I do not have a scope therefore, it might not be possible in the circuit because as I mentioned in my earlier post, the ramp generated by the integrator (the first half of KA358A) is fed into an external voltage cut-out device (VCO) which trips at some preset voltage. The output from the VCO is proportional to the ramp input until it trips. When the ramp reaches the preset value, the output is dropped to 735 - 750mV.

This output from the VCO is fed into the inv. amp (the 2nd half of KA358A). As long as this VCO-output is below the preset value, the output of the inv. amp is 0V. And when VCO-output is dropped to 750mV, the output of the inv. amp should rise to 12.7V (instead of 1.4V).

You don't need a scope. Why would you need anything other than the same tools that you used to measure the input and output for the data you've already given?

 

Perhaps, I could not explain you clearly.

The input to the VCO is a ramp 5V - 29V (the output from the integrator).

The VCO cuts-off the output at about 14.4V. Hence, the output from the VCO (which is actually Vin) is also a ramp from 4.2V to 14.4V. During this period the Vo remains 0V.

Right after 14.4V, the output of the VCO (Vin) drops to 735-750mV. And hence, Vo should go high to 12.7V ut I am getting 1.4V.


The start-up values, Vin=4.2V - Vref=1V - Vo=0V
During the ramp, the Vin=4.3V-14.4V - Vref=1V - Vo=0V
At the end, Vin=750mV - Vref=1V - Vo=1.4V

What other readings shall I get you and how? I am sorry

I attach a graph, which might give you a better idea.

 

Perhaps, I could not explain you clearly.

The input to the VCO is a ramp 5V - 29V (the output from the integrator).

The VCO cuts-off the output at about 14.4V. Hence, the output from the VCO (which is actually Vin) is also a ramp from 4.2V to 14.4V. During this period the Vo remains 0V.

Right after 14.4V, the output of the VCO (Vin) drops to 735-750mV. And hence, Vo should go high to 12.7V ut I am getting 1.4V.


The start-up values, Vin=4.2V - Vref=1V - Vo=0V
During the ramp, the Vin=4.3V-14.4V - Vref=1V - Vo=0V
At the end, Vin=750mV - Vref=1V - Vo=1.4V

What other readings shall I get you and how? I am sorry

I attach a graph, which might give you a better idea.

Don't use hyphens to separate values like this. The following:

Vin=750mV - Vref=1V - Vo=1.4V

Says that Vin is equal to (750mV-Vref), which is equal to (1V-Vo), which is equal to 1.4V (and, hence, Vin = 1.4V, Vref = -0.65V, and Vo = -0.4V).

In other words, don't use mathematical symbols as separators in mathematical expressions. Use a comma or a semicolon instead.

The graph you included is what? A simulation? The actual physical measurements (I though you said you said you didn't have a scope)?

How are you taking the physical measurements that you have been reporting?

What is preventing you from swapping out your Vin source to a poteniometer that will allow you to vary Vin?

What is preventing you from varying Vr and putting together a comparable table?

 

don't use mathematical symbols as separators in mathematical expressions. Use a comma or a semicolon instead.

Hmm, you are right. It can be confusing. I shall avoid in future.

The graph you included is what? A simulation? The actual physical measurements (I though you said you said you didn't have a scope)?

How are you taking the physical measurements that you have been reporting?

Yes, the graph is a simulation. For any new project, first I check my design on the simulator. When everything is okay, I go for physical assembling of the components. I don't know if it is a good practice or not but I feel comfortable with it.

Physical measurements were taken with a digital multimeter

What is preventing you from swapping out your Vin source to a poteniometer that will allow you to vary Vin?

What is preventing you from varying Vr and putting together a comparable table?

Do you want me to take out the chip from the circuit and then record Vo with variable values of Vin and Vref?

Or you want me to induct the POTs within the circuit? If so, how can the POT stabilize the constantly increasing voltage at the Vin source? It can be used when the voltage stops increasing. Perhaps you want me to get the readings just before the Vin becomes 735mV.

 

Do you want me to take out the chip from the circuit and then record Vo with variable values of Vin and Vref?

I'm not sure what you mean by this. Taking the chip out of that circuit and then making the measurements with that circuit doesn't make sense.

I'm suggesting that you test this part of the circuit in isolation. Implement just this amplifier on a breadboard and see how it behaves. Use a pot or other variable voltage source (can even be just putting different valued resistors in a voltage divider).

What is Vo connected to right now? It is possible that the reason that Vo isn't rising above 1.4V is because it is being loaded down by something?

Or you want me to induct the POTs within the circuit? If so, how can the POT stabilize the constantly increasing voltage at the Vin source? It can be used when the voltage stops increasing. Perhaps you want me to get the readings just before the Vin becomes 735mV.

If you only have a multimeter, then you need to take static measurements. You can't have Vin or Vr changing while you are taking the measurement. So you would DISCONNECT the output from the VCO from the input to this circuit and REPLACE it with an alternate signal source, such as a pot, that will allow you to apply static voltage levels to the inputs.

 

I'm not sure what you mean by this. Taking the chip out of that circuit and then making the measurements with that circuit doesn't make sense.

I'm suggesting that you test this part of the circuit in isolation. Implement just this amplifier on a breadboard and see how it behaves. Use a pot or other variable voltage source (can even be just putting different valued resistors in a voltage divider).

What is Vo connected to right now? It is possible that the reason that Vo isn't rising above 1.4V is because it is being loaded down by something?



If you only have a multimeter, then you need to take static measurements. You can't have Vin or Vr changing while you are taking the measurement. So you would DISCONNECT the output from the VCO from the input to this circuit and REPLACE it with an alternate signal source, such as a pot, that will allow you to apply static voltage levels to the inputs.

Got it, you want me to take out the chip from its base and build the same inv. amp on a breadboard and get you the readings. I'll be back soon.

I really appreciate your patience and help. Thanks.

 

It sounds like you are falling prey to a common mistake that people make -- you want to make a circuit that does X and so you build a circuit that does X and then turn on the power and see if it works. Since it usually won't, you are then stuck trying to figure out why and it is a nightmare because you seldom have a really good idea of how much different parts of the circuit impact each other.

Divide and conquer is your friend.

In general, you want to take an incremental build/test strategy in your work, be it hardware or software. Build up small portions of the design that can be tested independently and then verify that they behave as you expect. Then tie some of these tested sub-blocks together to create a functional block at the next higher level and see if the collection as a whole behaves as it should. If it doesn't, check to see if each sub-block is behaving as expected while interfaced to the other blocks. This will help you identify which subset of the sub-blocks are not playing nice with each other and you can then focus on correcting that problem. Once you have the new functional block working it then becomes a new sub-block for use at the next higher level.

At each step you want to integrate only a few sub-blocks together, say two to five, and you should design your system to facilitate this.

It might sound like you are going to spend this ungodly amount of time doing nothing but testing blocks. But the reality is that you will probably spend considerably less total time doing testing and fixing than before because each test is much simpler and narrowly focused. In addition, by building each block from just a few already-tested blocks you drastically increase the likelihood that the new block will work on the first attempt and also make it much more likely that you will be able to immediately spot and correct the problem if it doesn't.

Furthermore, it's an approach that scales well to larger designs. Let's say that you have design that you break down into 64 lowest level blocks. That's already a pretty big system. You have to do 64 build/test cycles to get those blocks tested, but each one is probably going to take a fraction of 1% of the time that it would take to test the whole system. Now let's say that your hierarchy takes four sub-blocks to build up a new block. Well, just 16 more build/test cycles and you have all the second-tier blocks, 4 more and you have the third-tier, and just one more and you have the full system.

 

It sounds like you are falling prey to a common mistake that people make -- you want to make a circuit that does X and so you build a circuit that does X and then turn on the power and see if it works. Since it usually won't, you are then stuck trying to figure out why and it is a nightmare because you seldom have a really good idea of how much different parts of the circuit impact each other.

Divide and conquer is your friend.

In general, you want to take an incremental build/test strategy in your work, be it hardware or software. Build up small portions of the design that can be tested independently and then verify that they behave as you expect. Then tie some of these tested sub-blocks together to create a functional block at the next higher level and see if the collection as a whole behaves as it should. If it doesn't, check to see if each sub-block is behaving as expected while interfaced to the other blocks. This will help you identify which subset of the sub-blocks are not playing nice with each other and you can then focus on correcting that problem. Once you have the new functional block working it then becomes a new sub-block for use at the next higher level.

At each step you want to integrate only a few sub-blocks together, say two to five, and you should design your system to facilitate this.

It might sound like you are going to spend this ungodly amount of time doing nothing but testing blocks. But the reality is that you will probably spend considerably less total time doing testing and fixing than before because each test is much simpler and narrowly focused. In addition, by building each block from just a few already-tested blocks you drastically increase the likelihood that the new block will work on the first attempt and also make it much more likely that you will be able to immediately spot and correct the problem if it doesn't.

Furthermore, it's an approach that scales well to larger designs. Let's say that you have design that you break down into 64 lowest level blocks. That's already a pretty big system. You have to do 64 build/test cycles to get those blocks tested, but each one is probably going to take a fraction of 1% of the time that it would take to test the whole system. Now let's say that your hierarchy takes four sub-blocks to build up a new block. Well, just 16 more build/test cycles and you have all the second-tier blocks, 4 more and you have the third-tier, and just one more and you have the full system.

Totally, agreed. And I follow the same rule when assembling.

My project consists 4 blocks:

1) RAMP GENERATOR -----> Working fine
2) LATCH CIRCUIT No. 1 --> Working fine
3) LATCH CIRCUIT No. 2 --> Working fine
4) INV. AMP -------------> Not working as expected

Now, coming on to the readings; I first worked on your tip of removing all connections to the output of the Inv. Amp. but, found no difference.

Next, I recorded the readings as you said. Please find the table in attached file.

 

What does it look like if you plot your results? What is the slope of the line in the active region?

 

What does it look like if you plot your results? What is the slope of the line in the active region?

The graph is attached.

 

What is the slope in the active region?

 

What is the slope in the active region?

Isn't that the slope you are asking for. I mean, the second pdf file?

 

Isn't that the slope you are asking for. I mean, the second pdf file?

But what IS it?! 200V/V? 0.076V/V? -42V/V? What?

 

But what IS it?! 200V/V? 0.076V/V? -42V/V? What?

Here is the plot V/V against Vin.
I am sorry, I have limited knowledge therefore, might be my answer is not as per your expectation. But if you guide me, I will get you correct reply.

 

The slope of the line is NOT the same as taking Vout and dividing by Vin.

The slope is how much the output CHANGES for each 1V of change in the input. So identify the portion of the Vout vs Vin graph that is in the active region and that is reasonably linear. Then draw you best estimate of a straight line through that region. Then pick two points (such as where the line crosses the axes) and find ΔVout, the difference in the output voltage (and it doesn't matter whether the points you use represent actual input/output values that the circuit can achieve -- you are merely using them to ffind the slope of the line that passes through the points that can be achieved), and divide that by ΔVin, the corresponding difference in input voltage.

What value results.

 

The slope of the line is NOT the same as taking Vout and dividing by Vin.

The slope is how much the output CHANGES for each 1V of change in the input. So identify the portion of the Vout vs Vin graph that is in the active region and that is reasonably linear. Then draw you best estimate of a straight line through that region. Then pick two points (such as where the line crosses the axes) and find ΔVout, the difference in the output voltage (and it doesn't matter whether the points you use represent actual input/output values that the circuit can achieve -- you are merely using them to ffind the slope of the line that passes through the points that can be achieved), and divide that by ΔVin, the corresponding difference in input voltage.

What value results.

I am sorry, it seems that this discussion is becoming more and more complicated for me. It is better to put forward the block diagram of my circuit so that you may get me the solution of my actual problem.

The ramp generator uses half of IC KA358A. This module working fine.

The green latch circuit is working fine.

The blue latch circuit is working fine.

The other half of IC KA358A is not working as expected.

By the way, earlier readings are also attached in .XLS format

 

And what happens if either of the blue or green relays is energized and the output of the opamp tries to go above 12V (or 12.7V)?

 

And what happens if either of the blue or green relays is energized and the output of the opamp tries to go above 12V (or 12.7V)?

This schematic is for VCO reader, which reads the type (PNP or NPN), and the internal preset cut-off voltage of the VCO. The Vref of op amp is fixed 1V. The initial output voltage from the ramp generator is 5V.

PNP-VCO CONNECTED
If a PNP-VCO is connected, the output would be about 4.4V. This would energize the green relay thru the green latch circuit. The energized green relay would shift the connection of the output of the op amp from the blue latch circuit to the brown relay.

The ramp is started with the start switch. The VCO-out increases with VCO-in as long as VCO-in is below the preset cut-off voltage of the VCO. Once the VCO-in crosses the cut-off limit, the VCO-out drops to 735mV.

The VCO-out is the Vin for the op amp. As long as VCO-out higher than 1V, the Vo would be 0V. And the moment VCO-out drops to 735mV, the Vo should go high to 12.7V which will energize the brown relay. The relay connectors would pause the ramp and that paused ramp output voltage would be passed on to the DVM.


NPN-VCO CONNECTED
On the other hand, if an NPN-VCO is connected, the VCO-out would be about 0.6V. This would make the Vo go higher to 12.7V and thus the blue latch circuit is activated and the blue relay is energized. The blue relay would disconnect the VCO-out from the green latch circuit and connects the +12V to the red relay.

The VCO-out remains at 0.6V as long as VCO-in is below the preset cut-off voltage of the VCO. Once the VCO-in crosses the cut-off limit, the VCO-out shoots up to the preset cut-off value.

The moment VCO-out shoots up, the Vo should go to 0V which will energize the red relay. The relay connectors would pause the ramp and that paused ramp output voltage would be passed on to the DVM.

 

WBahn

Is there something wrong with my earlier post?
Awaiting response.

 

I don't understand your schematic.

You've got these two boxes that just say,"Latch circuit for ...." What is in those? The top line that comes out of the right side of each of those boxes, what is it? Is the coil in the green box the relay coil for the switches that are also in green boxes? Since your system only appears to have positive voltages, what is the purpose of the circled LEDs that require the top wire to be negative in order to illuminate?

You say the output of the VCO goes to 0.735V. Where does that magic value come from? Why can't if go to 0.5V or 0.9V instead?