Measuring a flyback's current / High Voltage / Forums

Forums

4hv.org :: Forums :: High Voltage

« Previous topic | Next topic »

Measuring a flyback's current

previous 1 2 3 next

Move Thread

LAN_403

buxtronix

Tue Oct 04 2011, 11:12PM

Registered Member #4078 Joined: Tue Aug 30 2011, 12:53PM
Location: Sydney, Australia
Posts: 19

Lots of interesting ideas here!

As I was racking my brain further, I realised that I had at my disposal a sizeable capacitor - 0.25uF @ 35kV. Surely I can use this by measuring the time it takes to charge to a given voltage?

Give the RC time constant equation (t = RC), and given that R = V/I, then it makes sense that t = VxC/I and reworking this, it should turn out that I = (VxC)/t

So for example, if it takes 5s for my driver to charge 0.25uF to 15kV, then the average current is:

I = (15kV * 0.25uF) / 5

Which is 0.75mA. Sound reasonable?

Patrick

Tue Oct 04 2011, 11:35PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

buxtronix wrote ...

Lots of interesting ideas here!

As I was racking my brain further, I realised that I had at my disposal a sizeable capacitor - 0.25uF @ 35kV. Surely I can use this by measuring the time it takes to charge to a given voltage?

Give the RC time constant equation (t = RC), and given that R = V/I, then it makes sense that t = VxC/I and reworking this, it should turn out that I = (VxC)/t

So for example, if it takes 5s for my driver to charge 0.25uF to 15kV, then the average current is:

I = (15kV * 0.25uF) / 5

Which is 0.75mA. Sound reasonable?

Yes that sounds reasonable, and as its so simple id try it first before any of the other means.

Sulaiman

Wed Oct 05 2011, 09:19AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

If you are using a fixed duty cycle at a fixed frequency
to turn on and off a transistor
that switches the primary of the flyback,
then you are transferring pulses of fixed energy
(E = 1/2 x L x I^2) and (I = Vpri x Ton / Lpri )
at a fixed rate, i.e. constant power output.

Charging a capacitor from an initially discharged state
will result in longer duration current pulses at low voltage
rising to shorter duration current pulses at high voltage.

The output current pulse shape looks like a triangle with a steep front edge and a less steep trailing edge, but smoother.
The peak of the output current triangle is (ideally) the peak current at which the primary was switched off, x ( Np/Ns )
or
primary (Amp.turns) goes to secondary (Amp.turns)

P.S. if you 'scope all of this you will see all kinds of 'ringing' due to various capacitance/inductance resonances,
the description above 'filtered' ringing out.

P.P.S. based on the above,
using a flyback to charge an initially discharged capacitor
the transistor must be off long enough
for the energy in the flyback transformer
to be transferred to the capacitor.
So if the off time is not long enough,
each cycle will add excess energy to the flyback circuit
until something fails.
(or gets hot, misbehaves briefly, then carries on, if you're lucky)

PresentTeck

Wed Oct 05 2011, 08:41PM

Registered Member #3922 Joined: Thu Jun 02 2011, 06:24AM
Location:
Posts: 23

couldn't you measure the amps and voltage being drawn from the wall, and get wattage being drawn.

then take the amperage off the HV output and using Watts=Volts * amps get the voltage?

i wouldnt want to use a in line ampmeter, but one of the ones you can put on the outside of the wire, i think that would work right?

astralhighway

Wed Oct 05 2011, 09:48PM

Registered Member #4107 Joined: Sun Sept 25 2011, 07:30PM
Location: London
Posts: 53

Hi Present Tack, snap! That's another way of saying what I said back in my post on Oct 04 2011, 01:31PM

PresentTeck

Wed Oct 05 2011, 10:00PM

Registered Member #3922 Joined: Thu Jun 02 2011, 06:24AM
Location:
Posts: 23

sorry, i'll stop reading the forums at my school, it doesn't always load everything... but at least i know my idea would work

Mattski

Thu Oct 06 2011, 04:57AM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

wrote ...
couldn't you measure the amps and voltage being drawn from the wall, and get wattage being drawn.

then take the amperage off the HV output and using Watts=Volts * amps get the voltage?

i wouldnt want to use a in line ampmeter, but one of the ones you can put on the outside of the wire, i think that would work right?

It will be very approximate, since not all of the input power makes it to the output arc. Also the voltage and current waveforms are not constant, and the average output current times average output voltage is not necessarily equal to the average output power.

Xray

Fri Oct 14 2011, 04:59AM

Registered Member #3429 Joined: Sun Nov 21 2010, 02:04AM
Location: Minnesota, USA
Posts: 288

Everyone who suggests putting a meter in series with the load presumes that the meter should be placed (floating) in the high voltage output wire of the flyback, but there's a much better and safer way to measure the load current of a high voltage transformer. Disconnect the GROUNDED (low) secondary wire, and stick your ammeter between that wire and ground! So long as the meter stays connected in that manner, the voltage on the meter will be very low, typically only a few volts or less.

This is a very common method that was used to measure the tube current in the older AC type of X-ray machines. By AC type, I mean the X-ray tube is wired directly across the hv secondary windings, and the hv secondary "low" lead is conected to a terminal that allows monitoring the mA (tube current). The biggest difference between an AC X-ray transformer, and a flyback transformer (besides the obvious difference in core material) is the waveform. Any cheap milliammeter that you can buy from Radio Shack can measure the nice clean sinewave that your power company provides, but would have a very difficult time measuring the hashy, ringing, horribly noisy waveform that your hobbyist flyback circuit produces.

If you want somewhat of an accurate current measurement, probably the best way to do it would be to place a resistor between the "low" secondary lead and ground, and measure the waveform across it with an O-Scope. By integrating the voltage under the curve (you can smooth the hash with a capacitor) you can then calulate the power using Ohm's Law.

Comments?

teravolt

Fri Oct 14 2011, 03:57PM

Registered Member #195 Joined: Fri Feb 17 2006, 08:27PM
Location: Berkeley, ca.
Posts: 1111

i agree with Xray. the current viewing resistor(CVR) on the ground side should be any were from 1 to 10 ohms then you can rectify the voltage and filter it and connect a Amp meter. your cvr acts like a shunt and the value will detrmin the scaleing. you could add a series trimpot with amp meter to calibrate it.and the size of your filter cap will determin the response of the meter.

Xray

Fri Oct 14 2011, 10:56PM

Registered Member #3429 Joined: Sun Nov 21 2010, 02:04AM
Location: Minnesota, USA
Posts: 288

teravolt wrote ...

i agree with Xray. the current viewing resistor(CVR) on the ground side should be any were from 1 to 10 ohms then you can rectify the voltage and filter it and connect a Amp meter. your cvr acts like a shunt and the value will detrmin the scaleing. you could add a series trimpot with amp meter to calibrate it.and the size of your filter cap will determin the response of the meter.

I think in that situation you would use a voltmeter rather than an ammeter. You measure the voltage drop across the "CVR", and good ol' Mr. Ohm's Law will give you the current.

previous 1 2 3 next

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob