View attachment 113051

Thanks for simplifying the schematic, it makes it easier to understand. I made a new schematic with the help of the one you made:



My question is; did I make all the connections right? I have doubts where does the collector connect to? at the input or output of the voltage regulator? I would say at the output, but if i look at the schematic i get confused. Or do I have to switch the capacitors? 150uf at the output and 1500uf at the input?

I didnt bring the lamp in the schematic, dont plan on using it for the new box we are planning to make.

Thanks!

 

I would expect the 1500uF to be on the input and 150uF on the output of the regulator. The collectors of the phototransistors should connect to the regulator output. The phototransistor emitters go via resistors to ground and to each pole of the switch. You have the two poles of the switch linked when they shouldn't be. One of the millivolts terminals goes via a resistor to the pole of the switch and the other goes to ground.

 

I would expect the 1500uF to be on the input and 150uF on the output of the regulator. The collectors of the phototransistors should connect to the regulator output. The phototransistor emitters go via resistors to ground and to each pole of the switch. You have the two poles of the switch linked when they shouldn't be. One of the millivolts terminals goes via a resistor to the pole of the switch and the other goes to ground.

Ok I changed it now:


What I need to know in order to make the new box, is what is being measured at both the voltmeters.

On the voltmeter called volts, the voltage across the resistor connected to the ir-led is being measured.

On the other one called milivolts the voltage across a 1k+1k resistor is being measured. Why the 1k resistor connected to the plus side multimeter? Does that have any effect on what is being measured?

 

The 1k resistor wouldn't affect the measurement. I suspect it is there to protect the phototransistors if the meter terminals were shorted.

The circuit around the transistor is wrong. You had it right earlier. The left hand side of the pot should connect to ground instead of +12V and the collector of the transistor should connect to +12V.

 

The 1k resistor wouldn't affect the measurement. I suspect it is there to protect the phototransistors if the meter terminals were shorted.

The circuit around the transistor is wrong. You had it right earlier. The left hand side of the pot should connect to ground instead of +12V and the collector of the transistor should connect to +12V.

Sorry I don't get you, I drew the 10k pot both times the same. Do you mean theft hand side of the milivolts probe above? That would be strange because the plus side of the millivolts probe has a 1k resistor attached in real life, so i wouldn't change that.

 

Look at the connections around the transistor below.
Correct version:


Latest Version:

 

Look at the connections around the transistor below.
Correct version:
View attachment 113114

Latest Version:
View attachment 113115

Yes I see it now, I will change it. But why ground the base?

 

The base has a 1K resistor to the 10k pot and the other end of the pot is grounded. The +12V is onnected via 1k to the wiper of the pot.
When the wiper is at the grounded end of the pot there is no voltage on the base and so the transistor is off. When the wiper is at the other end of the pot the +12V is fed via the two 1k resistors to the base and so the transistor can conduct. As the pot is adjusted it will control the conduction of the transistor and so control the current through the LED.

 

Please ignore the fuse on the neutral side, it should not be there.

As always, connect the green wire to the chassis and use a GFI.

Thanks for the schematic , looks like a good way to test optos.

 

Today I made a new schematic:




I want to make a test box that has the same functionality, but with a better interface so to say.

At the milivolts probe, you are supposed to measure a voltage at FC2 between 0.01-0.05v bigger than FC1.
In the old test box, a multimeter is connected to read the output voltage of FC1 and FC2. When you have FC2 bigger than FC1 but within the given tolerance, you can move to next step of the proces.

What I drew is a op-amp comparator with FC1 as Vref and FC2 as the Vin. When the comparator see's a voltage at Vin bigger (between 0.01-0.05v ) than Vref, then a lamp or led lights up.

My question is, how does the comparator know that when he see's a higher voltage (between 0.01-0.05v) at Vin, that he should do something?

Another thing I changed, are the multimeters. I actually only need one, because the lamp or led lights up when you have the 0.01-0.05v difference at FC2.

After you do that you hit the triple pole switch so that the FC2 value (can also be FC1) is connected to the voltmeter. Then you adjust the voltage till you get 70mV on the voltmeter (part of the test instrcutions), hit the swtich again, and read the value across the resistor, that one should be between 0.15-0.45v.

Am I going the right way here?

Thanks!

 

how does the comparator know that when he see's a higher voltage (between 0.01-0.05v) at Vin, that he should do something?

If you must detect that the difference lies between two values then you need a 'window comparator' (comprising essentially 2 comparators with different respective reference voltages).

 

If you must detect that the difference lies between two values then you need a 'window comparator' (comprising essentially 2 comparators with different respective reference voltages).

Ok, I made a window comparator schematic to detect the voltage difference I want:


First I put a differential amplifier, with equal resistances so in that way, A=1, so Vout= V2-V1.
The inputs are FC1 and FC2, the voltage difference is the ouptut that becomes the input voltage of the window comparator.

The window comparator schematic I got from here:
http://www.learningaboutelectronics.com/Articles/Window-comparator-circuit.php

They say that the 4049-inverter is needed in order to get it working.

For the Vref, (0.01v & 0.05v), do I have to make two separate voltage supplies therefore? or is there another way around it perhaps?

 

For your window comparator to work you will need a negative supply (as well as the +5V) for the 741 opamps, because they can't handle input voltages close to 0V. An LM324 (quad opamp) would be a better choice than the 741's.
If you used the first stage to provide a gain of, say, 10 then the reference voltages for the second stage could be correspondingly larger (which would make them easier to check). A 3-resistor voltage divider from the 5V rail could be used to provide both reference voltages.

 

You could use a spare op-amp instead of the '4049 and then you won't need the 5V supply - everything can work from the 12V already available. The first op-amp needs to be a low input offset device as it is required to work accurately with a 10mV difference signal. The LM324 is 2mV.

 

Ok, I changed the 741's to 324's:


From what I read the 324's dont need a negative supply, so that makes it easier.

In the first stage I use a gain of 10 with the help of the transistor voltage amplifier. I thought i could use the differential amplifier to provide the gain, but looking at the formula Vout= -R2/R1 (V2-V1), I notice that i cant apply it because V1 and V2 are not constant. (they are between 2v-4v). So that's why I decided for the transistor amplifier to provide the gain.

Is this the best/easier way to do it?

The supply voltages and ref voltages i get from the voltage divider from the 12v rail. I have yet to calculate the resistor values.

And at the last stage i removed the 4049 inverter, i don't get why you have to invert the signal and looking at this example, i see it is not used.
http://www.learningaboutelectronics.com/Articles/Modified-window-comparator-circuit.php

 

You cannot use the transistor amplifier. As it is capacitor coupled it will only work with AC signals but here the signals are DC.
You can get the gain from the differential amplifier.

Do I understand correctly that you also need to measure the actual voltage on FC1 or FC2?
If so how is this arranged with this circuit?

[EDIT] You can use the 12V supply for the op-amps, you don't need the separate 5V supply if you are not using the '4049.

 

Do I understand correctly that you also need to measure the actual voltage on FC1 or FC2?
If so how is this arranged with this circuit?

Yes, to make it clear this is what the old test box does:

1. First you connect the sensor to the box. Then you adjust the voltage across the LED resistor until you read 4.5v on the multimeter.
   
   
2. Then you measure FC1&FC2 (with the help of a toggle switch). You have to adjust the backplate of the sensor until FC2 is 0.01-0.05v bigger than FC1. (this usually takes a while, you have to continually toggle the switch for the FC1 or FC2 position, look at the multimeter, check their values and do the subtraction. I want to do this with lights e.g green light for when FC2 is 0.01-0.05v bigger than FC1, orange when its 0.05-0.1v bigger, and red when the it's more than 1v bigger).
   
   
3. When FC2 is within 0.01-0.05v bigger than FC1, you have to adjust the voltage again until you read 70mv coming out of FC2 or FC1.
   
   
4. After you read 70mV on the FC multimeter, you read voltage on the multimeter connected across the LED resistor. This voltage should be between 0.15-0.45v.
   This is the old test box schematic:
   
   
   One of the changes i want to bring on the new test box is what i stated at step #2. I want a green light to burn when FC2 is 0.01-0.05v bigger than FC1. An orange light for when its 0.05-0.1v bigger, and a red one when it's more than 1v bigger.
   
   Right now im working on that circuit, this is where I'm a bit stuck at the moment:
   
   I want the differential amplifier to provide a gain of 10 (or more) like Alec said, that way the reference voltages of the window comparator are also larger which makes them easier to check.
   
   But how could i do that if the inputs of the differential amplifier (FC1&FC2), are not always the same?
   For the differential amplifier the gain formula is: Vout= -R2/R1 (V2-V1), I understand that if all the resistors are the same, the gain is equal to 1. In this case V2 and V1 are not constant, that would mean i would need to be changing the resistor values to be getting the same gain.
   
   With some sensors you get FC values of about 2.5v, other ones like 3.6v etc. I dont need to know how much exactly, what I want to know is thier difference and then amplify that difference so its easier to use in the window comparator.
   
   That's why I used the transistor voltage amplifier, but I wasn't aware that it was for AC signals.
   
   So how could i do that?
   
   
   
   
   

 

"Vout= -R2/R1 (V2-V1)"
The output is the ratio of the resistors multiplied by the bit in brackets. This is the difference between V2 and V1. So 1V and 1.!V will give the same output as 3.3V and 3.4V. That is the magic of that circuit.

 

You might need to null any intrinsic offset of at least the first stage opamp. Here's how you could do that :