How to Set Output Voltage to 3.9V and Choose Inductor for NCV890230 Switching Regulator

I require a switching regulator capable of delivering 2A peaks. The NCV890230 seems ideal, and very low cost. For the life of me I cant figure how the output voltage is set. I require a voltage of 3.9V. Also any suggestions of what inductor would be suitable would be appreciated. Thank you in advance. Daniel

You can use a zener diode of 3.9v accross the output volt

By setting the Feedback voltage divider to deliver 0.8V to the FB pin when the voltage across the filter capacitor (i.e. the regulator output voltage) is at the target voltage. The worst case FB bias current is 1uA so select the resistors accordingly.

OK,

Here's my recommendation for setting up the feedback voltage divider:

In my example, I will be using the component names used in the diagram in the On Semiconductor datasheet [http://www.onsemi.com/pub/Collateral/NCV890230-D.PDF].

Since the worst case FB bias current is 1uA, set the voltage divider current to at least 20 times that. In my example (below) I will choose 100 times. That would set it to 100uA, well below any concerns for power efficiency, and high enough that the FB bias will produce negligible error.

The nominal FB threshold voltage is 0.8V, so at 100uA, that would be a resistance value of:

Rfb2 = 0.8V/100uA = 8.0K

The voltage across the upper feedback resistor [Rfb1] is the target voltage minus the FB threshold voltage. For an output [target] voltage of 3.9V, Rfb1 would be:

Rfb1 = (3.9V - 0.8V)/(100uA + 1uA) = 31K

BTW: Notice that (3.9V - 0.8V)/100uA is also equal to 31K, so the bias current is essentially "swamped out" by the much larger voltage divider current (at least for an output voltage tolerance of +/-1.75%) -- so, you can ignore the bias current and just use the formula:

Rfb1 = (3.9V - 0.8V)/100uA = 31K

I have to ask... what is the rational for this suggestion?

Regarding the inductor, the datasheet specifies a 4.7uH inductor, so select a power inductor with a max current that is 1.5 to 2 times the peak current. To determine the Ipeak, use the following formula:

Ipeak = (((Vin - Vtran - Vout)Ton)/L) + Io

Where, Vin is the input voltage (in your case, probably the car's battery voltage), Vtran is the voltage drop accross the switching transistor when it's fully on (the on resistance is 650mOhm, so at 2A, the Vtran would be 2A * 650mOhm = 1.3V), Vout would be 3.9V. Ton is the time that the power transistor is on during one cycle--that is computed, below. Io is the max output current--in your case, 2A

To determine Ton we need to questimate the duty cycle that this chip will be using:

D = (Vout -Vd)/(Vin - Vtrans - Vd) where Vd is the voltage drop in the "free wheeling diode"

Once we know the duty cycle, we can apply that to the freq and get Ton:

D = Ton*f

Therefore:

Ton = D/f

So, if the minimum Vo is, say 11V (i.e. car ignition off, and battery a little bit run down), and the diode has a worse case forward voltage drop of 0.8V, then the Dutycycle is:

D = (Vout -Vd)/(Vin - Vtrans - Vd) = (3.9 - 0.8)/(11 - 1.3 - 0.8) = 0.34

So, since this IC runs at a worst case fixed frequency of 1.8MHz, Ton would be approximately:

Ton = D/f = 0.34/1.8 = 189ns

And, Ipeak will be approx:

Ipeak = (((Vin - Vtran - Vout)Ton)/L) + Io = ((11 - 1.3 - 3.9)189)/4.7) + 2 = 2.2A

So, select a Power Inductor with a max DC current around 3 to 4 Amp (saturation current) and a self resonate frequency well above 2MHz.

Of course, if the parameters that I used don't fit with your application, then be sure to recalulate using the formulas, before selecting an inductor.

And, anyone who wants to chime in and refine this (or correct it) please do. I'm not anywhere near as experienced at this as I'm sure some of my colleges, here, are

Steve, Thank you very much for taking the time and being so helpful. I understand now what to do. Its the first time I have ever posted on a board and I am amazed to find such valuable help from a complete stranger. Many thanks.

Well, build the thing and if it works, thank me then ;)

[And, also please tell me if I led you astray ]

Good Day Steve, would you mind giving me some advice on this as well. I paste from the data sheet ....
" OUTPUT PRECHARGE DETECTION Prior to Soft−start, the FB pin is monitored to ensure the SW voltage is low enough to have charged the external bootstrap capacitor (CBST). If the FB pin is higher than VSSEN, restart is delayed until the output has discharged."

My question is, as I will have a Lithium Ion battery on the output does this then mean that the battery needs to discharge prior to the regulator out-putting? Definitely not what I want!!! Thanks again for your help. Dan

First, let me say that I have no experience with this IC, so I can only guess, but that is what that sounds like. Looks like you'll need to isolate the output of the regulator from the battery until it reaches operating voltage. I imagine this could be done with a rectifier, but then the output voltage will be reduced by the forward drop of the rectifier. You could increase the regulator voltage to compensate, but because the rectifier voltage varies with current and temperature, the regulation would suffer.

Perhaps a relay or an active switch. Or, better yet, a different chip--perhaps one that is designed to charge Li-ion batteries.

Good Day Steve, Thanks for the response regarding the o/p. Ill get a better feeling for how the o/p behaves when I build the cct. I also had response from onsemi regarding the o/p calculation, which you helped me with. They took a little longer than you to respond, so thanks again for your help. Their response as follows:

You need to set up the resistors values in the resistor divider.
VFB = (RFB2 / (RFB2 + RFB2)) * Vout
When the RFB1 should be 4x RFB2

e.g.
RFB2 =26kOhm
RFB1 =100kOhm
VFB = (RFB2 / (RFB2 + RFB2)) * Vout
0.8=(26/(26+100))*Vout
Vout=0.8/0.206=3.88V

That's odd. I would think the ratio between RFB1 and RFB2 would depend on the output voltage. In this case its 4:1 but for different output voltages, that ratio would be different (unless I'm missing something).

Also, did they give you no information about inductor selection?