## Re: [Electronics_101] When do you use a diode on transistor / inductor ?

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• To clarify what the others said... When you put current through an inductor coil, the energy is stored as a magnetic field. If you cut off the power, the
Message 1 of 24 , May 3, 2011
To clarify what the others said...

When you put current through an inductor coil, the energy is stored as a
magnetic field.

If you cut off the power, the magnetic field is still there. It has to
go somewhere. In a single coil inductor like a relay or motor, it
forces current to continue through the single coil, which can induce a
large voltage spike and fry electronics.

In the circuit you show, the field has a second choice - it can force
current through the *other* coil, which can accept the current without
harm, and in fact is the purpose of this circuit.
• If there is no load on the secondary, then the kick-back voltage can be very high. There was a comment about 2.5 times the input voltage - good rule. I use 3.
Message 2 of 24 , May 4, 2011
If there is no load on the secondary, then the kick-back voltage can be
very high. There was a comment about 2.5 times the input voltage - good
rule. I use 3. Thus, if a 12V input, I would find a device with at least
36V Vce capability (for a bipolar transistor). This would mean a 40V or
50V transistor. Current rating is based on the ON time of the switch and
the inductance i=L * dv/dt, where dv is the input voltage, dt is the
ON-time of the switch. One could always throw a zener across the switch
to help in protection, but the self-capacitance of the zener/transorb
could affect the circuit operation as well.

Derek Koonce
DDK Interactive Consulting Services

On 5/3/2011 12:36 PM, jong kung wrote:
>
>
>
> > The secondary circuit relies on that inductive kickback to work.
>
> I suspect as much. I just wasn't sure.
>
> ===
>
> > If you put a diode across the primary, it would no longer work.
>
> How do you know which transistor is "strong" enough? Which spec in the
> datasheet do I look to find out?
>
> Thanks,
>
> Jong
>
> [Non-text portions of this message have been removed]
>
>
• This data sheet is weird. It talks about a NPN darlington, yet the pictorial is for a PNP. Did anyone notice this? The internal diode must be an adder since it
Message 3 of 24 , May 4, 2011
This data sheet is weird. It talks about a NPN darlington, yet the
pictorial is for a PNP. Did anyone notice this?

The internal diode must be an adder since it is not part of the bipolar
structure for that path. I suggested a zener or transorb for external
use. But, again, the self-capacitance would affect the circuit operation.

Derek Koonce
DDK Interactive Consulting Services

>
> http://www.datasheetcatalog.org/datasheet/siemens/Q62702-D246.pdf
>
> Vceo is a good place to start. This transistor is rated for 80V
> between the collector and emitter with the base open.
>
> Reverse voltage is handled by the internal diode and is clamped to
> 1.3V see spec Vf
>
> What's more important is understanding what the voltage at the
> collector will look like right after the transistor turns off. How
> much ringing will be present? In the reverse direction, the ringing is
> clamped by the diode in the transistor and the battery. But the
> transformer will probably ring in the forward direction as well and
> that is where we are concerned about Vceo.
>
> Richard
>
>
• I wasn t saying 2.5 times the voltage supply. I was talking about the voltage that will appear across the transistor from the kickback. That far exceeds the
Message 4 of 24 , May 4, 2011
I wasn't saying 2.5 times the voltage supply. I was talking about the voltage that will appear across the transistor from the kickback. That far exceeds the power supply voltage in a flyback circuit like this.

It will be limited by the peak voltage on the secondary, reduced by the turns ratio.

So 1kVrms means 1.4kV peak. The specification is for an impedance ratio of 8 ohms to 1k ohms, and the impedance ratio is the square of the voltage ratio, so we need the square root of that.

sqrt(1k/8) = 11.2
So 1:11.2, primary to secondary.

We need to err on the side of caution, so we'll say that the reflected voltage on the primary is 1/10th of what is on the secondary. 1.4kV/10 = 140V. So multiply that by 2.5 to 3, but you need to add the 12V power supply voltage to that because one end of the primary is attached to it. So 2.5 x 140 + 12 = 362V, or 3 x 140 + 12 = 432V.

That's where I'd start looking for a transistor of suitable power and current rating, with a Vceb of 400 to 450V. Or just look for a 500V rating or better.

Of course, the BD679 is only rated at 80V Vceo (in a pinch, it'll do in place of Vcer rating) but this would not be the first nor last time I've seen underrating on transistors in hobby circuits. Hm. Nor, for that matter, in commercial circuits.

Steve Greenfield AE7HD

--- In Electronics_101@yahoogroups.com, Derek <derek@...> wrote:
>
> If there is no load on the secondary, then the kick-back voltage can be
> very high. There was a comment about 2.5 times the input voltage - good
> rule. I use 3. Thus, if a 12V input, I would find a device with at least
> 36V Vce capability (for a bipolar transistor). This would mean a 40V or
> 50V transistor. Current rating is based on the ON time of the switch and
> the inductance i=L * dv/dt, where dv is the input voltage, dt is the
> ON-time of the switch. One could always throw a zener across the switch
> to help in protection, but the self-capacitance of the zener/transorb
> could affect the circuit operation as well.
>
> Derek Koonce
> DDK Interactive Consulting Services
>
>
>
> On 5/3/2011 12:36 PM, jong kung wrote:
> >
> >
> >
> > > The secondary circuit relies on that inductive kickback to work.
> >
> > I suspect as much. I just wasn't sure.
> >
> > ===
> >
> > > If you put a diode across the primary, it would no longer work.
> >
> > How do you know which transistor is "strong" enough? Which spec in the
> > datasheet do I look to find out?
> >
> > Thanks,
> >
> > Jong
> >
> > [Non-text portions of this message have been removed]
> >
> >
>
• ... Beginning to make sense.  Sort of like queezing a tube of toothpaste.  You squeeze it and that pressure has to go somewhere - and in this case it is
Message 5 of 24 , May 4, 2011
> If you cut off the power, the magnetic field is still there. It has to

> go somewhere. In a single coil inductor like a relay or motor, it

> forces current to continue through the single coil, which can induce a

> large voltage spike and fry electronics.

>
> In the circuit you show, the field has a second choice - it can force

> current through the *other* coil, which can accept the current without

> harm, and in fact is the purpose of this circuit.

Beginning to make sense.  Sort of like queezing a tube of toothpaste.  You squeeze it and that pressure has to go somewhere - and in this case it is designed to go to the secondary.

====

Long time ago, I think I remember seeing a transistor with a diode on the primary.  It might have been used as an inductor (nothing hanging off the secondary).  I didn't know much about electronics back then (nor I know much more now).

Jong

[Non-text portions of this message have been removed]
• ... voltage that will appear across the transistor ... far exceeds the power supply ... Do you think it matters if the secondary is there to absorb some (or
Message 6 of 24 , May 4, 2011
> I wasn't saying 2.5 times the voltage supply. I was talking
voltage that will appear across the transistor
> from the kickback. That
far exceeds the power supply
> voltage in a flyback circuit like this.

Do you think it matters if the secondary is there to absorb some (or much) of the kickback ?

Or to induce the high-voltage on the secondary, there has to be an inductive kickback on the primary?

Jong

[Non-text portions of this message have been removed]
• I guess I didn t cover that clearly enough in my response. I calculated the transistor voltage rating based on the voltage that appears on the secondary. Peak
Message 7 of 24 , May 4, 2011
I guess I didn't cover that clearly enough in my response. I calculated the transistor voltage rating based on the voltage that appears on the secondary. Peak voltage on the secondary, divided down by the secondary to primary turns ratio, then add that to the power supply voltage. Multiply -that- number by 2.5 and look for a transistor with that breakdown voltage or higher.

Actually "inductive kickback" isn't a very specific term.

The magnetic field, as it collapses, induces voltage on both primary and secondary windings. The voltage on each winding stays in proportion to the number of windings, so with 11.2 to 1 ratio, the primary will have 1/11.2 of the voltage that appears on the secondary.

Since the circuit is designed specifically so that the secondary circuit absorbs that energy of the collapsing magnetic field (via the voltage and current induced by it) and the primary circuit is specifically designed not to absorb any of it, yes, the secondary is there to absorb all that energy.

Yes, the high voltage on the secondary relies on this (for lack of a better word) inductive kickback. You could even do this with a single coil, no transformer, but then the transistor must be rated at a much higher breakdown voltage.

BTW, since that circuit does not send a symmetrical waveform to the secondary, you might find one phasing or another gives you a higher output voltage.

Steve Greenfield AE7HD

--- In Electronics_101@yahoogroups.com, jong kung <jongkung01@...> wrote:
>
>
>
> > I wasn't saying 2.5 times the voltage supply. I was talking
> voltage that will appear across the transistor
> > from the kickback. That
> far exceeds the power supply
> > voltage in a flyback circuit like this.
>
>
> Do you think it matters if the secondary is there to absorb some (or much) of the kickback ?
>
> Or to induce the high-voltage on the secondary, there has to be an inductive kickback on the primary?
>
>
> Jong
>
>
>
>
>
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
• On 05/04/2011 03:25 PM, jong kung wrote: (snip) ... There is another possibility than a diode, for a circuit, like this, that needs to produce voltage in each
Message 8 of 24 , May 7, 2011
On 05/04/2011 03:25 PM, jong kung wrote:
(snip)
>> In the circuit you show, the field has a second choice
>> - it can force current through the *other* coil, which
>> can accept the current without harm, and in fact is the
>> purpose of this circuit.
>
> Beginning to make sense. Sort of like queezing a tube of
> toothpaste. You squeeze it and that pressure has to go
> somewhere - and in this case it is designed to go to the
> secondary.

There is another possibility than a diode, for a circuit,
like this, that needs to produce voltage in each direction,
alternately across the secondary. You might get a better
efficiency and a lower peak primary voltage by resonating
the primary and secondary windings with parallel capacitors.

Switching the transistor on into a capacitor load will
increase the loss during that transition, but switching the
transistor off, while the inductance is paralleled with a
capacitor lowers the loss during that transition, because
the inductor current detours onto the capacitor, so the
transistor doesn't have to carry it until it is all the way
off and the voltage is rises in a smooth LC sine wave

You might tune one winding for third harmonic of the pulse
frequency and the other for the fundamental. This sort of
thing is done all the time for the step up supplies for the
cold cathode fluorescent lamps that back-light notebook
of these transformers, available from Digikey or elsewhere,
they often include the resonating capacitors in the
suggested driving schematic. Eliminating the higher
harmonics from the transformer operation can greatly
increase their efficiency and output voltage.

--
Regards,

John Popelish
• ... Can you explain that a little more (if you got the time).  Or just point us to a website that does explain it. Thanks, Jong [Non-text portions of this
Message 9 of 24 , May 7, 2011
> You might tune one winding for third harmonic of the pulse

> frequency and the other for the fundamental.

Can you explain that a little more (if you got the time).  Or just point us to a website that does explain it.

Thanks,

Jong

[Non-text portions of this message have been removed]
• ... Better efficiency but how about peak V out (V out on secondary).  I am assuming this is a stun gun - so peak V out is more important than efficiency. Jong
Message 10 of 24 , May 7, 2011
> You might get a better

> efficiency and a lower peak primary voltage by resonating

> the primary and secondary windings with parallel capacitors.

Better efficiency but how about peak V out (V out on secondary).  I am assuming this is a stun gun - so peak V out is more important than efficiency.

Jong

[Non-text portions of this message have been removed]
• ... http://en.wikipedia.org/wiki/CCFL_inverter -- Regards, John Popelish
Message 11 of 24 , May 8, 2011
In Electronics_101@yahoogroups.com, jong kung <jongkung01@...> wrote:

> > You might tune one winding for third harmonic of the pulse
>
> > frequency and the other for the fundamental.
>
> Can you explain that a little more (if you got the time).  Or just point us to a website that does explain it.

http://en.wikipedia.org/wiki/CCFL_inverter

--
Regards,

John Popelish
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