Antenna ferrite loopsticks verses air core?

Yes, that was a first iteration, nothing wrong with it, but it wasn't as stable (two wires with a weights on the end) as my tiny piece of double sided pcb. Mikek

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Is that gain in a following amp? The amp shown here is a follower with no gain.

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Rick C

Oh, so the free hanging copper is the old cap and the thing I can't see is the PCB cap which you like better?

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Rick C

When you say can't see, do you mean it small and your looking side on or do you mean you didn't see a link?

btw the 0.3pf cap is a approximately 1/8" disc of Teflon pcb 0.32" thick. PCB material is not critical. Easy enough to set size, check the gain of your amp without the input cap, then reduce the size of your input cap until gain equals 1. Mikek

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Yes, this is just the input, look back at Kleijers site, you will see this without bootstraping and the rest of the amp.

Mikek

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I just looked at my 15 inch square (air core) loop antenna on a scope. The amplitude was about 1 volt P-P. I connected a 150K resistor across the loop and the amplitude dropped to about half that. So, the impedance looks to be about 150K. The scope input is 1 Meg and 30pF and I used a 10X probe. The Q of the antenna was measured at 300 at the low end of the band (600Khz) and

200 at the high end. That was using a Boonton Q meter. So, it doesn't look like connecting a scope has much effect on Q in this case. But possibly doing the same test on a short ferrite loop would give different results. Actually, the Q of the loop should be about 100 or less if you want a 10Khz bandwidth.

Sounds about right, the coil I measured has a Q of 1000 to 1400 at

800kHz. If we say it is Q = 1200 and the Reactance at 800kHz is 1,166 ohms, the resonant peak will be 1.33Megohms. btw, I also measured my Qs with a Boonton 260. I had to reduce the injection voltage (Multiplier) way down in order to make that high Q measurement, Mikek

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I just meant it is small and I couldn't see what it was. Initially I thought it might be a small value ceramic cap you put in place to check out the amp before you optimized the cap. Now I realize it is a very small piece of PCB and *is* the optimized cap. I also didn't know which image was earlier and which was later.

Why would you adjust the overall circuit gain by trimming the cap? The amp gain can be trimmed, no?

What frequency range is this amp intended to be used for? Nearly all the stuff I've seen where Q matters is LF or MF. They use loop antennas for higher frequencies, but with too high Q the bandwidth gets so narrow even voice won't get through, just CW.

Some of the antennas Kleijer uses I would think would limit the bandwidth of his circuits with Q of over 1000.

If the frequencies are not so high, why not use an opamp for the final stage rather than the much more complex push-pull stage? I assume that is intended to be in essence a "power" output stage and does not otherwise contribute significantly to the characteristics of the rest of the circuit.

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Rick C

I'm sure the amp can be changed, but the end game was to make the input cap small, I couldn't go much smaller.

Kleijer says it's flat from 10kHz to 10 MHz.

Most of the stuff I'm looking at is crystal radio, by the time you start driving a headphone your Q gets lowered a lot. I never hear the crystal radio guys complain about lack of bandwidth.

Again not a common problem, but easy to fix with a load resistor.

I'm sure it could be built with a couple of opamps. It is a 50 ohm output. Could probably do better, this starts to distort at I think

8 Vpp. But I don't have any antenna that need much that headroom.

Mikek

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I'm not looking for bandwidth. I don't think I can get the Q high enough to cause a bandwidth problem.

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Rick C

Right.

Could be higher I guess. Depends on the FET's Ig. I was just piling onto the existing circuit.

Raises the input impedance, yes. Much? Maybe 5x for this quickie mod.

I didn't pick it, but I'm sure the designer picked that FET for low input capacitance and low feedback capacitance.

He chose well. I might consider a 2n4117, though, depending on the bandwidth requirements (I don't know how fast this thing has to be). But Mike already has the BF256C on hand, and it's pretty good.

Cheers, James Arthur

If the headphones or load resistor limit the Q, what is the bleeping point of all that Litz wire and custom tuning capacitor stuff? I thought the whole point was to get the Q as high as possible. If you start with a Q over 1000 and end up with a few hundred, I wouldn't expect that to be distinguishable from a coil with a Q of some hundreds which is where Kleijer started in his coil investigations.

I designed an opamp circuit to output an 8 volt signal with a 12 volt supply into 50 ohm cable. I didn't quite get 8 volts before it starts to distort, but it was close. I believe the gain-bandwidth product of the amp was over 30 MHz, so it should do ok at 10 MHz and no gain. But then you need a little gain I guess. Opps, no, the GB was 15 MHz.

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Rick C

Yep.

I suspect a lot. But you could always measure & be sure.

Yes.

No, this applies to the whole scheme. The more closely all the nodes follow the input signal voltage, the better our result.

Both.

Yes. I was just sketching the relevant part of the front-end.

You've got it perfectly. I don't expect a very large improvement from bootstrapping the 20M alone though--a resistor's capacitance is pretty low already, and two in series, even lower.

Your author's figures are inconsistent. He starts saying the input capacitance is 1.4pF and the input coupling cap is 0.3pF, but then he says the 0.3pF and FET T1's capacitances form a 17:1 divider. That can't all be true--0.3pF should form a 5.7:1 divider with a 1.4pF input, not

17:1.

When I guesstimate a 5x improvement, I'm banking on the 17:1 being true, c.in(eff) being 5pF, and getting that down to 1pF, roughly, with the circuit I sketched.

If you're already really at 1.4pF the improvement will only be 1.0pF/1.4pF, and not 1.0pF/5pF.

As I said before, a better buffer could do better--you could tweak the bootstrap to perfect null--but then chances are you'd have an oscillator.

What I posted seemed like a reasonable compromise for a first try.

Cheers, James Arthur

Yes, it is. But you still need to get the maximum power out to drive your headphones. In a perfect system, you would drop your Q by half to get maximum audio. But the diode gets involved and that I can't decipher.

This is radio dxing, grab every db of signal!

That's 8Vpp open circuit, 4Vpp loaded.

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Well then, eazy peazy. I used a circuit that used positive feedback to make the output impedance look higher than the resistor in the circuit. I think I used 12.5 or maybe it was just 12 and got multiplied by 4. This makes the calculation of the gain a little more complex so you would not be able to adjust the gain with a pot or anything... unless it was on an earlier stage.

The middle stage with the single transistor, anything special about that? It has a connection with caps I'm not familiar with from the emitter to the input. That would be a low level of positive feedback I believe. Any idea why?

Wait a minute!!! That bootstrap of the input resistors won't work. The

470 ohm source resistor is there to set a bias point with the source higher than the gate. The resistors are there to set the DC voltage of the gate to ground. Perhaps a cap could be used to feed the point between the two resistors?

I just realized that is the purpose of the two caps on T2 emitter to the

47 ohm resistor. It allows the resistors to set the DC operating point but bootstraps their capacitance. No?

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Rick C

Hmm, I don't know offhand!

Looks like magnets are moderately sensitive:

I'm not sure offhand how many n/cm^2 corresponds to what Sv/Gy/rem/etc. rate, or of course how that compares to beta or gamma radiation. But strong reductions in the 10^15 range seems pretty sensitive, like a knock-the-electron-out-of-place-and-she's-done ~quantitative kind of rate.

Hmm, I wonder how much that depends on the boron content. Ferrite magnets wouldn't need to worry about that.

Anyway, no, you wouldn't need to bring the magnet with, not unless you need that as a semi-permanent lock signal (as it's used in commerce).

There's also Wiegand wires, which I think are embedded (or omitted) in an array, to give a pattern of blips as they slide past the pickup coil and become magnetized.

Tim

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I thought about this a bit and came up with an improved follower.

The main limitation of the previous circuit was the FET's poor performance as a voltage-follower. Unaided, the T1 has a gain of about 0.6. That hits our bootstrapping from all sides. First, c(gs) (the largest capacitance) is only bootstrapped by 60%, leaving 40% of the BC547C's ~5pF c(gs). Next, we use that voltage to drive our less-than-unity Q2, which drives less-than- unity Q1. This all adds up.

Changing T1's load to a current sink makes T1 into a much better follower, increasing voltage gain from 0.6 to about 0.95. The better 'follower' action now bootstraps away nearly all of c(gs) (T1's largest capacitance), and gives us a better signal to drive the drain bootstrap as well. Good, good, and good. And not terribly much trouble to do, either.

Vdd Vdd -+- -+- | | | [22k] R5 Q1 \| | BC547B |---+-------. .

Mind that the probe acts like (9M || 11pF) + (100pF || 1M), but the cable and scope input aren't actually 100pF but are lossy, having probably around

50 ohms ESR, and all the distributed L and C implied by a transmission line.

If you look at (10pF + 50) ohms, the parallel equivalent is 5M (at 1MHz, to ballpark it), so the overall resistance is about 3M.

Still pretty big, but only 20 times bigger than 150k, not a huge amount. :)

Tim

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Thanks for the time. As it is now constructed the enclosure is the shield. Is that good or bad? ie. Should the enclosure be isolated from the shield?

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