Calculating combined feedback (feedback from multiple rails)
First of all, if you didn't know it already: I suck at math. But i know basic electronics theory, and when something bugs me i take out the pen & paper and stay on it till i've figured it out. If only i'd be treating homework like this as well... 
Anyway. I've been wondering for a long while how the values of the resistors in the combined 5v/12v feedback in group regulated ATX supplies were selected. When only one rail is in the feedback loop, you have a simple voltage divider. When two or more rails are involved however, you can't just calculate two voltage dividers and stick them together: it won't give the expected results because currents are also involved.
So, i decided to look into it a bit. And i quickly discovered it's nothing more than plain ole Ohm's Law, nothing fancy.
Say we have a power supply with one 5v and one 12v rail, coming from the same transformer, and with the output inductor on a common core. Typical ATX PSU situation. We want to include both rails in the feedback loop. The voltage we need to present to the error amplifier is 2.5v.
The first step is to fix R3, the bottom resistor in the feedback divider. Its value should be selected so that the current flowing thru it is at least 100 times the input bias current of the error amplifier. For the TL494 this is 1uA, so we would need at least 100uA flowing thru R3. I set R3 to 1k to keep things simple, i could have gone quite a bit higher in value - as a matter of fact, i could have used 10k. Anyway, moving on.
Once you have set R3, you need to find out the current flowing thru it. If you selected it based on the current, you have that already, but chances are if you did that, you're gonna end up with an ugly value, so best pick a standard resistor first and then calculate the current. Apply Ohm's Law, I= U / R. 2.5 volts over 1kOhm is 2.5mA.
Now, we need to fix R1 and R2. This is actually a lot simpler than it looks. Let's see. First, since both the current thru R1 and the current thru R2 will be going thru R3, it occurs that
To keep things simple, i assumed that the input rails will be loaded equally, ie the feedback is split 50-50 between 5v and 12v. Thus,
We know the currents. Now, we also need the voltages to calculate the resistance of R1 and R2. But look at it - we have those already! Since we want to have 2.5v at the junction of the feedback resistors, the voltage drop of the upper resistors must be the rail voltage minus 2.5v.
So we have:
That's all folks!
I hope that clears up some confusion that some of you might have had when looking at such a feedback divider.
If for example, you wanted to have 60% of the power on 12v, and 40% on 5v, you would make I(R2) = 0.6*I(R3) and I(R1) = 0.4*I(R3). You can add more rails as well - the current thru the bottom resistor of the divider will always be the sum of the currents flowing thru the resistors coming from the rails, so once you've picked your ratios it's the same calculation.
Anyway. I've been wondering for a long while how the values of the resistors in the combined 5v/12v feedback in group regulated ATX supplies were selected. When only one rail is in the feedback loop, you have a simple voltage divider. When two or more rails are involved however, you can't just calculate two voltage dividers and stick them together: it won't give the expected results because currents are also involved.
So, i decided to look into it a bit. And i quickly discovered it's nothing more than plain ole Ohm's Law, nothing fancy.
Say we have a power supply with one 5v and one 12v rail, coming from the same transformer, and with the output inductor on a common core. Typical ATX PSU situation. We want to include both rails in the feedback loop. The voltage we need to present to the error amplifier is 2.5v.
The first step is to fix R3, the bottom resistor in the feedback divider. Its value should be selected so that the current flowing thru it is at least 100 times the input bias current of the error amplifier. For the TL494 this is 1uA, so we would need at least 100uA flowing thru R3. I set R3 to 1k to keep things simple, i could have gone quite a bit higher in value - as a matter of fact, i could have used 10k. Anyway, moving on.
Once you have set R3, you need to find out the current flowing thru it. If you selected it based on the current, you have that already, but chances are if you did that, you're gonna end up with an ugly value, so best pick a standard resistor first and then calculate the current. Apply Ohm's Law, I= U / R. 2.5 volts over 1kOhm is 2.5mA.
Now, we need to fix R1 and R2. This is actually a lot simpler than it looks. Let's see. First, since both the current thru R1 and the current thru R2 will be going thru R3, it occurs that
- I(R3) = I(R1) + I(R2).
To keep things simple, i assumed that the input rails will be loaded equally, ie the feedback is split 50-50 between 5v and 12v. Thus,
- I(R1) = I(R2) = I(R3) / 2 = 1.25mA.
We know the currents. Now, we also need the voltages to calculate the resistance of R1 and R2. But look at it - we have those already! Since we want to have 2.5v at the junction of the feedback resistors, the voltage drop of the upper resistors must be the rail voltage minus 2.5v.
So we have: - U(R1) = 5 - 2.5 = 2.5v,
- R1 = U1 / I1 = 2.5v / 1.25mA = 2k
- U(R2) = 12 - 2.5 = 9.5v
- R2 = U2 / I2 = 9.5v / 1.25mA = 7k6.
That's all folks!
I hope that clears up some confusion that some of you might have had when looking at such a feedback divider.If for example, you wanted to have 60% of the power on 12v, and 40% on 5v, you would make I(R2) = 0.6*I(R3) and I(R1) = 0.4*I(R3). You can add more rails as well - the current thru the bottom resistor of the divider will always be the sum of the currents flowing thru the resistors coming from the rails, so once you've picked your ratios it's the same calculation.
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