Solar Tracker Circuit Is Giving Me Issues - Appreciate Anyone's Help

[mneary]

Do you have any other way to keep the motors stopped in darkness?

 

[rs14smith]

Do you have any other way to keep the motors stopped in darkness?

What do you mean? Do I have any other ideas?

 

[mneary]

Well, right now everything would work except that the circuit doesn't have a definitive state when in darkness. Is there a signal elsewhere in your system that could cut motor power off when there's no solar power?

 

[mneary]

hmm interesting, so your saying from this schematic, Last One, since they are both showing the same voltage that really the motor is OFF, or is the motor working against itself? I mean my logic behind it is like both sides (polarities) of the motor are fighting against each other like they are in a tug of war match lol ?

So if that is true, will that damage the motor over time? My logic is surely wrong but that's what I think of when I actually see a voltage reading, and not two 0 voltages for both digital multimeters when the solar tracker is getting equal sunlight.

The motor doesn't know what the voltage is on any of its terminals. It only know the difference between them.

 

[rs14smith]

The motor doesn't know what the voltage is on any of its terminals. It only know the difference between them.

Ohh okay I think I understand now.

So the motor will never react until it senses a difference in voltage from either side, and if both sides have the same voltage, rather that's 50v from the right side and 50 v from the left side, or 0v from the right side and 0v from the left side, the motor will not move?

 

[rs14smith]

Well, right now everything would work except that the circuit doesn't have a definitive state when in darkness. Is there a signal elsewhere in your system that could cut motor power off when there's no solar power?

Not really I mean, you know as much as I do about the whole system at the moment, but we could surely create something to do that if you know of a way I could integrate that into the circuit...?

At the moment I just don't have enough experience with electrical components to think of what else I could possible do to achieve this ya know?

I'm better at programming than having to use electrical components to do the job:

example:
if (A == 0volts and B == 0volts)
{
turn off;
}

lol, I have another semester until I start taking Micro. classes, and probably would have had this project done in no time


I'm starting to wonder if we could some how make use of another portion of the OP Amp and use another phototransistor who's primary job would only be to detect if there is light or not, and some how make that part of the OP Amp act as a switch for the entire circuit. I have like 2 more logics I can use from this 1 OP Amp but I'm not sure how to fully go about implementing that in my circuit

 

[john1]

Obligatory words to permit post

 

[rs14smith]

Obligatory words to permit post

Yeah for the past 30min I've been trying another idea of using a FET as a switch some way to control or act as a signal to determine rather or not to power the rest of the circuit.

The only problem I keep running into is, for most FETs I've ran into, they need a lot of voltage to open the gate, and for my condition, I'll only be using a phototransistor which has a max of 1V so, that wont be enough to even open the gate if I wanted to. I'm not sure if they make other FETs that allow a low voltage to open the gate and has a low forward voltage/current drop?

 

[mneary]

What is the circuit you're using as the input voltage source? You mention phototransistor?

 

[rs14smith]

What is the circuit you're using as the input voltage source? You mention phototransistor?

That's what A and B are, my phototransistors. I'm just going to replace both of those power sources with two phototransistors which detect where the sun is at.

So I don't have a whole separate circuit for just those two components, they are basically in all the circuits I've uploaded here, but we've just been manually been creating what state each phototransistor would be in.

Hope that makes sense

 

[mneary]

Reason I ask is, phototransistors don't usually generate a 0-1V signal.

 

[rs14smith]

Reason I ask is, phototransistors don't usually generate a 0-1V signal.

When I hooked mine up to my digital multimeter and placed full light on it, it gave me a reading of 1V.

I used the black ones from this set from radioshack: Matched Infrared Emitter and Phototransistor Detector - RadioShack.com

 

[mneary]

OK, according to some of the feedback at the radioshack, buyers say the 'phototransistor' is actually a photo diode. Photo diodes do generate noticeable current in sunlight, and look like open circuit when dark.

It means we can't depend on continuity in the diode during darkness. This isn't good or bad, but it influences the strategy for detecting darkness.

 

[rs14smith]

OK, according to some of the feedback at the radioshack, buyers say the 'phototransistor' is actually a photo diode. Photo diodes do generate noticeable current in sunlight, and look like open circuit when dark.

It means we can't depend on continuity in the diode during darkness. This isn't good or bad, but it influences the strategy for detecting darkness.

Well I mean the only reason I used it as I knew I needed something to detect the light, but was not sure what was the best and easiest component to work with, so if using another type of diode or light capturing component will make our lives easier, I'm all for it

 

[nickelflippr]

I'm starting to wonder if we could some how make use of another portion of the OP Amp and use another phototransistor who's primary job would only be to detect if there is light or not, and some how make that part of the OP Amp act as a switch for the entire circuit. I have like 2 more logics I can use from this 1 OP Amp but I'm not sure how to fully go about implementing that in my circuit

One thought here is to use a "light activated switch" to turn on the motor ground thru a logic level Nfet transistor. No light, then no ground for the motors. Look on the net, there should be plenty of circuit examples.

The Shack sells a little assortment pack of light dependent resistors, or LDR's. Use one of those and a cheapo comparator like a LM393, a high value potentiometer (200KΩ), four resistors (I think) and a good sized logic level Nfet to handle the anticipated motor current.

 

[rs14smith]

Okay, I made some changes to the circuit, and from a series of test, it's working perfectly, but I believe someone said something about it was wrong, but from my test, the circuit works exactly like we want it to.

A > B


A < B


A = B (as you can see the voltage is the same. That's what we want)


A = B (dead zone, again, the voltage is the same. That's what we want)


Lastly, I even did a test using LED's as it's a lot easier to see what is going on, and again, the test passed with flying colors!
**broken link removed**

Both when A = B (dead zone & when the sun is directly in the middle of the phototransistors/diode) the voltage to the motor is so low that it will not even turn on (saving power) which is good, and as you can see from short clip I made, the LEDs DO NOT turn on .

So my question is, what exactly about this current circuit creates an issue where we need all the other components I had before?

 

[mneary]

The third case where there's sun on both photodiodes is a problem. You'll find that Q3 and Q4 are both on, which would be OK, but they drive the gates of Q1 and Q2. The reason you're measuring 832mV on both sides of the motor is the N-channel transistors are stronger than the P-channel. But the P-channel transistors are still fighting really hard and are sending a lot of current through. Put an ammeter in the drain of Q1 and/or Q2. You should see 50 amperes going through each IRF5305.

Both nodes being at 11V+ is not a problem because this leaves the P-channel MOSFETs off.

Another problem is logical. The motor will run whenever the light isn't exactly the same, and moves to track the smallest of illumination changes. That's why I designed a range of 0.2V (we can adjust it later) where it's "close enough" and the motor isn't hunting. If there's no stopped range, the motor will jump back and forth whenever the light on one of the diodes changes just 0.05%. This is what the 10K resistors were there for.

 

[rs14smith]

The third case where there's sun on both photodiodes is a problem. You'll find that Q3 and Q4 are both on, which would be OK, but they drive the gates of Q1 and Q2. The reason you're measuring 832mV on both sides of the motor is the N-channel transistors are stronger than the P-channel. But the P-channel transistors are still fighting really hard and are sending a lot of current through. Put an ammeter in the drain of Q1 and/or Q2. You should see 50 amperes going through each IRF5305.

Both nodes being at 11V+ is not a problem because this leaves the P-channel MOSFETs off.

Another problem is logical. The motor will run whenever the light isn't exactly the same, and moves to track the smallest of illumination changes. That's why I designed a range of 0.2V (we can adjust it later) where it's "close enough" and the motor isn't hunting. If there's no stopped range, the motor will jump back and forth whenever the light on one of the diodes changes just 0.05%. This is what the 10K resistors were there for.

ahhh okay...back to the drawing board again I guess lol, but before I convert it back, can you tell me what I'm doing wrong when I'm trying to check the currect on Q1 as I know you have to break the circuit when you check the current, but I can't seem to get it to display the 50A you are referring to, so I know I did something silly in the schematic below:

 

[rs14smith]

Okay it seems my multimeter was turned in the wrong direction I'm assuming, but now it's showing like 113A LOL!!:



I'm going to switch the circuit back to the original and check that part again, as I had no idea what you were doing was effecting those two FETs.

 

[nickelflippr]

What if you run the each of the Op amp outputs thru (two) two input Nand gates. Attach opposing Op amp outputs for the second input of each gate.

Logic:
Inputs  Output
A    B    X
0    0    1
0    1    1
1    0    1
1    1    0