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Old 20th Aug 2018, 7:05 pm   #21
ukcol
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Default Re: Help me understand this circuit better. (Please)

Thank you again to Lawrence for the links. Thank you to also David and Peter for the technical explanations and in particular to Peter for taking the trouble to write the detailed analysis of the circuit.

I think I've nearly got it but I am now going to be a complete a s by asking you to convince me that we are indeed dealing with "normal" (positive) inductance rather than negative inductance.

First lets start with a definition of negative inductance so that we are all reading from the same hymn sheet so to speak.

1] With negative inductance the reactance is positive (as it is with positive inductance) that is, the reactance increase with frequency.

2] With negative inductance (this time unlike positive inductance) the voltage/current phase relationship is the same as that for a capacitor, that is, the current lags the voltage by 90 deg.


So the first question I would like to ask.

Are those 2 statements above valid?

Assuming for now that they are:

Following Peter's analysis, the voltage at the anode is "wriggling" due to the influence of the signal from (in this case) the local oscillator.

The voltage at the grid lags this voltage by 90 deg.

The anode current is forced to follow this voltage and so is also behind the grid voltage by 90 deg.

The anode current therefore will be 90 deg behind the anode voltage.

So my final question.

Is that not consistent with the valve behaving like a negative inductance?

This argument only holds true, of course, if my definition of negative inductance above is also true.
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Old 20th Aug 2018, 8:39 pm   #22
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Default Re: Help me understand this circuit better. (Please)

Re: Statement 2, I thought with capacitance the current leads the voltage.

Lawrence.
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Old 20th Aug 2018, 8:42 pm   #23
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Default Re: Help me understand this circuit better. (Please)

You need to establish a polarity convention to truly get your head round it. It isn't helped by me thinking in terms of electron flow, not conventional current, almost always! Puts me out of phase with most folks. However...

1) is correct, reactance increases with frequency, for an inductor (positive or negative).

2) If I have an inductor and put some volts on it, a bit later the current rises and catches up. So current lags in an inductor. Conversely, if I have a capacitor and shove some current through it, voltage catches up as it charges; so voltage lags in a capacitive circuit, or turn it round and say current leads if you prefer. (If I have a resistor and put volts on it, current shoots up immediately; and if I force current through it, volts shoot up immediately. They're in phase.)

In the reactance valve as drawn Post #1, if I suddenly step up the anode voltage, the grid-cathode capacitor will slowly charge, so the grid voltage will slowly rise, and so will the anode current. So this behaves like an inductor, not a negative inductor.

Of course, my 'slowly' is slow in terms of a microsecond clock, with the picofarad-valued capacitors shown, but you get the idea hopefully.

Last edited by kalee20; 20th Aug 2018 at 8:46 pm. Reason: Clarity
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Old 20th Aug 2018, 9:13 pm   #24
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Default Re: Help me understand this circuit better. (Please)

It's a case of practising until you get comfy. There's an awful lot of capacitors and inductors around so it's worth getting a feel for what they do.

It may seem too complicated to start with, but when you make the jump to accepting complex numbers (with real and imaginary parts) it suddenly cements a lot of things together and you may not believe it now, it actually becomes simpler.

One of the guys a couple of decades ago on the rec.radio.amateur group used to say that his answerphone greeting was "The number you have reached is imaginary. Please hang up, rotate your telephone through 90 degrees and dial again."

David
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Old 20th Aug 2018, 9:23 pm   #25
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Default Re: Help me understand this circuit better. (Please)

Yes, complex numbers are great, not only as a super tool for analysing AC circuits, but as an elegant mathematical completeness, the set of complex numbers is algebraically closed.

But, as a sanity check, any equation I derive, I try and check it by applying a thought-experiment with step changes of V or I, and visualising what happens. If it moves in the same was as an equation predicts, then it's likely the sums are right. If not, then I've slipped a minus sign somewhere.

The above experiment convinces me that the reactance valve as shown is inductive, and not negative-inductive (which as you correctly say is capacitive in phase, but changes the wrong way with frequency).

PS love the phone message!

Last edited by kalee20; 20th Aug 2018 at 9:24 pm. Reason: PS added
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Old 20th Aug 2018, 11:07 pm   #26
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Default Re: Help me understand this circuit better. (Please)

Quote:
Originally Posted by ms660 View Post
Re: Statement 2, I thought with capacitance the current leads the voltage.

Lawrence.

Quote:
Originally Posted by kalee20 View Post

.............2) If I have an inductor and put some volts on it, a bit later the current rises and catches up. So current lags in an inductor. Conversely, if I have a capacitor and shove some current through it, voltage catches up as it charges; so voltage lags in a capacitive circuit, or turn it round and say current leads if you prefer. (If I have a resistor and put volts on it, current shoots up immediately; and if I force current through it, volts shoot up immediately. They're in phase.)..........
I can't believe I made such a fundamental error, of course current LEADS voltage in a capacitor.


Basil Fawlty, "It would be quicker to train a monkey."


Quote:
Originally Posted by kalee20 View Post

.....In the reactance valve as drawn Post #1, if I suddenly step up the anode voltage, the grid-cathode capacitor will slowly charge, so the grid voltage will slowly rise, and so will the anode current. So this behaves like an inductor, not a negative inductor.....
This has finally convinced me.
Peter - you have the patience of a saint.

Quote:
Originally Posted by Radio Wrangler View Post
One of the guys a couple of decades ago on the rec.radio.amateur group used to say that his answerphone greeting was "The number you have reached is imaginary. Please hang up, rotate your telephone through 90 degrees and dial again."

David
Fantastic, I love it.
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Old 21st Aug 2018, 8:40 am   #27
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Default Re: Help me understand this circuit better. (Please)

Quote:
Originally Posted by Radio Wrangler View Post
One of the guys a couple of decades ago on the rec.radio.amateur group used to say that his answerphone greeting was "The number you have reached is imaginary. Please hang up, rotate your telephone through 90 degrees and dial again."
But there really should be a second message, following that: "Sorry, the number you just dialled is now negative. You rotated your phone 90 degrees the wrong direction. Please do a half-turn and try again."

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Peter - you have the patience of a saint.
Thanks - but one of the benefits of this Forum is there are people (such as Radio Wrangler in this post) who can pick holes in my own reasoning at times, which is why I tend to spell it out... all my life I've been learning...

...speaking of which, are there any comments on my thoughts (post 12) about the C47, C48 connection conundrum that Colin asked about in his kick-off post? It could be that the designer is a genius who's compensated for something subtle. Or he could be a clot. But it's intriguing!
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Old 21st Aug 2018, 8:57 am   #28
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Default Re: Help me understand this circuit better. (Please)

And Peter is one of the ones who can spot any holes in my reasoning.

Together, we all manage to cover each other's gaps (and keep each other honest )

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Old 21st Aug 2018, 6:33 pm   #29
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Default Re: Help me understand this circuit better. (Please)

While the imaginary number plane and complex numbers are very helpful in the analysis of AC circuits containing L, C and R and will give you a good answer about power factors, phase angles and if the circuit is overall inductive or capacitive etc.

But there is an element of it that should always be remembered; the imaginary number plane is purely imaginary. For example it is only the component of a current in phase with applied voltage that can do any real work that will manifest itself in our reality. The imaginary component doesn't, it is as though it doesn't exist in reality.

By analogy it is like two forces which act on an object that are at 90 degrees to each other, they act independently of each other, but at least in that case, one of the forces is not imaginary.

So it is necessary to distinguish between work done and a helpful mathematical tool in the analysis of AC circuits. This is why when certain AC circuits are analysed for their transient response, depending on the damping, an equation solution may fail because it has no real roots (an imaginary solution) and a new equation must be selected.
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Old 21st Aug 2018, 9:08 pm   #30
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Default Re: Help me understand this circuit better. (Please)

At the beginning of this thread when I asked "why an inductor in the grid circuit rather than a resistor?" Peter replied:

Quote:
Originally Posted by kalee20 View Post
I can see the reasoning behind an inductor rather than a resistor to maintain the DC path, g1-k. A resistor would stop the grid circuit looking purely capacitive. This would reduce the overall 90 degree phase shift between Va and Ia, so the thing would look like a more lossy reactance than with the inductor as grid leak.
Perhaps the above is a clue as to why C47 is returned to the cathode of V8 and not to the junction of R37/R38 as is the case with C48.

1] Obviously L25 must be returned to the junction of R37/R38 in order to give the correct bias conditions for grid G1 of V8.

2] C48 is likely returned to the same point for the convenience of taking one wire from the can containing C48 and L25. (As has already been suggested).

3] The disadvantage of the strategy in 2] above is that the resistance (R37) is in series with C48 reducing the phase angle below 90 deg. But perhaps this can be compensated for by a suitable choice of value for L25?

4] The larger value capacitor used on the lower LW frequencies (C47) can be conveniently taken to V8 cathode and no further compensation for phase angle is required as a result.

I do realize that this is pure conjecture on my part and comes from relative ignorance.

If I am talking complete nonsense I apologies but if there is anything above that merits discussion I will be interested to hear what others think.


I have repeated the circuit below for convenience.
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Old 22nd Aug 2018, 8:16 am   #31
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Default Re: Help me understand this circuit better. (Please)

With regards to point 3 in the post above, that is what I had wondered too, except I had thought R37 rather than R36 might have caused more phase shift. If you have the circuit working it would be easy to see with a two channel scope but putting a resistor say 1M across L25's connections for the grid return, connecting and disconnecting L25 and looking at the phase of the signal on the grid compared to the plate. Or maybe just set it up in a Spice engine to check that issue at the operating frequency.
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Old 22nd Aug 2018, 8:48 am   #32
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Default Re: Help me understand this circuit better. (Please)

Quote:
Originally Posted by Argus25 View Post
...there is an element of it that should always be remembered; the imaginary number plane is purely imaginary. For example it is only the component of a current in phase with applied voltage that can do any real work that will manifest itself in our reality. The imaginary component doesn't, it is as though it doesn't exist in reality.

By analogy it is like two forces which act on an object that are at 90 degrees to each other, they act independently of each other, but at least in that case, one of the forces is not imaginary.
I'd say that both the forces are real! It's when they come to move an object that angles are important.

If a force moves an object in the direction of the force, then work is done on the object (like the spring of a ball-point pen pushing out the point, when I press the button). If the object moves against the direction of the force, then work is done by the object on the force (like me pushing it back in again). But if the object moves at 90 degrees to the force, no energy exchange happens at all.

Your analogy with two forces at 90 degrees - I'm thinking about a railway locomotive pulling a truck on a straight rail, while there is a strong sidewind. The wind is no help at all and doesn't act to accelerate the truck. The locomotive, on the other hand does. If the sidewind moves slightly so that it is anything other than 90 degrees to the direction of the rail, then it will either assist the locomotive (so the truck accelerates faster) or hinder it (so the locomotive actually does work on the wind).

Quote:
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...perhaps the above is a clue as to why C47 is returned to the cathode of V8 and not to the junction of R37/R38 as is the case with C48.
You really want the grid-cathode impedance to be as near a perfect capacitance as possible, the cathode is the best place to take C47 to.

Quote:
Originally Posted by ukcol View Post
...the resistance (R37) is in series with C48 reducing the phase angle below 90 deg. But perhaps this can be compensated for by a suitable choice of value for L25?
It would be nice if it could! But it can't. A capacitor in parallel with an inductor has a phase shift which is inductive below the resonant frequency, and capacitive above. Very near the resonant frequency, the phase shift slews from the one to the other, passing through purely resistive on the way. If the L and C were perfect, this slew would be abrupt at the resonant frequency. With real-world components, it's fairly rapid (depending on Q). But the phase shift never goes 'the other side' of capacitive, which is what you'd need to do, to compensate for a bit of resistance in series. Only yet another trick (such as positive feedback) can do that.
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Old 22nd Aug 2018, 10:42 am   #33
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Default Re: Help me understand this circuit better. (Please)

Quote:
Originally Posted by Argus25 View Post
With regards to point 3 in the post above, that is what I had wondered too, except I had thought R37 rather than R36 might have caused more phase shift.
I understood that R36 with C48 (In the MW case) were the components giving the desired 90 deg phase shift in the circuit. What I was trying to say is that R37 would (perhaps?) reduce this phase shift below 90 deg. The reason I think R37 may have this different effect to R36 is that R36 is between anode and grid G1 of V8 whereas R37 is in series with C48 across G1 - cathode. Putting it another way, the g1-k voltage would be a function of the current in a capacitor AND a resistor.



Quote:
Originally Posted by ukcol
...the resistance (R37) is in series with C48 reducing the phase angle below 90 deg. But perhaps this can be compensated for by a suitable choice of value for L25?
Quote:
Originally Posted by kalee20
It would be nice if it could! But it can't. A capacitor in parallel with an inductor has a phase shift which is inductive below the resonant frequency, and capacitive above. Very near the resonant frequency, the phase shift slews from the one to the other, passing through purely resistive on the way. If the L and C were perfect, this slew would be abrupt at the resonant frequency. With real-world components, it's fairly rapid (depending on Q). But the phase shift never goes 'the other side' of capacitive, which is what you'd need to do, to compensate for a bit of resistance in series. Only yet another trick (such as positive feedback) can do that.
Thank you, the (late) education continues.
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Old 22nd Aug 2018, 11:34 am   #34
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Default Re: Help me understand this circuit better. (Please)

Quote:
Originally Posted by ukcol View Post
I understood that R36 with C48 (In the MW case) were the components giving the desired 90 deg phase shift in the circuit. What I was trying to say is that R37 would (perhaps?) reduce this phase shift below 90 deg. The reason I think R37 may have this different effect to R36 is that R36 is between anode and grid G1 of V8 whereas R37 is in series with C48 across G1 - cathode. Putting it another way, the g1-k voltage would be a function of the current in a capacitor AND a resistor.
Bang on Colin! The last sentence encapsulates it.

Strictly, the extra resistance in series with C48 is actually the parallel combination of R37 and R38 (we are neglecting the effect of C50 because it's comparatively huge).

The phase shift even so will not be quite 90 degrees, because the current through R36 is a function of both itself and the reactance of C48 and we really want it to be just dependent on R48. So the best reactance valve circuits use a large value for C48 (so the error is small) coupled with a high-gm valve (to make use of the correspondingly small Vg1).

It's instructive to calculate values - C48 at 45pF has a reactance of 3.5k at 1MHz. R36 is 90k. So the error is arctan (3.5/90) or 2.2 degrees (or another way, the phase shift is really only 87.8 degrees). This is approximate, as I have neglected the effect of L25... and its 6 ohm resistance...
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Old 22nd Aug 2018, 2:57 pm   #35
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Bang on Colin! The last sentence encapsulates it.
Perhaps this atones for me stating that current lags voltage in a capacitor?
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Old 22nd Aug 2018, 5:02 pm   #36
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I couldn't help taking things further. With R36 at 90kΩ and C48 at 45pF, 1MHz gives a reactance of C48 of 3.53kΩ (as stated already).

Total impedance of R36 and C48 is thus 90.07kΩ as they add in quadrature - virtually 90kΩ.

So at the 1MHz, 1V on the anode of the valve gives 0.039mV on the grid (in near-quadrature, of course). Now the AC/SP1 has a data-book value of 2.65mA/V for gm so the 39mV gives rise to 0.104mA of anode current. So 'reactance' = 1V/0.104mA = 9.6kΩ.

This is exactly what you'd expect for a 1.53mH inductor. Of course, it depends on the actual value of gm, which depends on bias, voltage on g2 and g3, etc. But it gives an idea.

The loss resistance, as calculated above, comes from the 2.2 degrees error. You can represent this as either 370Ω in series, or 250kΩ in parallel. (Effective Q is thus 26)

There are other parasitic losses... the valve's ra at 120kΩ (which thus has a bigger effect than the not-quite 90 degrees in the phase shift network). And finally there's the phase-shift network itself which is directly across the valve, this is the worst spoiler as it puts 90kΩ of resistance across the thing. Ultimate parallel loss resistance is then (250kΩ, 90kΩ, 120kΩ in parallel) ie 42.7kΩ which at 1MHz gives the circuit's Q as a rather disappointing 4.4.

But don't forget that the oscillator coil itself is going to be around 90uH for a typical radio, so the parallel loss resistance of the reactance valve circuit is going to be correspondingly diluted.

Last edited by kalee20; 22nd Aug 2018 at 5:04 pm. Reason: Add last sentence
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