Close

The General Design Problem

A project log for 350 Watt True Sine Inverter

Functional design revision of the prototype TS350.

Brian CornellBrian Cornell 05/09/2019 at 17:590 Comments

Twelve volts is a ridiculous standard.  Most of industry, sans automotive, abandoned (or never adopted) it more than a half century ago.  Automotive just didn’t have an incentive and so the consumer world and all the stuff in its orbit are stuck with it.  Hybrid & EVs are changing this but not soon enough.

Conceptually an inverter is simple:  boost the voltage and modulate it in some way to obtain the desired AC output.  There are several ways to do this depending on the application but all have in common the necessity of high currents if the input voltage is to be low.

A first order analysis is instructive.  The inverter is to deliver 350 watts at 120 V AC RMS, 60Hz.  Assume that only constant impedance, non-reactive (power factor of 1), loads will be used.  The RMS current is:

    Po(rms) / Vo(rms) = Io(rms)    ->    350 / 120 = 2.91 amps        (1)

Also assume the power conversion is 100% efficient.  With a 12 volt (constant) input the required RMS current is:

    Po(rms) / Vi = Ii(rms)        ->    350 / 12 = 29.16 amps        (2)

But the inverter must be capable of delivering power to match the load impedance throughout the sine cycle which is highest at peak.  So for 120 V AC RMS the peak voltage is:

    Vo(rms) * √2 = Vo(pk)        ->    120 * 1.41 = 169.2        (3)

The impedance of the load is:

    Vo(rms) / Io(rms) = Rload    ->    120 / 2.91 = 41.24 Ω        (4)

At the sine peak the load current is:

    Vo(pk) / Rload = Io(pk)        ->    169.2 / 41.24 = 4.1 amps    (5)

And the power consumed by the load is:

    Vo(pk) * Io(pk) = Po(pk)    ->    169.2 * 4.1 = 694 watts        (6)

Hence, the inverter’s peak input current is:

    Po(pk) / Vi = Ii(pk)        ->    694 / 12 = 57.83 amps        (7)

Almost 58 amps.  That’s a lot of juice and, since no machine can be 100% efficient the actual currents will be higher.  At 90% efficiency the RMS input current increases to:

    (Po(rms) / n) / Vi = Ii(rms)    ->    (350 / 0.9) / 12 = 32.4 amps    (8)

Additionally, it’s expected that the inverter’s power supply impedance will be much lower and that the interconnecting supply wiring is absent nasty parasitics.  Sure.

So the general design problem is big current and questionable source impedances.  How it is solved for is the primary determinant of the machine’s overall efficiency and performance.

Discussions