Stabilising wienbridge sine wave using FET

[GeoffC]

and what are they are supposed to achieve ?

To: davenn

Hi,
I now have a better idea. Use only one more op-amp, and convert the AC voltage to variable DC voltage using a pot/divider into one of the inputs. This must give the LED a sine wave out!
This I'm sure will give me my sine wave. Regards, GeoffC.

PS - is the signature OK!

 

[BobK]

You could possibly achieve a sinusoidal output from the LED by varying the voltage a small amount just below the nominal forward voltage, but the output signal would be considerably smaller than using a full wave, which let's you go from zero to full brightness.

IR remotes use a square wave and they are very sensitive, I built one, and you pretty much had to completely enclose the transmitter before it would not be picked up by the receiver even though pointed away from it.

Bob

 

[BobK]

To: davenn

Hi,
I now have a better idea. Use only one more op-amp, and convert the AC voltage to variable DC voltage using a pot/divider into one of the inputs. This must give the LED a sine wave out!
This I'm sure will give me my sine wave. Regards, GeoffC.

PS - is the signature OK!

Nope, here is what you get if you use a shifted sine wave:



The green is the voltage across the LED, the blue is the current through the LED, which is roughly proportional to the light output (but not exactly).

The problem is that an LED is not a linear device. You can see that it puts out no light up to about 2.4V then rapidly increases with increasing voltage.

Bob

 

[davenn]

PS - is the signature OK!

yup

 

[GeoffC]

Nope, here is what you get if you use a shifted sine wave:

View attachment 25238

The green is the voltage across the LED, the blue is the current through the LED, which is roughly proportional to the light output (but not exactly).

The problem is that an LED is not a linear device. You can see that it puts out no light up to about 2.4V then rapidly increases with increasing voltage.

Bob

To: BobK

Hi Bob,
I see what you mean. The green trace is hitting 0 volts, and there is no light until 2·4 volts or more come in to play. I think it might be possible to have a smaller P/P voltage ie (3·6 – 2·4 = 1·2 volts P/P), and the off-set voltage raised between 2·4v and 3.6v. would this work, or have I gone off on a limb.

All of this seems to be happening quite fast lately, I need to knock up a small circuit or two on the bread-board to see what works, or what doesn’t work. - Regards, Geoff.

 

[(*steve*)]

The light output is reasonably linear with current (although it's not perfectly linear)

if you used a shifted sine wave as the reference to a constant current source you could achieve a close to sinusoidal variation in intensity.

assuming your detector is also reasonably linear you could capture a (noisy) sine wave.

 

[GeoffC]

Nope, here is what you get if you use a shifted sine wave:

View attachment 25238

The green is the voltage across the LED, the blue is the current through the LED, which is roughly proportional to the light output (but not exactly).

The problem is that an LED is not a linear device. You can see that it puts out no light up to about 2.4V then rapidly increases with increasing voltage.

Bob

To: - BobK – duke37 – davenn - LvW – (*steve*)

Hello - All

I think I’m going off on a limb. The idea of the sine wave in the beginning was twofold. Firstly: was for the use of a frequency. This enabled the receiver to obtain only light of interest that was oscillating at 50KHz. Secondly: To switch the LED on slowly, and not on instantly, as a square wave would. I do not think the LED needs to be on all the time in its cycle, but somewhere around 50% or near. So, my last idea, was to off-set or shift the sine wave in to variable DC. Volts. This would give a string of camel humps on the oscilloscope, but this should not matter, as long as the humps are all the same. As to the print out that Bob kindly sent me - I think this is best done, - unless…………………..?

I found a couple of filter circuits in ‘Martin Hartley Jones’ a - Band Pass & a Precision rectifier. These should come in useful after the ‘Photo Diode’ But I still have a long way to go yet.

Regards, GeoffC.

(Be away for a day and a half, back Saturday.)

 

[duke37]

Why not switch the LED quickly?

You could use a synchronous detector, locked to the source frequency. This reduces the need for filters and will do the precision rectifier job. I do not have further details but radio signals of 1W or so have been transmitted from UK to Australia. In this case, the frequency reference is obtained from GPS.

No extreme frequency stability is required since the detector follows the source.
This type of circuit was used in LVDT measurements. (Linear variable differential transformer)

 

[BobK]

A square save at 50KHz and peak voltage P actually has more energy at 40KHz than a sine wave with the same peak voltage. In fact, the 50KHz sine component of the Fourier transform has 4/pi times the amplitude of the square wave! See here:

http://mathworld.wolfram.com/FourierSeriesSquareWave.html

So a detector tuned to 50KHz will see a larger signal in a 1V square wave than it would in a 1V p-p sine wave.

You have given no compelling reason for using a sine wave when pretty much every consideration (complexity, accuracy and stability of frequency, signal amplitude) all favor the crystal + divider square wave as the signal.

Bob

 

[GeoffC]

Why not switch the LED quickly?

You could use a synchronous detector, locked to the source frequency. This reduces the need for filters and will do the precision rectifier job. I do not have further details but radio signals of 1W or so have been transmitted from UK to Australia. In this case, the frequency reference is obtained from GPS.

No extreme frequency stability is required since the detector follows the source.
This type of circuit was used in LVDT measurements. (Linear variable differential transformer)

To: duke37 - Bobk

I did say in one of my threads quote:My electronic knowledge is somewhat limited, and I tend to be more at home with Op-Amps and sine waves.

So, I know a little in the use of - NE555 (Square Waves) – TTLs – CMOS etc, mainly from a B-Tech I did some years back. I’ve not used or made a crystal oscillator. I did Google up a circuit using a crystal oscillator, and I think I could get my head around it. I’ve not used a frequency divider before, other than a 7493 TTL bit.bin. counter.

Square waves are ‘what I call rather thumpy’- A Little while back I was putting together a circuit for a RD department for their in-house use, and was asked to insert a capacitor to round off the corner of the square wave!!!

My sine wave circuit works, and am now looking towards a comparator to bring my sine wave in to position of play you might say, and drafted out the other two circuits that are needed. At the moment time is not on my side. I’ll come back to you when I have some results.

Regards, GeoffC

 

[GeoffC]

Bob[/QUOTE]

A square save at 50KHz and peak voltage P actually has more energy at 40KHz than a sine wave with the same peak voltage. In fact, the 50KHz sine component of the Fourier transform has 4/pi times the amplitude of the square wave! See here:

http://mathworld.wolfram.com/FourierSeriesSquareWave.html

So a detector tuned to 50KHz will see a larger signal in a 1V square wave than it would in a 1V p-p sine wave.

You have given no compelling reason for using a sine wave when pretty much every consideration (complexity, accuracy and stability of frequency, signal amplitude) all favor the crystal + divider square wave as the signal.

Bob

To: Bobk & Duke37
Hello,
A circuit enclosed.
Might this be the type of circuit you have in mind for me to oscillate the LED? - (It looks quite easy!)
As for a frequency divider, is this a chip or a circuit?
Is there any difference between a ‘Crystal’ & ‘Quartz Crystal’?
Sorry for the questions.
Regards, GeoffC.

 

[BobK]

Yes, that is a typical crystal oscillator.

There is not difference between "crystal" and "quartz crystal" in this context.

A divider consists of using a binary counter, with some logic to reset it after a certain count (an AND gate taking all the bits that make up the divisor and feeding the output to reset.)

The beauty of the crystal oscillator is that it is very accurate, though it will drift a little bit with temperature, though probably not enough for you to worry about.

Bob

 

[duke37]

The CMOS 4060 is an oscillator/divider which divides by two several times. The minimum division is 16 so with a 4MHz crystal the output is 250kHz. Another divider of 5 will give 50kHz. It may be better to finish with a divide by two to get an equal on/off waveform.If the frequency is not critical, the 4060 can divide by 64 giving an output of 62.5kHZ.
The 4MHz crystal was chosen because it is common, other crystals can be used with different division ratios.

As far as I know, all frequency determining crystals are quartz but other materials such as boron titanate can have similar properties.