Is it possible to modify this voltage booster to regulate input voltage instead of output voltage?

Thread Starter

rudyauction8

Joined Jan 27, 2012
250
OK here it is. The basic concept is based on the BQ2031 white paper I posted earlier. I adapted it to the LM2577 converter you are using.


- U1 has to be a rail to rail op-amp. I indicated an LMC6482 but you can use any well behaved slow r2r op-amp.
- You get to select the components that make the LM2577 work.
- R11 and D2 set the maximum voltage out. Vmax = 1.23V + Vzener. So, the 1N4743a is a 13 Volt zener and will give you a Vmax of about 14.2 volts. Good for a lead acid battery
- R13 sets the nominal current output. Iout = 1.23/R13.
- R4 sets the max power point. The sensed panel voltage is scaled to 50% by R1 and R2. So, your MP point is scaled at 50% also. Look at it as a pivot point. Panel voltage above this point will tell the regulator to output more current. Panel voltage below this point will tell the regulator to output less current.
- R8 sets how much the output current will change as the panel voltage changes. R9 will effect this also. If you can't get what you want by adjusting R8, you can lower R9 for more effect and increase R9 for less.

The components that are to the left of C2 (plus R9) make up the simple MPPT controller. All that circuit does is produce a inverted and scaled voltage based on the panel voltage. The components to the right of C2 is your simple switcher but it uses feedback from a current sensing resistor, so it will output a constant current. The reference for the LM2577 is 1.23 volts. If R 13 was a 1.23 ohm resistor the switcher would output a constant 1 amp. The MPPT control adds or subtracts to that feedback voltage. (Pin 2 of the LM2577 is the summing point for the current feedback and the MPPT control.) As the panel voltage drops, the MPPT controller produces a higher voltage. This voltage is added to the current feedback and will cause the LM2577 to output less current. Use the BQ2031 white paper as reference.

OK, there you go. It took me two beers and a cup of coffee to design this up. Feel free to ask questions, I will assist. I want you to keep us updated on this project. I especially want you to show Papabravo and tcmtech what a positive attitude can achieve. :D
Thanks a bunch, I'll have all the parts ordered from digikey tomorrow. I'll post updates when done.
 

Thread Starter

rudyauction8

Joined Jan 27, 2012
250
upload_2014-10-15_13-33-40.png


How's this look? I already have hundreds of resistors and capacitors. Specifically the inductor, op-amp and zener diodes - are they the correct components?
 

Lestraveled

Joined May 19, 2014
1,946
OK, looks good.....except (just want to check your numbers)......
The second item, the .5ohm, 1 watt resistor, I assume is your current sense resistor. .5 ohms X 4 = 2 ohms implies .75 amps out nominally. If this is what you are thinking, looks good.

The op amp is a dual, so you will need to take the unused section and make it a voltage follower (- input connected to output) and ground the + input.
 

Thread Starter

rudyauction8

Joined Jan 27, 2012
250
OK, looks good.....except (just want to check your numbers)......
The second item, the .5ohm, 1 watt resistor, I assume is your current sense resistor. .5 ohms X 4 = 2 ohms implies .75 amps out nominally. If this is what you are thinking, looks good.

The op amp is a dual, so you will need to take the unused section and make it a voltage follower (- input connected to output) and ground the + input.
I got 4 of the .5 ohm resistors so I could do .5 to 2 ohms at .5 ohm intervals to find the right value. Thanks for the tip for the opamp, I've only used one a couple times so I didn't know about that trick.
 
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