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Author Topic: Phase Shift Sinewave Oscillator Construction  (Read 33766 times)

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I need to try and build a sinewave oscillator that will work from a 100V supply and give me a large amplitude sinewave.

I did not want to use LC networks in case external magnetic fields affect the L component.

Stability is not an issue but harmonic content maybe.

I have found a great article that explains the design steps involved.
Operating frequency will be 35kHz

http://home.earthlink.net/~doncox/wec/Oscillators.html

Here's a picture of the circuit
   

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Working through the steps one by one i hope to build the circuit.

Step 2
Quote
Select a transistor. Almost any NPN bipolar transistor with decent gain will work. The 2N2222 is one of the most commonly available for small-signal circuits, and has a current gain of at least a hundred for currents around 150mA.

My Transistor of choice is the ZTX857

Technical/Catalog Information   ZTX857
Vendor   Zetex Inc
Category   Discrete Semiconductor Products
Transistor Type   NPN
Voltage - Rated   300V
Current Rating   5A
Package / Case   E-Line
Packaging   Bulk
Lead Free Status   Lead Free
RoHS Status   RoHS Compliant
Other Names    ZTX857
http://www.diodes.com/zetex/?ztx=3.0/3-3-2b@rid~9

Hfe of 100-300
   

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Step 3
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Select power supply voltage. Supply voltage must be enough to turn the transistor on (0.7V) with overhead to work with (low voltage reduces your amplifier gains). Too high a voltage and we risk overheating components unnecessarily. Anything in the 5-15V range is the most common. For our example, we will use 10V.

I will use 25V

Step 4
Quote
Select your high-end power limits. This would be due to your collector-emitter current. Most discrete small-signal components are typically capable of dissipating 100-250mW. We will choose 50mW in the transistor. It is a good idea to allow a factor of two or more in power to reduce the chance of melting or burning if the power dissipation conditions are less than ideal. (You may also find yourself with an upper limit on your power supply current which will force you to go back and adjust your values. These trade-offs are the bread-and-butter of much of practical design work.)

I will go for 50mW dissipation limit.

Step 5
Quote
Calculate power, voltage, and current in transistor output to stay within our power limits. Imagine a subcircuit which is just Rc, Re, and the transistor's collector-emitter side. The highest power dissipated in any one place is likely at the output of the transistor. Power transfer from our collector and emitter resistors (our "source") will be maximized when their sum is the same as that of the transistor (our "load"). (We are ignoring the loading effect of the first RC stage right now--remember, this is just to get a reasonable power dissipation.) The following relationships are reasonably true:

    Rce = Rc + Re
    Vce = VRc + VRe
    Pce = PRc + PRe
    IRc = IQce = IRe

Since we know the transistor gets half the power, it is like giving it half the supply voltage for the same current

    P = I * E
    I = PQ1 / Vce
    I = 50mW / 5V = 10mA

I = 50mW / 12.5V = 4mA

Step 6
Quote
Calculate lowest Rc and Re based on maximum power allowed. Let's narrow our selection problem with Rc and Re. Having an emitter resistor stabilizes a common emitter transistor circuit, but must not hog too much of the available output voltage. Today, let us choose Rc = Re. This makes a very stable circuit, increasing the sensitivity to base bias from around +/-10% to about +/-50% and reducing our dependency on a specific transistor gain, but we have just limited our DC gain hFE to 1 (=Rc/Re). Later, we will steal back our lost gain in the AC domain with an emitter bypass capacitor.

Since we know current is 10mA, power for each is half of transistor's quiescent dissipation, and voltage dropped across each is half of Vce, we have more than enough information:

    Rc = Re = VRe / I = 2.5V / 10mA = 250 ohms each (249 ohms is a standard 1% value)
    Ic(max) = Vcc / (Rc + Re) = 10V / (250 + 250 ohms) = 20mA

25mW is well within the 100mW of a typical 1% metal film resistor that is so common nowadays.

Rc = Re = VRe / I = 6.25V / 4mA = 1K5 ohms each (1K5 is a standard value)

Ic(max) = Vcc / (Rc + Re) = 25V / (1k5 + 1k5 ohms) = 8.33mA

He has this statement
Quote
25mW is well within the 100mW of a typical 1% metal film resistor that is so common nowadays.
Pe = VRe / I = 6.25V * 10mA = 250mW dissipation so i am ok to use film resistors which are 600mW
« Last Edit: 2012-02-22, 20:37:57 by Peterae »
   

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Step 7
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Adjust RC input impedence to take best advantage of available transistor signal output. Too low an impedence and voltage is dragged too low, too high and available power is thrown away. This circuit needs all the voltage it can get, so we'll allow throwing away some power to maximize signal voltage. At the least, we won't go lower than the 3dB point (equal impedences for maximum power transfer). So, we will keep the RC impedence > 250 ohms to guarantee modest attenuation. (Note most of the change in voltage across R will occur when XC is within a factor of several times that of R; many first order solutions then can be arrived at by assuming R = XC.) For this first-cut we will select R >= 250 ohms, which will make RC impedence roughly twice our lower limit. (There is some discussion concerning loading effects in the sections "Collector Capacitor Loading Effects" and "Emitter Capacitor Loading Effects" below.)

Step 8
Quote
Select frequency of oscillation. For this example, we will choose 1kHz. In practice, circuit dynamics will cause frequency to be roughly twice as much. I describe the effect of the transistor stage on this in the next section after the first-order analysis. Since noise is being amplified and filtered, frequency lower limit is established by gain allowed by bypass capacitor value, and upper limit by phase shifts available in RC's and transistor. High frequencies will pass will negatively feed back through the base virtually unimpeded to ground, while very low frequencies won't find enough phase shift to reinforce themselves.

I have chosen 35kHz

Step 9
Quote
Select C for the RC's. Typically, capacitors come in much more limited values than resistors. There are series which only have two or three values per decade, and other series will generally include these values. The most common values in use, and which embody small size versus low noise, are the simple decade values of 0.1uF and 0.01uF. We will choose 0.1uF.

I will try 2n2 or 0.0022uf for an initial trial. Not sure what the rules or limits are for choosing this.


Step 10
Quote
Calculate R to complement C at oscillation frequency. There is no trivial way to estimate the oscillation frequency better than within a factor of two or so short of using the derived equations. A 3-section RC uses f = 1 / (2 * pi * SQRT(6) * R * C), but we have a four section RC, so we use the following: *

    f = 1 / (2 * pi * SQRT(10 / 7) * R * C) (for a 4-section RC)
    R = 1 / (2 * pi * SQRT(10 / 7) * f * C)
    R = 1 / (2 * pi * SQRT(10 / 7) * 1kHz * 0.1uF)
    R = 1332 ohms (1.33k is the closest 1% standard resistor value.)

1.33k ohms is comfortably above our 250 ohm lower limit while still not throwing away too much power

R = 1 / (2 * pi * SQRT(10 / 7) * 35kHz * 0.0022uF)

R = 1.729K 1K8 is the nearest preferred value giving me 33.625 kHz

given the resistors have a 1% tolerance i can selectively choose them to get closer on the frequency as the calculated value is on the top boundary of 1% plus i will have the capacitor tolerance to play with as well.
« Last Edit: 2012-02-22, 21:08:23 by Peterae »
   

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Step 11
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Select DC operating point of transistor. When the transistor is off, all the voltage is across it, and when it is fully on, Vce is about 0.3V. So, the collector voltage (Ve + Vce) can swing from a low of 4.85 + 0.3V at saturation to a high of 10V when off. For maximum usage of this range, quiescent operating point of the collector should be:

    Vqt = ((Vmax - Vmin) / 2) + Vmin = ((10V - 4.85V) / 2) + 4.85V = 7.4V

Vqt = ((Vmax - Vmin) / 2) + Vmin = ((25V - 12.35V) / 2) +12.35V = 18.675V

Step12
Quote
The transistor amplifier stage must be biased into its active range, so some base current must be provided by Rb. The R of the final RC stage acts with it as a voltage divider, providing a reference voltage for the base of several times the needed base current. As Vbe approaches 0.7V, the transistor begins to turn on. Most of the voltage drop across Re will come from the collector current = Ib * Current Gain of the transistor, pulling the emitter voltage up to about 0.7V below that of the base where it begins to pinch off and reach a stable state. Thus, the emitter current is effectively controlled by Re and the base quiescent voltage.

First, we know we have 10mA quiescent current across 250 ohms for Rb. While a good first guess of current gain for an unknown transistor would be 100, if we use the classic 2N2222A NPN transistor, the hFE gain is typically rated 100 to 300 (precise gain would have to be measured directly.) A first-order estimate of base current required to control our quiescent emitter current (using hFE = 200) is

    Ib = Ic / current gain = 10mA / 200 = 50µA

Base voltage divider current for a stiff reference voltage would be typically about 10 times base current:

    Ib(divider) = Ib * 10 = 50µA * 10 = 0.5mA

With the emitter quiescent DC voltage needing to be 2.5V, the base quiescent voltage is 0.7V higher

    Vb = 2.5V + 0.7V = 3.2V

The trouble is we already have the R of the final RC stage setting our base divider requirements with 1.33k ohms. This is over 20 times lower than we could get away with, but we go with it for now. We can ignore the load which is only about 1% of our source:

    Ib(divider) = Vb / RRC = 3.2V / 1.33k ohms = 2.41 mA
    VRb = Vcc - Vb = 10V - 3.2V = 6.8V
    Rb = VRb / Ib(divider) = 6.8V / 2.41 mA = 2.82k ohms

using hFE = 200

 Ib = Ic / current gain = 8.33mA / 200 = 41.65µA

Base voltage divider current for a stiff reference voltage would be typically about 10 times base current:

    Ib(divider) = Ib * 10 = 41.65µA * 10 = 0.4165mA

With the emitter quiescent DC voltage needing to be 25V, the base quiescent voltage is 0.7V higher

    Vb = 6.25V + 0.7V = 6.95V I think i have it correct that i want 6.25V emitter quiescent DC voltage

Ib(divider) = Vb / RRC = 6.95V / 1.8k ohms = 3.861mA

    VRb = Vcc - Vb = 25V - 6.95V = 18.05V
    Rb = VRb / Ib(divider) = 18.05V / 3.861 mA = 4.674K
« Last Edit: 2012-02-22, 21:23:12 by Peterae »
   

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Step 13
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Estimate equivalent resistance through the base to ground. From the above calculations of quiescent base voltage and current, we have:

    Rb-e = Eb / Ib = 3.2V / 50µA = 64k ohms

   Rb-e = Eb / Ib = 6.95V / 41.65µA = 166K ohms

Step 14
Quote
Calculate an emitter bypass capacitance which allows AC gain at our oscillating frequency to be high enough to sustain oscillation. The emitter bypass capacitor allows us to establish an AC operating point which is a bit different from the DC operating point of the transistor. Re creates a great deal of transistor gain degeneration (AC gain hfe here is Rc / XCe). If Ce is added in parallel with Re and is suitably large, the AC signal will find a low-reactance path on the emitter through the capacitor rather than the resistor and will return to us most of our lost gain (for our AC signal, not DC). If we want most of our gain back (say 100), we need a 100 times smaller reactance for the bypass capacitor than Re:

    XCe = Re / 100 = 249 / 100 = 2.49 ohms
    XCe = 1 / (2 * pi * f * C)
    Ce = 1 / (2 * pi * f * XCe)
    Ce = 1 / (2 * pi * 1kHz * (2.49 ohms * 0.01)) = 64µF

XCe = Re / 100 = 1K5  / 100 = 15 ohms
    XCe = 1 / (2 * pi * f * C)
    Ce = 1 / (2 * pi * f * XCe)
    Ce = 1 / (2 * pi * 35kHz * (15 ohms * 2n2)) = 137µF

Step 15

Quote
Estimate base reactance to the AC signal.

    Xbe = XC * hfe = 2.49 * 200 = 500 ohms

Xbe = XC * hfe = 15 * 200 = 3K ohms

Reading further on he goes on about the operating frequency being twice the calculated values GRRRR, i will need to build it and see.
« Last Edit: 2012-02-22, 21:32:16 by Peterae »
   

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Well the simulation does not seem to work with my calculated values, i guess i better try his calculated values to see if his works, maybe i made a mistake.
Strange why his diagram does not even use any of his calculated values. I wonder whats going on.
   

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It's not as complicated as it may seem...
How much current does it need to drive Peter?
   

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Not really sure right now because i have not designed the device to be driven, low mA would do it, the plan was to build a buffer stage anyway to match this.

There seem to be other people using all sorts of values/ratio's for biasing & Rc Re

The values i have calculated above do not seem to be in the ratio that others use for instance he said keep Rc & Re the same yet on the attached diagram there 1/10 ratio
   

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It's not as complicated as it may seem...
I don't think you want Re and Rc the same; you would have no gain then.

In the first circuit the ratio is 2:1.

Another approach to this requirement, would be to use a high voltage operational amplifier, then use my "cheap and simple oscillators" document to build what you need. No guarantee it would work, but I can't see why it wouldn't at the moment.

.99
   

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I am not even sure 100V op amps are available, could be wrong. I guess i could build one out of transistors LOL

I know i can get this working should not be too hard really.

The gain is achieved by using the emitter bypass capacitor.

It would appear because the bypass capacitors reactance is a lot less than the emitter resistor's value we get our gain back.
   

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It's not as complicated as it may seem...
OK, with the bypass cap yes. I didn't see that.

HV op-amps exist, that's why I suggested it. ;)

.99
   
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You could do it easily with a couple of 12AX7's and a split-rail +/- 300VDC supply  O0

 :D

-----

I'll recant the above statement. I just remembered the output voltage was still very low. It'll never put out +/- 100V.

Maybe a garden variety low voltage circuit driving a class AB amplifier?

   

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You could do it easily with a couple of 12AX7's and a split-rail +/- 300VDC supply  O0

 :D

-----

I'll recant the above statement. I just remembered the output voltage was still very low. It'll never put out +/- 100V.

Maybe a garden variety low voltage circuit driving a class AB amplifier?


You know i have been very tempted to go down this route but i know even less about valves than i do transistor theory LOL
I guess i could easily read up, but then there's the grid transformer and the 300V rail to build, it just seems to get more complicated, and my next stage is to build a sine 0-360 Degree phase shifter as well which i have not looked at regarding valves but i am pretty sure i can do it using 3 jfet's.

I do also have 5 of the valves SM mentioned cannot remember the valve code 12b??? dual triodes.

I could also just use low voltage circuits and add an amp stage on the end to get my HV i guess but i could not really see any reason why i shouldn't built it all to work at 100V apart from working out how to convert existing circuits to work at higher voltages.

I am jumping the gun a little because i must get the sine going first but here's the phase shifter circuit

Darren
Sorry i never realized you can get op amps that high a voltage, i will look to see what i can find.
   

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I am wondering if i could drag a 32kHz crystal up to 35kHz, but also not sure what the max working voltage of a crystal is, just looked at a datasheet and it does not say. Probably would not like 100V across it

It would certainly give me a stable sine.
   
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Peter, I think you should not abandon this oscillator just yet, there's a lot to be learned.  I'm surprised the manual indicates the frequency could be off by a factor of 2, so much for precise engineering!  Lol

Here's my suggestions.

1). First get the amplifier biased at the mid range of the voltage (50 volts).  Make the gain a bit lower by including a resistor in series with the emitter cap. Gain is -RC/Re , and Re will be this series resistance.  Chose a gain of 10 to begin with.

2) Then, design and test the RC shift network separately by itself and make sure it produces a -180 deg phase shift at the frequency you want.

3) when you put it all together notice that you will have two 10 k resistors in parallel at the base, one from the shift network and one from the biasing on the transistor, so remove one of the resistors, or the frequency will deviate, also account for the other resistor connected to V+, in the design.   anyway some tweaking will be required on the resistor value, maybe use a pot.

4) personally, I would choose a lower voltage for the oscillator, and the high voltage in the final driving stage.

5) most importantly, have fun!  :)

EM
« Last Edit: 2012-02-20, 05:54:09 by EMdevices »
   

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Hi EM
I will try what you say thanks.

I should also be able to do a bit more reading on the theory design yet as there is quiet a bit written about this type of oscillator.

PS i have it working in LT spice, not sure how good that sim is with regard to having a working circuit though, anyway i think i have enough info to get some bits to build it.
« Last Edit: 2012-02-19, 21:18:19 by Peterae »
   
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Here's a simulation of just the phase shift network.

It's a 3 section RC network, and if you place it in the feedback path of the amplifier, from the collector to the base,  it will oscillate at the frequency where the phase is 180 deg,   because the amplifier is inverting, or 180 out of phase, so the network adds another 180 deg phase and you get 360 deg phase, or zero phase, so the signal can now build up in strength, but only if the gain is sufficient or greater then one, which is the other criteria for oscillation.

On the diagram I highlighted the 180 deg phase spot, then we look at the gain of the RC network and notice that the gain is -30 dB, but this is voltage level plotted in dB, so the attenuation is about 0.032,  so we need a gain of at least 32 from the amplifier to compensate, not 10 like I said before, sorry about that, that was just a number I pulled out of the hat.  Maybe to really get it to oscillate right away make the gain 50.

The values of the capacitor and resistor can be adjusted from what I have shown, and it's desirable to increase the resistance and make the capacitance smaller by the same factor, so 2K resistance and 1 nF capacitance will keep the same frequency,  or 20k and 0.1 nF will be another pair of values.


I might build two of these myself, and offset them by 60 Hz,    lol      :-X

EM
« Last Edit: 2012-02-20, 15:28:24 by EMdevices »
   

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I am glad you have taken interest, many chips around these days and it's very easy to bypass a simple cheap transistor circuit.
Do you feel the stability of the circuit will be good enough to maintain a 60Hz difference between the 2 osc circuits.

3 other things i picked up along the way

The bits can be soldered into a small blob and epoxy glued together to form a little 3 leg encapsulated osc, the guy painted it with green paint also  ;D

4 RC networks means you need less gain on the transistor

and it can be beneficial to swap the R & C over, this way the cap is to ground, i understand this can be better for harmonic rejection.

Late last night i did a bit of research into the jfet phase shifter circuit and found it's virtually impossible to find a jfet that works over 50V so it now looks like i will be running at a lower voltage and then use a final stage to shift up to HV.

   

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The calculations above work out on the 4 stage RC simulation using 3n3 and 1k for each stage i get 180 degrees phase shift at the end of the chain with an input of 35kHz, dividing the input level against the output level it would appear i need a gain of at least 23.94
   

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I Re did the calcs above for a 25V supply and built the osc using 5% resistors and 20% caps and a BC547C

The results are pretty poor, it runs fine but precision parts are needed to guarantee the 180 degree phase shift to get a clean sinewave.

You can see from the waveform on my scope the sine has a steeper rise because of the mismatch RC values causing an imprecise phase shift through the 4 RC networks.

Because the frequency is also unpredictable not what you calculate i will move on to another type of oscillator, maybe wein.

The main problem also is that you cannot just use pots in the RC chain to make a variable frequency because C also needs to be changed with each change of R to maintain the 180 degree phase shift.

The calcs for biasing and gain seem to work as above.

It's been very interesting going through the above process and i learned quiet a bit.  O0

Green trace is collector, Yellow trace is base.
   

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OK this is looking good for my next try, it has amplitude stabilization and frequency control.

http://www.redcircuits.com/Page92.htm
   

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For some reason this does not function in LTspice or Multisim  :'( i don't know enough about the sim to understand why.

Looks like i will need to breadboard it first.
   

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OK i have this circuit working good in both ltspice and multisim now.

Anyone know how you simulate a pot in ltspice?
   

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Built the circuit on breadboard tonight, got it to work well, there was no easy way of adjusting the frequency apart from dropping caps in, the trimmer only sets the distortion level, you adjust just after the sine stops and then back a smidgen, so i was ready to give up until i realized the circuit works well up to 30v without component change with a low current being drawn under 1ma @ 30v but the consequence of an adjustable voltage rail is that it adjusts the trimmer position which in turn alters the frequency of a stable sineus, so i managed to adjust to 35kHz no problem, when i checked out the waveform using the spectrum it was really clean, -43db harmonic distortion, so i am really happy, just need to tart it up on vero now.

   
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