10 mH inductor (28R)

[J4e8a16n]

Hi,

What is the meaning of 28R ?

JP

[Jebnor]

Non reactive resistance?  as in 28ohms?

[Mandelbrot]

28 ohms does seem like a reasonable resistance of a 10mH inductor. That could be the case.

[J4e8a16n]

Maybe it is a reactance?

[c4757p]

No, that timer will run around 2 kHz, and the reactance of a 10mH inductor at 2 kHz is 126 ohms.

[TerminalJack505]

That schematic originally comes from here, by the way.  That's Colin Mitchell's Talking Electronics site.  The 28R is the inductor's ESR.  The site eludes to this here.

[vlf3]

The DC metered resistance would depend on the wire gauge used, an example I have, the DC resistance comes out at 390 ohms ?

Interesting to find the 555 frequency is near 2.220 Htz and reactance is 139.42 ohm... that's assuming a 50% duty cycle, due pin 3 output provides about that percentage duty.

[J4e8a16n]

Hi,

Should I set the oscilloscope to dc?

I have build the circuit here are the results.
The inductance resistor from the multimeter reading is 15 Oms. I bought the inductance it is supposed to be 10mH.
1451 Hz.
The power supply delivers 23 mA
The zener uses about 7.5 mA.

I used a zener witch is readed  at 2.5 volts.

If I measure the output with a multimeter, I get 97 volts!

[alm]

Try finding a resource that explains the difference between (DC) resistance and reactance without introducing stuff like complex math. I don't have one handy, AllAboutCircuits.com might have a good write-up.

A perfect inductor will have an inductance, but no DC resistance. A DMM set to resistance would measure zero Ohms (it uses DC), but an AC signal will be affected, because it has reactance (AC resistance). The reactance is caused by its inductance, which is measured in Henry. So a perfect inductor might have an inductance of 10 mH but a DC resistance of 0 Ohms.

In the real world, copper wire has a finite resistance. Especially many turns of fairly thin copper wire. This causes a DC resistance in addition to the current. This real inductor might measure 28 Ohms with a DMM (its DC resistance), but still have a 10 mH inductance. A DC signal would see it as a 28 Ohm resistor (since DC is not affected by inductance), but an AC signal will see it as a 28 Ohm resistor with an ideal 10 mH inductor in series (a decent approximation for low frequency work). Both the ideal and the imperfect coil will have the same inductance, but will behave slightly differently because of the difference in DC resistance.

This zener diode tester will deliver a fairly high DC voltage, but limited to a low current. A reverse-biased zener should start conducting as soon as the voltage across it exceeds it nominal voltage (eg. 2.5 V), thereby limiting the voltage across it. If you measure the voltage without a diode across the output, then you should expect to measure a very high voltage. If you measure the same with the zener connected, then the zener might be open circuit.

It looks like the DC resistance of the coil limits the current through it when BD679 is conducting, so you might get a higher maximum voltage/current with a lower series resistance. From the 63v, I would expect the maximum output voltage to be well below 100 V. You could try putting a ~12 Ohm resistor in series with the coil and see if that lowers the voltage.

[J4e8a16n]

Thank you for te answer. 

Does the diode split the frequency by 2?

Does the oscillating coil create an ac current?

Does the 555 create an ac curent at pin 3?

If I measure the voltage between the PS + terminal and the neg side of the diode. I get 3.3 volts.
Between the PS + side  and the diode + side,  I get 3.8 volts!

 

I took other measures.  The DC and AC makes things complex.  The multimeter and oscilloscope too !

JP

[J4e8a16n]

A DC signal would see it as a 28 Ohm resistor (since DC is not affected by inductance), but an AC signal will see it as a 28 Ohm resistor with an ideal 10 mH inductor in series (a decent approximation for low frequency work).
This zener diode tester will deliver a fairly high DC voltage, but limited to a low current. A reverse-biased zener should start conducting as soon as the voltage across it exceeds it nominal voltage (eg. 2.5 V), thereby limiting the voltage across it. If you measure the voltage without a diode across the output, then you should expect to measure a very high voltage. If you measure the same with the zener connected, then the zener might be open circuit.

It looks like the DC resistance of the coil limits the current through it when BD679 is conducting, so you might get a higher maximum voltage/current with a lower series resistance. From the 63v, I would expect the maximum output voltage to be well below 100 V. You could try putting a ~12 Ohm resistor in series with the coil and see if that lowers the voltage.

I know the maths.  My problem is understanding them in a real circuit.

Thanks for the explanations.

>when BD679 is conducting
the inductor build is MField

>when BD679 is closed
the inductor  discharge in the 1uF, 47K and the output?

[TerminalJack505]

Here's how I would explain that circuit...

When the transistor conducts a current will flow through the inductor.  This current will cause a magnetic field to be created.  This is why the current through the inductor builds-up slowly, rather than instantly--the energy goes into building the magnetic field.  Once the inductor saturates (as I assume it will due to the low frequency but that may not actually be the case) the magnetic field no longer builds-up and the current is at its maximum.

When the transistor turns off the magnetic field begins to collapse.  All the energy that went into building the magnetic field is returned to the circuit.  (Naturally, there's some loss.  We're just talking theory here.)  The current through the inductor wants to continue flowing because of the collapsing magnetic field.  If it can't find somewhere to flow to then the voltage across the inductor begins to rise.  The topmost terminal of the inductor is held at a constant 5.8V so the rising voltage is seen at the other terminal.  (The 1N4148 diode's anode.)

When you have a Zener diode in the circuit then it will clamp the rising voltage to whatever the Zener is rated at.  If you don't have a Zener diode in place then the voltage will rise as high as it can.  One of the biggest factors in how high it will go is the Q factor of the inductor.  (The DC resistance of the inductor is one of the factors in the inductor's Q factor.)  Everything else being equivalent, an inductor with a higher Q factor will generate higher voltages than one with a lower Q factor.  The inductor shown in the schematic with the higher DC resistance likely produces lower voltages.

Since you are seeing the voltage rise to 100V you probably want to put a permanent Zener in the circuit to limit how high it goes--to protect the components as well as the operator.  The transistor is rated only to 80V and the capacitor, as shown in the schematic, is rated at 63V.

[J4e8a16n]

If it can't find somewhere to flow ...

Thanks for your clarification.
It can always flow throughout the 47k resistor. Maybe not fast enough....

JPD

[TerminalJack505]

Correct.  I think the resistor is primarily meant to bleed the charge off the capacitor once the circuit is powered down.

[J4e8a16n]

I red 103 volts with a Fluke.

So, if the impedance is at 0 while discharging,  the current would be  103/ 28 =  A ?


The circuit draws 162mA from the power supply.  10mA goes throughout the multimeter and to the zener, 5 to the 555 , could the remaining xmA go for the Inductor?

JP

NB I have updated the file in the June 25, 2013, 06:38:05 AM  message

[c4757p]

Think of an inductor like momentum for current. You build up momentum when the transistor is on, then when you shut it off, the current slams through the diode instead. It's not going any higher than what the current through the inductor was at the moment the transistor shut off, which is determined by the inductance, series resistance and switching rate. I think I got the switching rate right (out of memory from when I calculated it a few days ago  ), so here's a rough model of the circuit testing a 51V Zener diode. Green is the transistor current, blue is the diode current.

The output diode and capacitor, which I did not include (lest they make the plot harder to understand) will take the edge off that, as well, because the capacitor will gulp the peak current.

[J4e8a16n]

Hi,

I have got ltspice. Would you attach the asc file?

Thanks.

JP

[c4757p]

I didn't keep it. Should take all of a minute to throw together. The only invisible property is the inductor's DC resistance - right click on it to add that.

[J4e8a16n]

Here is an asc file.  The curve is an envelope of a  jigsaw wave  forL1. 

[c4757p]

I didn't open the file, but from the screenshot, it looks almost perfect.

1. Isn't it easier to use the ground symbol to ground things, instead of a twisting ground trace? It would be much easier to read...
2. Turn the Zener diode around. Zener breakdown is reverse breakdown. You have it configured to measure forward breakdown.

[J4e8a16n]

Hi,

I sill dont understand why the voltage goes so high.
Without capacitor , resistor and zener:  1.4 KVolts
With cap only:  110 volts.
If I put a 1uf cap and a 220 Ohms resistor:  16 volts
If I put a 1uf cap and a 47k Ohms resistor: 77 volts

So that resistor controls the 'thing' .

JP

[c4757p]

Like I said, momentum for current. It tries to keep the current equal, and if it can't, it gives a voltage proportional to how fast the current was changed. Without those things, there is nowhere for current to flow, so it's just pushing against the breakdown voltage of the transistor.

Trap for young players: SPICE only supports breakdown if it's included in the particular component model, and it's often not. 2N2222 breaks down around 70-80V, so it wouldn't be able to go any higher than that.

[SeanB]

It is more an energy balance. You put energy into the magnetic field in the coil when the transistor is on, and when you turn off the transistor you have to dissipate the energy somehow. If there is no load the voltage rises until the energy is dissipated. Whether it goes into a resistor, capacitor or killing the transistor is immaterial, the energy is going to be going somewhere.

[J4e8a16n]

The voltage rises because it sees higher voltage (transistor voltage or else) and try to overcome it?
If the transistor breakdown would be at 200 volts , would the inductor voltage rises up to 200 volts?

JP

[J4e8a16n]

I noticed that the inductoor does not have enough time to discharge completely through the iu capacitor or 47k resistor.

So it discharche to ( lets say)  10 , 20, 30, 40, etc volts. Then it charges from 20, 30, 40 , volts climbing  a staircase.

SO,

It is more an energy balance. You put energy into the magnetic field in the coil when the transistor is on, and when you turn off the transistor you have to dissipate the energy somehow. If there is no load the voltage rises until the energy is dissipated. Whether it goes into a resistor, capacitor or killing the transistor is immaterial, the energy is going to be going somewhere.

Thank for your answer.

JP