 ### Network # Understand VCO

#### irq20xdfr

Aug 12, 2016
23
Hello,

Can someone explain me how this circuit works?

I know that a VCO changes oscillations based on a input voltage but i just can't understand how every part of the animation circuit is involved. What's the use of the capacitor? What's the use of resistors?

#### duke37

Jan 9, 2011
5,364
I got the text but not the circuit.
The description seems to have been done quite well.

#### irq20xdfr

Aug 12, 2016
23
Could someone explain me how the current is going? I mean, sorry for my english... the direction or the flow of the charges.... i just see that they going forward and then backwards and i don't get the play of every component in the circuit, for example i don't know why some opamp outputs are connected again as input to another components...

can anyone explain the whole flow of the circuit?

thanks

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510

What is your background with electronics?

#### davenn

Moderator
Sep 5, 2009
14,214
I got the text but not the circuit.
The description seems to have been done quite well.

check to see if your java is up to date works OK for me

#### irq20xdfr

Aug 12, 2016
23
Ok, some people is telling me that i'm not being specific with mi question... I'm going to reformulate it.

How an op-amp is involved in a VCO? I currently know that a op-am is a type of differential amplifier used like a differential in math... but, what is it its role in a VCO and in the circuit of the example? And yes, install Java JDK pls if you don't see the applet.

Thank you in advance to everyone... I am a software developer so i don't have too much context in electronics. Just basics.

Here's the link by the way to install Java JDK

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
A differential amplifier and a differential in math share all the same similarities that either do with a differential in a car.

The role of an op-amp is to provide gain.

The real magic in a VCO is the way an input voltage is used to alter the frequency response to a feedback element which determines the frequency of oscillation.

And some of us are on devices which simply don't support Java. It is best to post an image of the circuit.

#### davenn

Moderator
Sep 5, 2009
14,214
And some of us are on devices which simply don't support Java. It is best to post an image of the circuit.

it's an interactive image, Steve, doing a screen dump wouldn't show the operation here's a screendump you will just have to imagine the current dots oscillating back and forward
and the 3 scope views at the bottom in motion Dave

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
At least I could see the circuit.

I'm kinda OK at reading circuit diagrams that aren't multimedia enabled.

• davenn

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Ok, that's a cool circuit, and it is comprised of a few building blocks.

The first thing to concentrate on it the left op-amp with a capacitor in the feedback loop. This is an integrator (google it for more information). Basically, the output voltage is proportional to the input voltage and time. So a constant input voltage will give you a ramp, either positive or negative depending on the sign of the voltage.

So what is the input voltage? Well, typically you have an input and a reference, and the input voltage is related to the difference between them. In this case the input signal is its own reference!

Imagine that the mosfet is off (very high resistance). This results in the input voltage to the non-inverting (+) input being half that at the inverting input (-) if we ignore the feedback element.

The op-amp tries to make both inputs equal, and it does this by ramping down the output voltage at a rate sufficient to allow the current through the capacitor to drop the voltage at the inverting input to that of the non-inverting input.

So the rate at which the voltage ramps is proportional to the absolute difference, but the voltage difference is proportional to the control voltage. This implies that as you raise the control voltage, the ramp rate increases.

Ignoring the circuit elements on the right, which control it, imagine the mosfet turns on. This changes the relationship between the voltages applied to the inputs of the op-amp in a way that reverses the ramp on the output.

Now for the other op-amp. This is set up as a Schmitt trigger (google it). As the input voltage drops below a certain fixed value the output of the Schmitt trigger goes high, turning on the mosfet which causes the ramp to reverse. As the voltage now rises above another (higher) fixed value, the output of the Schmitt trigger goes low, turning off the mosfet reversing the ramp again.

One effect of this is that the output of the integrator ramps up and down between two fixed voltages.

Because the control voltage determines the ramp rate, a secondary effect is that the frequency changes.

Does that explanation work for you?

#### irq20xdfr

Aug 12, 2016
23
Ok thank you! and sorry for not thinking about posting an image Ok, so i'm kinda confused about the fed back thing, i'm reading more about it in wikipedia, but i get what you say about that it generates an oscillation.

Another thing that i don't get in the circuit is the ends where some resistors are... some are pointing to the bottom to some icons with three bars, i could remember that that is ground, what i don't understand is why are needed, i see them like pointless, can you clarify this please too?

One last thing? could you please clarify me the point about feeding back the output? what's the purpose of this and how does it work?

Thank you so much for your time and knowledge!

#### irq20xdfr

Aug 12, 2016
23
Ok, that's a cool circuit, and it is comprised of a few building blocks.

The first thing to concentrate on it the left op-amp with a capacitor in the feedback loop. This is an integrator (google it for more information). Basically, the output voltage is proportional to the input voltage and time. So a constant input voltage will give you a ramp, either positive or negative depending on the sign of the voltage.

So what is the input voltage? Well, typically you have an input and a reference, and the input voltage is related to the difference between them. In this case the input signal is its own reference!

Imagine that the mosfet is off (very high resistance). This results in the input voltage to the non-inverting (+) input being half that at the inverting input (-) if we ignore the feedback element.

The op-amp tries to make both inputs equal, and it does this by ramping down the output voltage at a rate sufficient to allow the current through the capacitor to drop the voltage at the inverting input to that of the non-inverting input.

So the rate at which the voltage ramps is proportional to the absolute difference, but the voltage difference is proportional to the control voltage. This implies that as you raise the control voltage, the ramp rate increases.

Ignoring the circuit elements on the right, which control it, imagine the mosfet turns on. This changes the relationship between the voltages applied to the inputs of the op-amp in a way that reverses the ramp on the output.

Now for the other op-amp. This is set up as a Schmitt trigger (google it). As the input voltage drops below a certain fixed value the output of the Schmitt trigger goes high, turning on the mosfet which causes the ramp to reverse. As the voltage now rises above another (higher) fixed value, the output of the Schmitt trigger goes low, turning off the mosfet reversing the ramp again.

One effect of this is that the output of the integrator ramps up and down between two fixed voltages.

Because the control voltage determines the ramp rate, a secondary effect is that the frequency changes.

Does that explanation work for you?

We posted almost same time, i'm reading your answer... it's pretty much complete, thank you! Where is the mosfet in the circuit i just see the triangules that represent op-amps..? Sorry for my ignorance, i seem to understand the rest about your answer, im going to get deep on some terms.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
The mosfet is the thing made of two disconnected bars (like a capacitor with one side longer than the other). It is under the left op amp and is directly connected to the right op amp. Think of it like a voltage controlled switch.

#### Herschel Peeler

Feb 21, 2016
401
So the rate at which the voltage ramps is proportional to the absolute difference, but the voltage difference is proportional to the control voltage. This implies that as you raise the control voltage, the ramp rate increases.

Ah! This helps me. Thank you. I never understood this myself.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,865
There is a problem with the following explanation, copied from the OP's link:

"The second op-amp is a Schmitt trigger. It takes the triangle wave as input. When the input voltage rises above the threshold of 3.33 V, it outputs 5 V and the threshold voltage falls to 1.67 V. When the input voltage falls below that, the output goes to 0 V and the threshold moves back up. The output is a square wave. It's connected to the MOSFET, causing the integrator to raise or lower its output voltage as needed."

The statements in red are incorrect. When the output of the first op-amp (integrator) rises above 3.33 V the second op-amp (Schmitt trigger) outputs 0 V. When the output of the first op-amp falls below 1.67 volts the second op-amp outputs 5 V. This is exactly opposite what the original statement states.

There are two threshold potentials, determined by the output state of the second op-amp. 3.33 V occurs when the output of the second op-amp is high (+5 V) and 1.67 V occurs when the output is low (0 V). The positive-going sawtooth input drives the output of the first op-amp in a negative direction when the MOSFET is NOT conducting, which means the second op-amp output must be low (to turn off the MOSFET). When the MOSFET is conducting, the second op-amp output must be high (to turn on the MOSFET) and the first op-amp output is driven in a positive direction.

Because the first op-amp output is applied to the inverting input of the second op-amp (connected as a comparator with positive feedback), whenever the first op-amp output is more positive than the threshold voltage on the non-inverting input of the second op-amp, the second op-amp will switch its output low. When the first op-amp output is more negative than the threshold voltage on the non-inverting input of the second op-amp, the second op-amp will switch its output high. This causes the triangle wave at the output of the first op-amp to oscillate between the two threshold potentials. The amplitude of the low-frequency sawtooth, applied to the input, simply controls how fast this switching occurs, higher amplitudes charging and discharging the integrating capacitor at faster rates.

Biasing the non-inverting input of the first op-amp at half the input voltage is clever. However, the explanation of why connecting the inverting input to ground through a 49.9 kΩ resistor causes the output ramp to reverse direction is rather obtuse IMO. Basically this integrator circuit is dynamic and the output ramp depends on a constant current flowing into or out of the capacitor. Note the output of the integrator is always a positive voltage, either linearly increasing or linearly decreasing.

When the output ramp is going negative, a constant current is provided by the current into the inverting input through the 100 kΩ resistor from the sawtooth voltage and by the negative-going output voltage of the op-amp removing this same current through the capacitor. Thus the capacitor has a constant current charging it. It is constant because the negative feedback produced by the op-amp will maintain the inverting input at the same potential as the non-inverting input. Therefore there is a constant voltage drop across the 100 kΩ resistor equal to Ein-0.5 Ein = 0.5 Ein no matter what the value of Ein happens to be.

Now, what happens when the MOSFET adds 49.9 kΩ from the inverting input to ground? Because of the negative feedback, the op-amp will still drive its output to maintain the inverting input at 0.5 Ein, which means this is the voltage drop across the 49.9 kΩ resistor too. But, since 49.9 kΩ is half the value of 100 kΩ, twice the current must flow through the 49.9 kΩ as flows through the 100 kΩ resistor. Moreover, it must flow out of the inverting input to ground because its polarity is the same as the voltage applied to the non-inverting input. Half of the current through the 49.9 kΩ resistor cancels the current through the 100 kΩ resistor, and the other half of this current discharges the integrator capacitor, the op-amp output going positive in order to do so.

If this tedious explanation hurts your head, try using Ohm's Law and Kirchoff's Laws to analyze the currents in the 100 kΩ and 49.9 kΩ resistors and the current in and out of he capacitor. Assume the inverting and non-inverting inputs are always equal (because of negative feedback) and the op-amp inputs draw no current, which is true of an ideal op-amp with negative feedback. And remember the voltage change across a capacitor is dV/dt = I / C. If I and C are constant, the rate of change of voltage across the capacitor is constant, leading to a linear ramp in voltage across the capacitor. Also, remember one end of the capacitor is at a fixed positive voltage above ground equal to 0.5 Ein. Pay attention to polarity signs.

#### irq20xdfr

Aug 12, 2016
23 Thank you for all of your responses.

Can you please explain me what is the purpose of the resistors in a negative inverse opamp? Do they decrease the current or voltage ?

I'm trying to understand I everything from scratch since I don't have too much background.

Specially in finding that I don't have a proper understanding about resistance and other terminology.

https://postimg.org/image/x1xgu8ow9/

Thanks

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,865
I am a software developer so i don't have too much context in electronics. Just basics.
Can you please explain me what is the purpose of the resistors in a negative inverse opamp? Do they decrease the current or voltage ?
It appears that your lack of understanding of op-amps might be part of the problem. Many years ago, in the 1960s, when I first encountered operational amplifiers and negative feedback circuits, I had a difficult time trying to understand what was going on. Then one day I ran across a simple explanation of what an ideal op-amp does, and more importantly how to use that explanation to analyze circuits involving op-amps. If you know basic circuit analysis involving resistors, capacitors, and inductors, in other words how to calculate current through and voltage drop across these components using Ohm's Law and Kirchoff's Laws, adding op-amps to the mix is fairly easy.

The key to understanding op-amps is the characteristics of an ideal op-amp. No such thing exists of course, but actual op-amps come close enough to ideal to make circuit analysis possible, with some residual error in the resulting analysis. Even this residual error can be removed by substituting the equivalent (more complicated) circuit of a real op-amp instead of using the ideal op-amp. So, your first step is to go Google some explanations of what an ideal op-amp does.

In a nut-shell: ideal op-amps are three-terminal devices that have two infinite input-impedance differential inputs (inverting and non-inverting) and one bi-polar output with zero source impedance. Between the two differential inputs and the bipolar output is a gain element that has characteristics of infinite gain, infinite bandwidth, infinite voltage sourcing capability, and infinite current output capability. It is these four "infinite" characteristics of the ideal op-amp that simplifies circuit analysis using real op-amps.

You, amigo, rock back on your heels and say, "How can this be, Kemo Sabe? My computer deals with numbers, not infinities! Infinity is not a number!"

Good question. But it is not the infinities that cause the problem in circuit analysis; indeed, it is those infinities that simplify the analysis. Consider the characteristic of infinite input impedance of the two differential inputs. This means neither of those two inputs either sources or sinks any current. From a circuit analysis point of view these inputs are just points on a schematic diagram at which nothing happens. So, for your inverting op-amp example, whatever current goes through the input resistor R1 must "find" some other path to ground. The only other path is through the output or feedback resistor R0. The point where these two resistors connect, the inverting input of the op-amp, is often called a "summing point" because the sum of the currents entering this point must equal the sum of the currents leaving this point. That's Kirchoff's current law: can't have charge (the flow of charge is current) accumulating at any point in a circuit. Whatever current goes into a point must also leave that point.

So how can we compute that current through R1 (and the equal but opposite current through R2) since we don't know what the voltage at the summing point is?

This takes a bit of imagination. Imagine that the input to R1 is a positive voltage with respect to ground. Notice that the non-inverting input is connected to ground. When this positive potential is applied to the inverting input through R1 the op-amp output will be driven in a negative direction by the infinite-gain element. How far negative? Well, since the gain element has a "value" of infinity, then the output must be infinitely negative. A gazillion volts (or more) negative with respect to ground! Obviously that ain't gonna happen, even with an ideal op-amp, and here is the reason why: the negative going op-amp output is applied to R0, forcing a negative current through R0 into the summing junction. This current counters the current through R1 and decreases the potential at the inverting input from a positive value towards a zero value. Why zero? Because any other potential will cause the op-amp output to be infinite, either positive or negative. If the initial current through R0 was too high, driving the inverting input negative, then the op-amp output would go positive, producing a positive potential at the inverting input through R0 which would drive the output negative which is back where we started from. This is negative feedback and it is often a difficult concept to visualize because of the circular cause-and-effect reasoning that I just offered.

What you need to realize is the negative feedback (the current through R0) is driving the inverting input to be at the same potential as the non-inverting input, which in your example inverting op-amp configuration happens to be zero or ground potential. With nearly zero differential input voltage between the inverting and non-inverting inputs, the op-amp output could be any value between a positively infinite or a negatively infinite voltage, the polarity depending on whether the inverting input is slightly more positive than the non-inverting input (ground) or slightly more negative than the non-inverting input. In this case, we notice that under steady-state conditions the current in R1 entering the summing point has to be the same as the current in R0 leaving the summing point. So, the op-amp output has to be negative to make this happen. Since the inverting input is now virtually at ground (it is actually infinitesimally positive with respect to ground to produce a negative output) we can declare the summing point to be at ground potential with very little error. That means the current through R1 is calculated to be (Vin) / R1. The current through R0 is the same value but negative, calculated to be (-Vout) / R0. Equating the two currents and rearranging terms, we calculate the "gain" of this inverting op-amp circuit as (Vout) / (Vin) = - (R0) / (R1).

None of the above analysis applies unless the op-amp is configured for negative feedback. With positive feedback, the op-amp steady-state output will always be at either the positive power supply rail or the negative power supply rail, except when the input signal causes a transition between those two states.

I hope some of the above explanation helps you to understand op-amps.

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