Hi,

Ok that makes more sense now.
However, I do not think you can set the pot at 0 percent to make that 100k pot zero Ohms. That's because then you are left with just the 200.4 Ohm resistor. Is that what you are doing for the highest frequency?
If you only have 200.4 Ohms there then the current into the following op amp circuit would be:
i=15/200.4=74.85ma
which is probably too high for the op amp.
Remember that for the circuit to work, the current through the capacitor has to be the same as the input current through the input resistor, and that means the output current from the op amp has to be about 75ma also, and I do not think that op amp can do that high of a current. You could check that though.
If that's the case, you will have to lower the value of the capacitor to get to the higher frequency.

Now if the maximum output of the op amp was 20ma, can you calculate the minimum value of the potentiometer?
That will help to determine the maximum size of the capacitor.
[LATER]
I just checked the data sheet, it says that the max current on the output would be closer to 15ma when the op amp swings negative. That means you should design for less than that instead of 20ma. The maximum positive output current is a little more but it has to work when the output goes negative too so that 15ma looks like the max.
Another point is that the outputs will not likely go all the way up to plus and minus 15 volts either. There's always some drop in the internal transistors. A better design target point might be 13v or something (check data sheet).

I never thought about checking the maximum current an op-amp could output. I believe I've resolved the issue by lowering the supply voltage to 3V. The only remaining step is to examine the drop caused by the transistors to fine-tune and precisely adjust the frequency.

 

How will lowering the supply voltage help compensate for an op-amp having limited output current capability?

 

I never thought about checking the maximum current an op-amp could output. I believe I've resolved the issue by lowering the supply voltage to 3V. The only remaining step is to examine the drop caused by the transistors to fine-tune and precisely adjust the frequency.

As this is homework
what do your other class mates come up with ?

 

How will lowering the supply voltage help compensate for an op-amp having limited output current capability?

Hi,

By getting at the root cause in this particular application.

I can't say right now if it is the best idea, but by lowering the power supply voltage the input current to the op amp is limited to a lower value and thus the output current will be less.
If 15v causes 75ma, then 5v will cause 25ma, 2.5v (if it was possible) would cause just 12.5ma.
I don't think that's the right way to do it though, especially since part of the problem statement was to use plus and minus 15 volt supplies.

 

I never thought about checking the maximum current an op-amp could output. I believe I've resolved the issue by lowering the supply voltage to 3V. The only remaining step is to examine the drop caused by the transistors to fine-tune and precisely adjust the frequency.

Hi,

Is that really a good idea though? For one thing, aren't you supposed to use plus and minus 15v supplies?
Also, can those op amps really work effectively with just plus and minus 3v supplies?

 

The supply voltage does not have much effect on the input current of an op-amp, At least not those from the past 30 years. What reducing the supply voltage does do is lower the linear output voltage range, which will increase distortion.
All of the triangle generation schemes shown here depend on controlling a constant current to charge and discharge a capacitor. That method is limited a bit, so other schemes are likely to allow a wider control range, which was the goal of the TS. Controlling the gain of a loop may have a wider sweep range ability.

 

The supply voltage does not have much effect on the input current of an op-amp, At least not those from the past 30 years. What reducing the supply voltage does do is lower the linear output voltage range, which will increase distortion.
All of the triangle generation schemes shown here depend on controlling a constant current to charge and discharge a capacitor. That method is limited a bit, so other schemes are likely to allow a wider control range, which was the goal of the TS. Controlling the gain of a loop may have a wider sweep range ability.

Hi,

This is an application dependent variation not a parts spec variation.

The point I was really making was that for some circuits it does because the input current depends on the power supply voltage, and that is because of the size of the input resistor. For example, if you have a +11v power supply and you are assuming the output voltage of one of the op amps is 10v (1v less for this example only), and that 10v feeds into a 10k input resistor for the next stage, then the input current for that second op amp is 10/10000=1ma, and so in the linear mode the output current (no load) will be 1ma. Now decrease that +11v to +6v and with that assumption of 1v loss the output of that first op amp will be just 5v, and 5v into that 10k input resistor is 5/10000=0.5ma, so the output current of the second op amp is now 1/2 of what it was before. It was 1ma with an 11v supply, now it is 0.5ma with a 6v supply.

Although it does not change the specs of the op amp (although sometimes it does that too), it changes the operating conditions of the circuit such that the op amp does not have to put out as much current anymore to keep it in the linear region.

This is the mechanism we see in this circuit, but as I said before going as low as 3v can't be a good idea. Can an op amp like this one even work down that low (data sheet would tell us that).

 

Hi,

This is an application dependent variation not a parts spec variation.

The point I was really making was that for some circuits it does because the input current depends on the power supply voltage, and that is because of the size of the input resistor. For example, if you have a +11v power supply and you are assuming the output voltage of one of the op amps is 10v (1v less for this example only), and that 10v feeds into a 10k input resistor for the next stage, then the input current for that second op amp is 10/10000=1ma, and so in the linear mode the output current (no load) will be 1ma. Now decrease that +11v to +6v and with that assumption of 1v loss the output of that first op amp will be just 5v, and 5v into that 10k input resistor is 5/10000=0.5ma, so the output current of the second op amp is now 1/2 of what it was before. It was 1ma with an 11v supply, now it is 0.5ma with a 6v supply.

Although it does not change the specs of the op amp (although sometimes it does that too), it changes the operating conditions of the circuit such that the op amp does not have to put out as much current anymore to keep it in the linear region.

This is the mechanism we see in this circuit, but as I said before going as low as 3v can't be a good idea. Can an op amp like this one even work down that low (data sheet would tell us that).

I wonder if the OP , as this is home work, has gotten to the bottom of this ?

 

I wonder if the OP , as this is home work, has gotten to the bottom of this ?

Hi,

Oh right, when he comes back maybe he will have some more news.
Personally, I think lowering the capacitor value would do it, but there could be other issues I have not looked into yet.

 

Now I am wondering if that 500:1 range must be without switching. If switching capacitors were allowed then it would be much simpler.

 

As this is homework
what do your other class mates come up with ?

Today I was told by one class mate that they had it working with the same circuit I'm trying to use. As for the current limitation, they asked the teacher about it and he told them to reduce supply voltage.

Hi,

By getting at the root cause in this particular application.

I can't say right now if it is the best idea, but by lowering the power supply voltage the input current to the op amp is limited to a lower value and thus the output current will be less.
If 15v causes 75ma, then 5v will cause 25ma, 2.5v (if it was possible) would cause just 12.5ma.
I don't think that's the right way to do it though, especially since part of the problem statement was to use plus and minus 15 volt supplies.

Yes, we are told to use \( \pm 15 \) V, but we can obtain other supply voltages by using other opamps with a suitable gain.

As for 3V being to low, I don't see any problem on simulations, maybe this Monday when building the circuit I'll see them. I'm finally using LM324 for all opamps on circuit. Reading its datasheet I saw that they can work with voltages from 3 to 30V. I'm on the lower side of this range, so maybe I need to increase supply voltage a bit (maybe 4V?)

 

I'm finally using LM324 for all opamps on circuit. Reading its datasheet I saw that they can work with voltages from 3 to 30V

Yes, it will work down to 3V.
But if you look at the common-mode input voltage and the output voltage swing limits, you will see that you won't have much more than about a 1.5V signal swing at 3V.

 

When using such a low voltage it is very easy to get into a very high distortion area of the op-amp, namely clipping. THAT is why all of the early ones were powered from 15 volts or +/- 15 volts. So that is the other side of the low voltage consideration.

 

Today I was told by one class mate that they had it working with the same circuit I'm trying to use. As for the current limitation, they asked the teacher about it and he told them to reduce supply voltage.



Yes, we are told to use \( \pm 15 \) V, but we can obtain other supply voltages by using other opamps with a suitable gain.

As for 3V being to low, I don't see any problem on simulations, maybe this Monday when building the circuit I'll see them. I'm finally using LM324 for all opamps on circuit. Reading its datasheet I saw that they can work with voltages from 3 to 30V. I'm on the lower side of this range, so maybe I need to increase supply voltage a bit (maybe 4V?)

Hello,

Well for the LM358 and similar LM324 you can reduce the voltage more than with a TL082 device. You may run into other problems though, which I am sure you will find out when you go to bench test it.

However, what is wrong with changing the capacitor? Does that limit something else you are working for?
When you lower the value of the capacitor, you can increase the value of the input resistor and thus reduce the required output current of the op amp.

 

I once had a physics instructor who would always answer questions with the instruction "Divide by seven." That teacher who told them lower the voltage may not have been intending to provide any benefit of the advice.