Hi MrAl,
Need your great help!
Few weeks later today I calculated all resistors, voltage and currents of this simple circuit designed by you. But I am being puzzle for Series Resistor (you wrote R=90 ohms). I tried a formula given by you but got different value.
Please can you show me this few steps of calculation please?
(I tried with simulation but puzzled more)
Hello Willen,
The forumla on the schematic (post #10) shows:
Vout=(1+R2/R1)*Vref
so when R1=R2 this comes out to:
Vout=2*Vref
The chip has a built in reference that is set to 2.5v, so the output will be two times 2.5v which comes out to 5 volts. So that's how you get the output voltage.
Now to set the series resistor R (shown as 90 ohms in the schematic) you have to know the load current Iout and the minimum input voltage Vmin, and of course the output voltage Vout. Assuming R1 and R2 are large (like 10k each) we use the load current to calculate the voltage drop across R (the 90 ohm resistor) and make sure that with that amount of voltage drop we still have enough voltage to reach at least 5 volts for the output.
The voltage drop is:
Vdrop=R*Iout=90*0.010 (90 ohms times 10ma)=0.9 volts
and since the minimum input voltage is 6 volts this gives us a non regulated output voltage of:
Vmin-Vdrop=6-0.9=5.1 volts.
Now since 5.1 volts is greater than what we want which is 5.0 volts, this value of resistor is ok and leaves a little room for error.
To calculate this value or R exactly, we would use this:
R=(Vmin-Vout)/Iout
so we would get:
R=(6-5)/0.010=100 ohms
so 100 ohms would be the more exact value, but 90 ohms allows a little room for error.
You also want to calculate the chip current when the input is at Vmax to make sure it does not exceed the requirements of that chip. The max for that chip is 100ma i think (double check this on the data sheet). And also, check the power dissipation in the series resistor R and the chip too.
So first lets calculate the current through the chip at max input of 9.5 volts...
Ichip=(Vmax-Vout)/R-Iout
so:
Ichip=(9.5-5.0)/90-0.010=0.040 amps
and this is only 40ma so the design works so far because this is less than 100ma.
Now the power dissipation in the chip is:
Pchip=Ichip*Vout
so:
Pchip=0.040*5=0.2 watts, which is 200milliwatts.
That sounds ok but check the data sheet to make sure the package can handle 200mw without a heatsink or else add a heatsink.
Now the power in the resistor R is:
Pr=(Vmax-Vout)^2/R
so:
Pr=(9,5-5.0)^2/90=0.225 watts which is 225 milliwatts.
Now a 1/4 watt resistor is 250mw and we have a possible 225mw in the resistor, so maybe we should use a 1/2 watt resistor instead which is 500mw which would be much better.
You may also want to think about the efficiency. We have here 5*0.010 or 50mw output and at times we might have 225mw dissipated in the resistor and 200mw in the package, so there will be times when the efficiency is quite low meaning a large part of the energy stored in the battery goes to waste. This is the basic nature of the shunt regulator.
Is this design process more clear now?