors. If you do not want your writing to be edited mercilessly, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource (see OpenCircuits:Copyrights for details). Do not submit copyrighted work without permission!

Cancel Editing help (opens in new window)