Hello to all,
This has already been said, but I'd like to stress the efficiency issue a bit more.
With an ideal converter, the input and output power relationship is simply:
Pin=Pout
and in terms of voltages and currents:
Vin*Iin=Vout*Iout
However, no converter is ideal which would mean 100 percent efficient. Instead, there is an efficiency factor that is always less than 1.
For a real converter, the power relationship changes to (Eff is the efficiency factor less than 1):
Pin=Pout/Eff
and since Eff is always less than 1, the power input is always greater than the power output.
In terms of voltages and currents again this would be:
Vin*Iin=Vout*Iout/Eff
Unfortunately, the efficiency Eff can be kind of low. To start, 80 percent, which means the efficiency factor Eff would be 0.80, and that means the input power will be a lot more than output power. This is easy to calculate too.
Let's start with an input of 5 volts and an output of 10 volts at 1 amp.
Since the input and output voltages are considered constant, and the output current also constant, the power relationship can then be written as:
5*Iin=10*1 ( where again 5v input, 10v output at 1 amp output).
That would make Iin equal to 2 amps in an ideal converter. Considering the efficiency factor Eff this would become:
5*Iin=10*1/0.80
or simply:
5*Iin=12.5
and so now that means Iin is higher than before by 25 percent:
5*2.5=12.5
so now the input current Iin is equal to 2.5 amps when in the ideal converter it was only 2.0 amps.
Now let's drop the efficiency a little more, down to 70 percent. We then have:
5*Iin=10/0.7
or:
5*Iin=14.29
and this means the input current went up to about 2.86 amps.
We can start to see how this messes up the original calculation with an ideal converter, and how much more current that we might see on the input than in the ideal case. Thus, the efficiency always has to be taken into account.
With 5 volts input and 12 volts output and 1 amp output and 75 percent efficiency the input current would be:
Iin=3.2 amps.
With only 0.5 amps out the input current would be:
Iin=1.6 amps.
With only 0.3 amps out the input current would be:
Iin=0.96 amps.
All of these Iin figures are the average current levels though. The peak levels depend on the inductor value and the pulse width.
If the inductor value is too low this will mean higher peak currents and that could mess things up also. If that is the case, the inductor value has to be increased. Because the inductor value is in the denominator of the current calculation, a larger inductor reduces the peak current. Because the pulse width is in the numerator, the shorter the pulse time the lower the peak current. Unfortunately, it is possible that this chip can skip a pulse which means the next pulse might be longer than usual. This would cause the inductor current to go up higher than usual, so the inductor value gets more critical.
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