Calibrator Driver & Coupling Considerations

M. Kesselman

8-5-02

The design review committee pointed out that the calibrator should
include provision to assure that it cannot injure the current
transformer if it should fail. The DC current associated with a current
amplifier failure causing transformer saturation or over-current in the
calibration winding was of particular concern.

Bergoz was contacted and a request made of transformer failure modes
(see appendix A). From this; one important parameter is the saturation
current of 0.01Coulombs. Using this number one can determine the maximum
current allowable before saturation for a sinusoid. This sets a peak
sinusoidal current allowable to 0.01πf.

1.  To assure that at no time could a failure in the calibrator drive
    amplifier cause saturation, AC coupling is required.

2.  In addition, to assure that sinusoids of low frequency do not cause
    saturation a limited driver power supply is required. This would
    require a double failure mode to cause harm to a transformer. A
    maximum of about 1 amp calibration winding current would simulate a
    10 amp beam current due to the 10 turns in the calibration winding.
    Limiting the power supply to about 1.2 amps would require operating
    the calibrator at a frequency of no less than 38.2Hz.

3.  Transient characteristics of the coupling circuitry must permit the
    output current to stabilize and decay back to zero within the
    inter-pulse period of about 15.7ms. This time is determined by the
    coupling capacitor, load impedance (near 1 ohm) and the source
    impedance of the current source. To achieve high accuracy, one wants
    a high source impedance, while to match the twisted pair 78 Ohm
    cable one needs a source impedance near 78 Ohms. One compromise is a
    DC source impedance near 1000 Ohms for near 0.1% accuracy, and an AC
    source impedance near 78 Ohms to match the transmission line at
    higher frequencies. Keep in mind that driving the load with a
    sinusoid will cause the AC source impedance to strip-off some of the
    current. The band pass characteristics can be somewhat controlled to
    achieve a compromise. (consider 1K DC source resistor, and an 82 Ohm
    in series with a 0.01uF AC load at the source (194KHz), and a 0.5uF
    coupling capacitor (0.5ms time constant)). The suggested parts will
    provide a recovery time constant of 0.5ms, and will stabilize within
    0.73% in 2.5ms, and will decay to 3E-5% in 7.5ms (15uA for 50 amps).
    This will allow a minimum of 10ms for the calibration to take place,
    leaving 5.7ms for additional timing activities.

4.  Operating at "reasonable voltage" levels is another requirement for
    practical amplifier selection and power considerations. The
    transformer looks to the calibration source like a 1 Ohm resistor in
    parallel with the calibration winding inductance of about 1mHy. The
    coupling capacitor will dictate the required drive voltage and
    therefore the power requirements. Operating at low frequencies (must
    use frequencies of about 1KHz and higher to assure the transformer
    is operating in the flat portion of its bandpass characteristics)
    will require higher voltage. A 30KHz signal will deliver 0.983 amp
    through a 0.5uF capacitor with 8.33 peak volts ( with the load AC
    terminated in 140 Ohms in series with 0.01uF (113KHz) to achieve a
    high frequency 50 Ohm termination for the output winding transient
    response). This will require a supply capable of a peak of 69.4 VA.
    Note the voltage and current are nearly 90 degrees out of phase,
    delivering only 1W peak to the load (1 amp peak into 1 Ohm). If the
    frequency is increased, the voltage can be reduced (reducing power
    requirements) however, the AC coupled matching networks will begin
    to draw more current away from the load and
    ![](media/image1.wmf){width="5.988888888888889in"
    height="1.479861111111111in"}introduce some more error.

Figure 1 -- Simulation Circuit

![](media/image2.png){width="6.0in" height="4.5in"}

Figure 2 -- Analysis results for 30kHz sinusoidal current (1.7% lost at
input, 99.5% efficiency to load, 0.5ms time constant)

APPENDIX A

Sent Mon 7/29/2002 6:59 AM

From Julien Bergoz \[bergoz@bergoz.com\]

To Kesselman, Martin

CC ian@gmw.com; brian@gmw.com; Cameron, Peter; Dawson, W. Craig;
\'Richard Witkover\'

Subject Re: Transformer failure question

Dear Martin,

Thank you for the news about the Design Review. Is BNL/SNS closer to a
decision

on the RING BPMs ? I.e. what position info is needed: baseband, 402.5MHZ
?

Now about the FCTs:

1a) A high-frequency current in the cal. winding causes core losses
(eddy currents \--\> Joule effect in the core alloy).

I don\'t have a formula to describe it. Let me give some examples:

\- 50 A.turns rms at 13.56MHz cause \< 2W core loss hence heating.

\- 500 mA.turns rms at 500 MHz and harmonics of unfiltered 50ps bunches

cause 40-50\'K heating of FCT core in air.

If you think you may be flirting with danger, let me know the frequency
considered.

1b) An AC or DC current will cause winding wire heating by Joule effect.

Depending of transformer size, resistance is from 0.3 to 0.6 ohms.

1c) The energy transfer from calibration winding to read-out winding is
limited (stops) when the core is saturated. The core will saturate with
ca. 10mC. Once the core is saturated, the read-out winding receives no
signal anymore.

2a) The FCT is very likely to be permently damaged when exposed to \>
210\'F for more than 1-2 minutes. We recommend not to heat it beyond
176\'F at any time.

2b) An external magnetic field will alter, then disable, reversibly (not
destroy) the FCT. Typically, core saturation due to external field
starts with 20 Gauss of radial field (transverse to the beam direction)
or 100 Gauss axial field (colinear with beam direction). Partial
saturation of core causes the droop figure to increase, lower cutoff
frequency to increase. Otherwise, a partially saturated core gives true
read-out value at higher frequency. When the core is completely
saturated, the lower cutoff reaches the higher cutoff, and no more
signal gets thru.

3\) To desaturate a core, pass an oscillating current thru any of its
windings. This can be done at any frequency. Common knowledge says the
demagnetization sequence should start with a current high enough to
saturate the core, and the current be progressively reduced until nil.
Our experience shows that a current too low to saturate the core will
still be able to desaturate it.

My guess is that the calibration current you will apply to the FCTs is
probably high enough to desaturate the FCT core \--in a few thousand
cycles\-\-- after it has been saturated. Note: to saturate it, make
10-turn winding around the FCT and pass 2A dc. It will be completely
saturated !

Best regards, Julien

\>Hi Julian,

\>We got through our Design Review this week OK.

\>One question that was raised during the review of the beam current
monitor

\>system has to do with failure modes.

\>They were interested to know if the calibrator failed and delivered
some

\>(maximum) current to the calibration winding, how much current would
be the

\>maximum allowed to assure the transformer does not fail?

\>This sounds confusing.

\> Let me rephrase;

\> 1- How much DC current (amplifier failure-delivers max current)
placed on

\>the calibration windings is permissible?

\> 2- Can you foresee any other failure modes?

\> 3- How would one demagnetize the core if it became saturated by a

\>calibrator failure? Do you have a recipe?

\>

\>Regards,

\>MartyK

\--

Julien Bergoz (Mr.)

Bergoz Instrumentation http://bergoz.com

Espace Allondon Ouest - 01630 Saint Genis Pouilly, France

Tel.: +33-450.426.642 Fax.: +33-450.426.643 GPS: N46°15\'28\" E6°1\'41\"

See you soon at :

LINAC\'02, Gyeongju Korea, August 19-23, 2002.
http://linac2002.postech.ac.kr

FEL\'2002, Argonne, Sept.9-13, 2002, http://www.aps.anl.gov/fel2002/

DIPAC\'03, Mainz,
http://www.esrf.fr/conferences/DIPAC/images/DIPAC2003.pdf

APPENDIX 2

Analysis of transformer calibration winding:

Equivalent circuit analyzed

For this circuit, it represents an impedance load on the current source:

Z=num/den

Where; num=Rs\[sLcal +RL +(SCc)(Slcal)RL\]

Den=s^2^Lcal\*Cc\*(Rs+RL)+s\*Cc\*Rs\*RL+s\*Lcal+RL

The denominator can be put in the form;

s^2^ +2\*ζ\*ω~n~+ω~n~^2^

with

ζ\*ω~n~ =(1/2)\*\[Rp/Lcal +1/(R~Σ~\*Cc)\]

Rp=Rs\*RL/(Rs+RL)

R~Σ~=Rs+RL

The poles are at;

-ζ\*ω~n~ {1+/- jsqrt(1-1/ζ^2^)}

Therefore, the transient response time constant is;

τ=1/ζ\*ω~n~