solar tracker

[foTONICS]

Hey all,

So I wanted to design a solar tracker for my final project and found this very helpful schematic online. I understand almost all of it except for a few things.

1. Why the feedback loop on IC1a and IC1c, is it some sort of gain control? and why would it need to be tied to 9 volts?

2. What is the purpose of IC1d, is it just inverting the signal from IC1c?

3. Caps C1, C2, C4, and C5, are these just smoothing caps for noise or do they serve a different purpose?

Any help would be greatly appreciated, here is the link where I found the schematic:

http://www.phoenixnavigation.com/ptbc/articles/ptbc55.htm

The schematic is attached:

Thanks again everyone!

 

[CDRIVE]

Hey all,

So I wanted to design a solar tracker for my final project and found this very helpful schematic online. I understand almost all of it except for a few things.

1. Why the feedback loop on IC1a and IC1c, is it some sort of gain control? and why would it need to be tied to 9 volts?

Yes, it's typical negative feedback. IC1a,b & c have a gain of 10. They're powered from the 9V rail because it's regulated.

2. What is the purpose of IC1d, is it just inverting the signal from IC1c?

Yes

3. Caps C1, C2, C4, and C5, are these just smoothing caps for noise or do they serve a different purpose?

Noise filtering

Does this help?
Chris

 

[foTONICS]

So the 9V part is where I get a little confused. I didn't know you could power the op amp like that. So the supply rails shown for that first IC aren't shown for clarification purposes or they are purposefully left disconnected. If this is the case we never went over that in my linear circuits class

 

[CDRIVE]

So the 9V part is where I get a little confused. I didn't know you could power the op amp like that. So the supply rails shown for that first IC aren't shown for clarification purposes or they are purposefully left disconnected. If this is the case we never went over that in my linear circuits class

The LM339 is an open drain quad comparator. It operates from a single ended supply. It's common practice to indicate Vcc to only one of the comparators because the same Vcc node powers all four comparators within the same chip. The same convention applies to the ground pin. You do realize that IC1, a, b, c & d are all the same chip?

Chris

 

[foTONICS]

Ya I understand that they are all on one chip. I've just never seen a circuit where their outputs are powered from a separate source. I'm used to seeing simple versions where only the inverting and non inverting inputs are powered

 

[duke37]

Comparators often have outputs that can only pull down so to get a rise inoutput voltage, it is necessary to have a resistor to the supply.
The outputs of more than one comparator can be connected together whereas two op-amps cannot.

 

[CDRIVE]

Ya I understand that they are all on one chip. I've just never seen a circuit where their outputs are powered from a separate source. I'm used to seeing simple versions where only the inverting and non inverting inputs are powered

Perhaps I misunderstood your question. Are you referring to what duke describes below? If so that's why I specified that this comparator is 'Open Drain' output. IE, it can't source current. It only sinks current. Hence the need for pull-up resistors on the outputs.

Chris

Comparators often have outputs that can only pull down so to get a rise inoutput voltage, it is necessary to have a resistor to the supply.
The outputs of more than one comparator can be connected together whereas two op-amps cannot.

 

[foTONICS]

OK this is making more sense now, thanks a lot for all the help. It made things so much clearer

 

[foTONICS]

So I did a little research and this is what I get from it. If the + input of IC1a is greater than the - input then the output goes into high impedance causing LED2 to turn on cause it sees the 9 volts. But if the opposite is true it sinks the current through R4 turning off LED2?

 

[CDRIVE]

So I did a little research and this is what I get from it. If the + input of IC1a is greater than the - input then the output goes into high impedance causing LED2 to turn on cause it sees the 9 volts. But if the opposite is true it sinks the current through R4 turning off LED2?

To be more precise .. If IC1a's output is low a current path from the 9V rail through R4 is sinked to ground. If IC1a output is high a current path from the 9V rail through R4 is routed though R21 to the base of Q4. This turns Q4's Collector - Emitter junction on which turns on LED2. The same logic applies to IC1c as it relates to R15, R19, Q3 and LED1.

Chris

 

[foTONICS]

So I was running just a small part of the circuit ( IC1a, C1, C2, R1, R2, R3, R4, VR1, the photoresistors, and the 3904 that the output controls) and was wondering what the purpose of the feedback is? I understand how and why feedback is used but when I take out the 1M feedback resistor I dont see a change, what's the point of it??