# An Engineer in Wonderland: What do volt.seconds look like?

Coulombs I can do.

They quantify charge and are proportional to the number of electrons shifted, and not too difficult to visualise as electrons are real little round things.

In a circuit they are being pushed up a gradient against their natural tendency to stay still, if you get my meaning.

Coulombs get an outing in CΔV=IT, that particularly handy formula for sizing reservoir and de-coupling capacitors.

Now, I do occasionally wrestle with LΔI=VT when picking inductors for switching converters.

And I know this equation is equal to volt.seconds.

But the trouble starts if I have to visualise volt.seconds as they are far too abstract.

Can anybody help?

Alice (special contributor)

By the way, the title An Engineer in Wonderland was inspired by the 1967 book ‘The Engineer in Wonderland’ by Professor Eric Laithwaite: champion of the linear induction motor, and a man who certainly knew his volt.seconds.

(Picture – wackyvorlon, under Creative Commons Attribution Licence)

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1. 'Alice'
August 20, 2010 16:18

Thanks Mr Butcher.
That does indeed help with the visualisation, if I concentrate sufficiently.
I suspect I am not a natural when it comes to thinking electromagnetically.
As for magnetic switching, is that the principle behind mag-amps?
‘Alice’

2. Bob Butcher
August 19, 2010 05:27

A very useful way to look at Volt-Seconds is as follows. The formula is
Volt*Sec = N*A*Delta B
where N is the number of turns on the core, A is the cross section area (m^2) and Delta B is the change in magnetic flux in Teslas.
This formula can be re-arranged to calculate the number of turns required on a transformer core for example to utilize the magnetic core material without saturating the core.
N = (Volt*sec) / (A*Delta B)
Another application is sometimes used in pulsed power applications, called “magnetic switching”, where the magnetic core is intentionally saturated by applying a voltage for long enough to reach the maximum magnetic flux (Bsat). If you recall the B-H curve is shaped like a letter S, and becomes nearly horizontal when saturated, and inductance is inversely proportional to the slope of curve. When the core saturates, the inductance effectively decreases from the iron core value to the air core value, a decrease of several orders of magnitude. Remember that di/dt = V / L
and when the voltage is first applied, the current rises slowly (di/dt is small). At saturation L decreases and a much higher current can flow through the inductor, behaving like a leaky switch.

3. Bob Butcher
August 19, 2010 01:13

One of the more useful formulas regarding Volt-Seconds is
Volt-Seconds = N*A*Delta B
where N is the number of turns on a magnetic core, A is the area in meters^2 and Delta B is the change in magnetic flux in Tesla. The implication then becomes that if a voltage is applied to the inductor (or transformer winding), the current begins to increase with time, and therefore the magnetic flux. If the magnetic flux reaches the saturation value for the magnetic material, the current will increase much more rapidly. This is the basis for what is called magnetic switching in the pulse power industry.
What has happened is the inductance of the coil has decreased from its iron core value to near its air core value, frequently several orders of magnitude decrease. Recall that
V / L = di / dt
since the inductance has decreased drastically at saturation, di / dt will increase by several orders of magnitude, a similar result to what happens when a switch turns on.
The same formula can be used to determine the number of turns on a transformer winding required to keep the magnetic flux well below the saturation level of the iron core.

4. Alice
July 30, 2008 14:22

Thanks Mr Collins.
All is clear.
‘Alice’

5. July 26, 2008 21:58

It might help to realize that a volt*second is identically a Weber, and that Webers/meter^2 are Tesla. So volt*seconds are equivalent to magnetic lines. Remember that a changing magnetic field is the basis of induction? Well a Weber/second is a volt. In a standard setup, the change of a magnetic field of one weber per second induces one volt. And a volt is just a Joule per coulomb. So a weber per second field change acting on a coulomb of charge requires one joule of energy. I do not know if that helps any. But switch to magnetic units and concepts when you come to volt*seconds. Finally, a Weber is 10^8 lines of magnetic something. You can get some conversion (including volt*seconds) at http://online.unitconverterpro.com/conversion-tables/convert-group/factors.php?cat=magnetic-flux&unit=1&val=

6. Ken Chicken
March 20, 2008 13:58

Volt-seconds to me are like hitting a block to slide up a plank…. with friction. A steep ramp needs a lot of clout to get the block to the top… Height is equivalent to volts. a long ramp needs a lot of clout to get the block to the end – even if a gradual incline…. equivalent to a lot of seconds. So Volt-seconds are an impulse (not the sort in girl’s aerosols). Make any sense? (I may be wrong, too!).

7. March 20, 2008 12:01

Hmmmmm !
Having just stuck my fingers across the terminals of a PP3 battery and counted 1-mississippi , I am still no further enlightened as to what a volt-second ( or 9 in my case ) is.
And I’m not sure I ever will be.
I think the root of the problem is the way you have grouped the terms of the equation.
When visualising inductor currents I always re-arrange the equation into the form ;
dI/dT=V/L
This to me makes much more sense.
The rate of rise of current throught the inductor dI/DT is equal to the applied voltage divided by the inductance value.
So 1 volt across a 1mH inductor gives 1000A/sec.
Easy :O)