HP 33120 confuse explanations?

[J4e8a16n]

Hi,


The quote comes from:  http://www.prc68.com/I/JouleThief.shtml#Fig6

Current Transformer

Using the Fair-Rite 73 material transformer used above with two windings of 59 turns each (see above) with the center tap where the power supply was connected not used.
To calibrate a single wire was passed through the center and driven by the HP 33120 with a 10 V pk-pk square wave.  Since it has a 50 Ohm internal impedance the calibration current is 200 ma.  The scope displays 171 mv so the cal factor is 1.17 Volts/Amp.

This plot is not showing a linear ramp down from the peak current but instead a step.  That may be because this is not an optimum current transformer.
3 Nov 2011 - Note:

The output of a current transformer needs to have a resistive load since it's output is a current, not a voltage.  By choosing the load resistor you can end up with a "nice" Volts/Amp conversion constant.  That's probably why the output looked wrong.

The current transformer is NOT DC coupled.  The DC level will average to zero.
The vertical channel has zero volts offset and is 100 mv per division. 
During the LED off time the current through T1B and the transistor C-E junction ramps up until the transistor is turned off.  That transition is the trigger point at the left edge of the scope.
Immediately after the transition the transistor current is zero which allows setting the DC levels on the waveform.  Adding 75 mv gives the DC levels.
0 ma at the left edge and ramping up to 180 ma.

Hcore = (0.4 * PI * Turns * Current)/ (effective length of core)
Hcore  = (0.4 * 3.14 * 7 * .18) / 2.18 =0.726 Oe 

Why isn't it 59 turns?

http://info.ee.surrey.ac.uk/Workshop/advice/coils/terms.html#strength
The power supply is showing a current of 74 ma.
The emitter current does not include the LED current so a seperate measurement was made of the total current which turns out to have the same peak to peak value but looks more like a sawtooth (does not have as pronounced a flat part).  A linear ramp from zero current to 180 ma has an average value of 90 ma.  But this is over 32 us and the other 5 us of the period there is no current so the average current would be 32/(32+5) * 90 = 77 ma which is pretty close to the meter reading of 74 ma.

Why the cal factor leads to 1.17 Volts?
Where does comes from this calculation ?

32/(32+5) * 90 = 77 ma

  The period is 36 us.

Thanks for your attention.

JP

[dacman]

200 mA is expected, 171 mV is the reading.  For a cal factor, 200 mA / 171 mV = 1.17 Amps / Volt.
To correct the reading to Amps, multiply the scope reading by the correction factor.

[J4e8a16n]

200 mA is expected, 171 mV is the reading.  For a cal factor, 200 mA / 171 mV = 1.17 Amps / Volt.
To correct the reading to Amps, multiply the scope reading by the correction factor.

He does not seem to have a picture of those  171mV on his page.  Is this 171mV   AC?   Or is it the +171mV side of a 342mV AC?

Am I dumb? 
(I know you cant really say yes  ;-)

JPD

[J4e8a16n]

200 mA is expected, 171 mV is the reading.  For a cal factor, 200 mA / 171 mV = 1.17 Amps / Volt.
To correct the reading to Amps, multiply the scope reading by the correction factor.

1.17 Amps / Volt.

I agree with you lthough on the page it is writen Volt/Amp...  Otherwise we would have V^2/A
Later in the page

"nice" Volts/Amp conversion constant.