Way to charge a capacitor linear over time

[Roff]

The input (control voltage), do you (by default) insert it into the circuit to "go to zero after the 5th ms", OR
the circuit does that (revert input to zero)?

I am assuming the circuit does it.

So this means everything is as I thought they could be.

Maybe I 'll spend a couple of spare weekends (maybe it takes more) building that darn thing

Do you by any chance can think of potential "problems" in reality that are not depicted into sim?
Like things I should expect to might not work, or to behave unstable, or even different.
Experience matters, I think, in analog circuit world.
Would be cool to share yours with us, imo

OK, I'm back.
The control voltage is not forced to zero by the circuit. It is independent. The current is forced to zero when the ramp voltage reaches the control voltage, but it is released when the ramp voltage reaches zero. if the control voltage is positive (non-zero) at that time, the ramp will immediately begin to rise.
I can't think of any potential problems, but that doesn't mean there aren't any.

Cheers,
Ron[/b]

 

[e-l-e-c-t-r-o]

The input (control voltage), do you (by default) insert it into the circuit to "go to zero after the 5th ms", OR
the circuit does that (revert input to zero)?

I am assuming the circuit does it.

So this means everything is as I thought they could be.

Maybe I 'll spend a couple of spare weekends (maybe it takes more) building that darn thing

Do you by any chance can think of potential "problems" in reality that are not depicted into sim?
Like things I should expect to might not work, or to behave unstable, or even different.
Experience matters, I think, in analog circuit world.
Would be cool to share yours with us, imo

OK, I'm back.
The control voltage is not forced to zero by the circuit. It is independent. The current is forced to zero when the ramp voltage reaches the control voltage, but it is released when the ramp voltage reaches zero. if the control voltage is positive (non-zero) at that time, the ramp will immediately begin to rise.
I can't think of any potential problems, but that doesn't mean there aren't any.

Cheers,
Ron[/b]

Hi!!

Welcome back.

Ok, I got it this time.

I also gave my assignment to my tutor, we was suprised by its direction, I think it got me an A+
So, thank you

We manage to gave it a look together, to see how it works and what could the applications could be (if any).

He told me that as an idea it is really nice, but kept questioning the control voltage's nature and role in the circuit.

He told me that if I shall use this kind of control voltage (i.e where it reaches its max value it stays there for ever) we could have the "problem" you mentioned:
The cap we continue to charge again :?

I found your "trick" with the current (probably) the best part of the circuit (the control voltage is there but it does not "pass through") 8) ,
but the control of the initial input has to be made as a "reset input" state

Otherwise I am afraid that this beatiful circuit would have minimal practical value; which results no motives to actually built it

Anyway, your help has been amazing, glad to have discussed with you 8)

PS:Any comments on a potential "work around" for this issue would be great!

 

[Roff]

I asked you specifically about the control voltage in a previous post:

I don't think you understand my question. Where does your input voltage come from? When the input voltage gets clamped, it doesn't go away, like the charge on the cap does when it gets discharged. The input voltage is still there - it is just "disconnected" from the input. If you release the input clamp as soon as the cap discharges, the cap will start to charge again. Then you have an oscillator, as we have discussed. If, on the other hand, you latch the input clamp, you will need some sort of external stimulus to reset the clamp.
I think the key point I'm trying to make is that the input voltage is (in my mind) an independent control voltage. If it goes to zero when you clamp it, how does it assume a non-zero value some finite length of time later?

Cutting off the current is equivalent to resetting the input voltage. It's what you do next that's critical.
In the last schematic, I reset the flip-flop when the ramp got to zero, because you gave me either that option or the option of using a one-shot.
A third option is the external reset (some sort of pulse input, or a switch) to reset the flip-flop, as I mentioned above. See the circuit below.

We could instead cut off the current by resetting the input voltage, but it would require an analog switch (two, to do it right), and we would wind up with the same functionality.
Speaking of functionality, I asked you several times about the usefulness of this, but you seemed to think your instructor would be dazzled by the rope tricks - never mind that we couldn't catch a steer with them.

PS V5 is still your control voltage. I just used a battery for the sim.

 

[e-l-e-c-t-r-o]

I asked you specifically about the control voltage in a previous post:

I don't think you understand my question. Where does your input voltage come from? When the input voltage gets clamped, it doesn't go away, like the charge on the cap does when it gets discharged. The input voltage is still there - it is just "disconnected" from the input. If you release the input clamp as soon as the cap discharges, the cap will start to charge again. Then you have an oscillator, as we have discussed. If, on the other hand, you latch the input clamp, you will need some sort of external stimulus to reset the clamp.
I think the key point I'm trying to make is that the input voltage is (in my mind) an independent control voltage. If it goes to zero when you clamp it, how does it assume a non-zero value some finite length of time later?

Cutting off the current is equivalent to resetting the input voltage. It's what you do next that's critical.
In the last schematic, I reset the flip-flop when the ramp got to zero, because you gave me either that option or the option of using a one-shot.
A third option is the external reset (some sort of pulse input, or a switch) to reset the flip-flop, as I mentioned above. See the circuit below.

We could instead cut off the current by resetting the input voltage, but it would require an analog switch (two, to do it right), and we would wind up with the same functionality.
Speaking of functionality, I asked you several times about the usefulness of this, but you seemed to think your instructor would be dazzled by the rope tricks - never mind that we couldn't catch a steer with them.

PS V5 is still your control voltage. I just used a battery for the sim.

Hi

It seems I did not pay to much attention to the role of the flip-flop.

Now I see the circuit can do what I thought it could. 8)

BTW, my tutor is indeed dazzled by tricks. :shock:

BTW, what sim are you using? I am on "workbench sim8" and I can't get it to work

 

[Roff]

I use SwitcherCAD III from Linear Technology. It is totally free, and is not castrated like most demo versions of simulators. However, it does require more knowledge of spice than you would need to run most simulators.

 

[e-l-e-c-t-r-o]

I use SwitcherCAD III from Linear Technology. It is totally free, and is not castrated like most demo versions of simulators. However, it does require more knowledge of spice than you would need to run most simulators.

Oh, ok.

I will try to simulate it with this program. BTW, can u post the file?If it is no problem, of course. Otherwise, don't bother. :roll:

One thing to note:

In your simulation of the last circuit, do you still set IC cap = 0 ?

In my sim, I can't do that for capacitors (or maybe I don't know how?), and I was wondering what the potential problems would be when I don't set that initial condition.
I mean, voltage cap when the sim starts is zero by default, right?

Last but not least, in my sim I use an "ideal" flip flop (not the the model u mention). Is this a potential problem?
During my simulation, the cap does not discharge when the ramp exceeds the control voltage; it continues to charge up to the point the circuit can handle.So, what could i be doing wrong?

Help appreciated.Really!

 

[Roff]

I use SwitcherCAD III from Linear Technology. It is totally free, and is not castrated like most demo versions of simulators. However, it does require more knowledge of spice than you would need to run most simulators.

Oh, ok.

I will try to simulate it with this program. BTW, can u post the file?If it is no problem, of course. Otherwise, don't bother. :roll:

One thing to note:

In your simulation of the last circuit, do you still set IC cap = 0 ?

In my sim, I can't do that for capacitors (or maybe I don't know how?), and I was wondering what the potential problems would be when I don't set that initial condition.
I mean, voltage cap when the sim starts is zero by default, right?

Last but not least, in my sim I use an "ideal" flip flop (not the the model u mention). Is this a potential problem?
During my simulation, the cap does not discharge when the ramp exceeds the control voltage; it continues to charge up to the point the circuit can handle.So, what could i be doing wrong?

Help appreciated.Really!

If I disable .IC v(cap)=0, the cap starts out charged to 8V, but it runs down to zero and then runs normally.
My flip-flop model is actually a behavioral model (no subcircuit with transistors, etc.) that is in the SwitcherCAD library. I changed the parameters (Vhigh=9, Td=100n, Trise=100n) to approximate the behavior of a CD4013.
Your problem with not discharging - I don't know. What op amp model are you using for U4? Do you have the neg. supply pin connected to a negative voltage, or to GND?
I'm going to attach two files, but the forum won't accept a .asc file (will it?), so I will zip them. I hope it accepts zips.

 

[e-l-e-c-t-r-o]

If I disable .IC v(cap)=0, the cap starts out charged to 8V, but it runs down to zero and then runs normally.
My flip-flop model is actually a behavioral model (no subcircuit with transistors, etc.) that is in the SwitcherCAD library. I changed the parameters (Vhigh=9, Td=100n, Trise=100n) to approximate the behavior of a CD4013.
Your problem with not discharging - I don't know. What op amp model are you using for U4? Do you have the neg. supply pin connected to a negative voltage, or to GND?
I'm going to attach two files, but the forum won't accept a .asc file (will it?), so I will zip them. I hope it accepts zips.

Ok, I will try this and we see what comes. :shock:

Than God that with the sim I cannot be electrocuted 8)

For U4, I used a LM 741 and LTC272 and LT 272. So, common used o-pamps, I guess.
The best result came from the sim' s "ideal" opamp, though. :?

U4 is connected appropriately. All things are. Yet it seems not to work due to wrong component selection? :roll:
Even when I use all "ideal" components, the circuit does not work as expected.The cap does not discharge.

So, can you suggest component selection :?:

Help greatly appreciated.

 

[Roff]

Can you post your schematic?

 

[e-l-e-c-t-r-o]

Can you post your schematic?

You mean the workbench file :?:

What schematic?

In case u mean the picture of circuit, I ll post it a few hour later that and the graph from the signals.I am not at home right now.

 

[Roff]

Can you post your schematic?

You mean the workbench file :?:

What schematic?

In case u mean the picture of circuit, I ll post it a few hour later that and the graph from the signals.I am not at home right now.

I meant the schematic of what you were simulating, i.e., the picture of the circuit (that's what a schematic is).

 

[e-l-e-c-t-r-o]

Can you post your schematic?

You mean the workbench file :?:

What schematic?

In case u mean the picture of circuit, I ll post it a few hour later that and the graph from the signals.I am not at home right now.

I meant the schematic of what you were simulating, i.e., the picture of the circuit (that's what a schematic is).

the schematic is exactly the same with the one you have in your post with the signals (in,out) graph.

The only difference is that all transistors, diodes, opamps and flip flop are "ideal".
The supply voltages are exactly as you have depicted.

So, what could be the case?

Anyway, I will load your files into the program you mentioned and will shall see. :roll:

 

[Roff]

When you copy a schematic to another simulator, there is always the possibility for error. I do that as part of my job, and believe me, it happens, and sometimes the errors aren't obvious.

 

[Optikon]

When you copy a schematic to another simulator, there is always the possibility for error. I do that as part of my job, and believe me, it happens, and sometimes the errors aren't obvious.

The obvious ones are spotted! The worst is copying from old, photocopied paper into some electronic form. Invariably, something gets a wrong value..

Have you guys figured out an application for this? I like the circuit but am confused about what it could really be used for. As I understand it, it started as some exercise for school?

 

[Roff]

Have you guys figured out an application for this? I like the circuit but am confused about what it could really be used for. As I understand it, it started as some exercise for school?

The voltage-controlled current source is useful as a ramp generator, VCO, variable load, etc., but I have no idea why our OP wants to stop the ramp when it reaches the control voltage. I think he was having trouble with the concept that the cap voltage has no relationship to the control voltage.