# Question about Np in transformer design

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.

#### indaco

Hello, I compared 2-3 formulas found on theory of designing transformers and I have a doubt on parameters on it. In particular this one about number of turns of primary:

Np = V*10^6/(4.44*Ac*f*B)

The first one is on V, is it the plate voltage on Ip(max)? I read in another pdf that refers to Vrms. And then B, can I assume =7.000 G (half of Bmax) for standard Fe-Si laminations?

Thanks

#### cerrem

That is the standard Flux density equation... Derived from Maxwell's equation... The V is rms voltage...... is is not the DC voltage, because it needs to have a rate of change... The AC voltage swing you can find easily from plotting a load line the the plate curves...

#### 45

If you measure B in Gauss, f in Hz, V in Volts rms and A in cmq (typical) then the factor is 10^8 and not 10^6.

V is the AC voltage (Vrms) applied to the transformer and B is the AC induction.

The formula is valid only if there is no DC current (like in a perfect push-pull).

7000G might be conservative in case of PP transformer with no DC imbalance but this also depends on the (low) frequency f and the efficiency you want.

For a single ended transformer the formula above is incomplete. You need to add the DC component. Without going into details this is:

Np = 10^8/(B*Ae) [(L*Idc) + V/(4.44*f)]

where you can recognise the AC term and the new DC term which depends on the anode current and the primary inductance. In this formula B is the total DC + AC induction.

In this 7000G AC induction is too much for standard lamination because you need to add the DC induction caused by the anode current. Because there is no zero crossing Bdc needs to be equal or greater than Bac. So at least Bdc needs to be at least another 7000G. This makes 14000G in total which is generally too much. You will see clear signs of saturation because this doesn't happen uniformly in EI laminations.

In other words from the formula above Bdc > Bac means that L*I > V/4.44*f.

A good starting point is a total of 8500 Gauss (4500G for Bdc + 4000G for Bac) at 30Hz. Of course at 20Hz it will be more for the same applied voltage....but still away from saturation.

#### Mark Tillotson

As usual this sort of calculation is much clearer done in SI units.

The induced voltage per turn is 2π * f * A * B
(using tesla, sq metres, Hz, all rms, assuming sinusoidal waveform)

But we want B to be replaced by B max, in which case a factor of sqrt(2) appears:
V/turn = 2π * f * A * Bmax/sqrt(2)

Thus primary turns is Vp / (V/turn) = Vp * sqrt(2) / (2π * f * A * Bmax)

No arbitrary constants and massive scaling factors, little chance to foul up the calculation, easier to memorize, easy to see the relation to the actual physics.

Notice that sqrt(2) / 2π = 1 / 4.443, which is where the 4.44 comes from in the original version, the 10^8 comes from there being 10^4 gauss/tesla and 10^4 cm^2 in a m^2.

The mistake with 10^6 for 10^8 is exactly the sort of error you avoid working wholely in SI units.

#### indaco

If you measure B in Gauss, f in Hz, V in Volts rms and A in cmq (typical) then the factor is 10^8 and not 10^6. <snip>
Thanks, you've been exhaustive...I'm learning about in particular from the Wolpert's book and the italian T.U. audiolibro

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.