AMX TX

[Carl-NC]

When I look at the schematic again... "Rsens" of 0.1 ohm seems to be the sweet spot for the current sensing point.
You didn't explain... or I didn't read it, but is that what the ADC1 and ADC2 samples are for?

I want to sample the flat-top at 2 points. The second point is used in the servo to set the buck regulator voltage for a particular peak current. Both points are used to determine slope for the slope correction PWM.

[ivconic]

Yes, you could probably design some kind of fast shoot-through sensor and, with a lot of extra circuitry, figure out a way to deal with the problem. But again, why not just take advantage of the micro and do it the easy way? And yes, I always include lock-up protection in the TX by capacitor-coupling the TX clocks. I haven't shown that yet, but it's in the actual design schematic, which is unfinished. The exact timing is TBD, I won't know that until I build the circuit.

Coil resistance almost solely determines power consumption so you want it as low as possible.

And "TBD" stands for "(T)remulant (B)lagger (D)riftometer", I assume?

(That's the Carl character I was talking about; everything in a sentence that can be shortened; he will cut it short! He is very sparing of letters!)
...

"...Coil resistance almost solely determines power consumption so you want it as low as possible..."



Low as possible... what? Power consumption or coil resistance?
Because lowest possible coil resistance will lead to highest possible power consumption and vice versa!
...
No, I was thinking; in order to protect fets: is safer if coil resistance is higher...
But than again; the current will be low(er).
And that's something, I assume; you don't want here?

[lucifer]

...
No, I was thinking; in order to protect fets: is safer if coil resistance is higher...
But than again; the current will be low(er).
And that's something, I assume; you don't want here?[/FONT]

Coil resistance won't help you protect the FETs. The shoot-through current is the current that flows from positive to negative terminal passing only through the FETs (and not through the coil). And since FETs have very low ON resistance that is quite the same as a short circuit. This is why you need the dead times in the clock pulses that drive the FETs.




As to the coil resistance, I think you need it as low as possible. The higher the coil resistance, the more current is lost as heat.

[ivconic]

Coil resistance won't help you protect the FETs. The shoot-through current is the current that flows from positive to negative terminal passing only through the FETs (and not through the coil). And since FETs have very low ON resistance that is quite the same as a short circuit. This is why you need the dead times in the clock pulses that drive the FETs.




As to the coil resistance, I think you need it as low as possible. The higher the coil resistance, the more current is lost as heat.

Now I understand, thanks!

[Qiaozhi]

And "TBD" stands for "(T)remulant (B)lagger (D)riftometer", I assume?

"To be decided" or "to be determined".

[ivconic]

"To be decided" or "to be determined".

I knew it, our good George! Just kidding Carl. He wouldn't be him if he didn't shorten the terms as short as possible!

[Carl-NC]

I have to use TLAs, I'm a slow typist.

Power consumption is proportional to coil resistance for a given TX current. Suppose I want a peak current of 500mA (peak-to-peak = 1A) and coil R=2Ω; also assume the FET switches are ideal, and there are no diode drops. VTX will need to be 1V (0.5A*2Ω) and power dissipated by the coil is 0.5W. Note that the power delivered by VTX is also 0.5W because there are no other losses in the circuit. Now suppose I wind a coil with R=4Ω... VTX needs to be 2V to achieve 0.5A and power dissipated is 1W.

In reality there are a couple of diode drops and the FET switches have a very minor amount of resistance. But coil resistance sets the pace. Keep in mind this is a constant-current TX circuit, not constant voltage. Also, Lucifer nailed the shoot-through issue.

[Willy Bayot]

So the correction granularity will depend on the micro clock. The plot above suggests that a 100MHz micro will get you within +/- 10uA of perfect.

About the error mesasurement, to get a precision of +- 10µA, we need to be able to measure the difference between the two end of the slope with enough precision.
A difference of 10µA generates a difference of ADC voltage of 10µV. This is 3.3V/10µV = 330000, so a part over 330000 or 3 ppm.
This needs an ADC precision of 15 bits.
If using standard integrated ADC of 12 bits, we shall need to apply a digital integration to gain the missing 3 bits. That means summing 4^3 = 64 consecutive periods before using the integrated error measurement.

[Olly]

About the error mesasurement, to get a precision of +- 10µA, we need to be able to measure the difference between the two end of the slope with enough precision.
A difference of 10µA generates a difference of ADC voltage of 10µV. This is 3.3V/10µV = 330000, so a part over 330000 or 3 ppm.
This needs an ADC precision of 15 bits.
If using standard integrated ADC of 12 bits, we shall need to apply a digital integration to gain the missing 3 bits. That means summing 4^3 = 64 consecutive periods before using the integrated error measurement.

Perhaps we would be better off by amplifying the difference between the two signals first and then presenting that to the ADC

[Willy Bayot]

Perhaps we would be better off by amplifying the difference between the two signals first and then presenting that to the ADC

No way, the two values to measure do not appear at the same time. Thus, it is not possible to make the difference with an analog circuit and then, amplify it.

[Olly]

No way, the two values to measure do not appear at the same time. Thus, it is not possible to make the difference with an analog circuit and then, amplify it.

I was talking about amplifying the difference between the two sampled and integrated values of course…

[Olly]

I was talking about amplifying the difference between the two sampled and integrated values of course…

To clarify, I’m suggesting a differential amplification of the 2 sampled signals TXV1and TXV2 from Carl’s schematic. No reason this should not be possible.

[Willy Bayot]

To clarify, I’m suggesting a differential amplification of the 2 sampled signals TXV1and TXV2 from Carl’s schematic. No reason this should not be possible.

Those signals do not appear at the same time, One is sampled at the beginning of the slope and the other one is sampled at the end. They could not be amplified in the same differential amplifier

[Carl-NC]

Willy is correct. The resolution of the 12b ADC is 0.8mV which corresponds to 0.8mA of tilt. If we want better tilt accuracy, more analog gain is needed. But we don't want to apply much more gain to the peak current reading; currently it is scaled to 1V/A with (again) a resolution of 0.8mA, which is about right for controlling the peak. It could probably be multiplied by another 2x (2V/A) but no more.

Olly is also correct; we can solve this by subtracting the samples and applying more gain to the difference. We do this all the time in PI by subtracting the EFE sample from the main sample, even though they don't occur at the same time. The only reason you would not want to do this is if you are trying to correct tilt at a pulse-by-pulse speed, which is probably not realistic. Like automatic ground tracking, you do it over a short span of time. So we can assume a sampling tau of at least 10s or 100s of half-cycles.

Here is a circuit that should solve this:


I've posted complete schematics here so I will make changes, pending feedback.

[Willy Bayot]

We do this all the time in PI by subtracting the EFE sample from the main sample, even though they don't occur at the same time.

Yes but this subtraction is made in PI from INTEGRATED samples which is not the case here. You just cut two separate windows. They would need to be integrated first by an analog integrator for this subtraction to work.

So we can assume a sampling tau of at least 10s or 100s of half-cycles.

If you make a digital integration over such a long duration, I fear that the reaction of the compensation system will not be fast enough to correct the fast ground variations and the variations of coil height over the ground.