MC33167 switching regulator

rwmoekoe

Joined Mar 1, 2007
172
it does affect the compensation filter.
if you need to put a voltage divider, just go for it, but take the paralelled impedance into consideration (a voltage divider consists of two resistors we consider as paralleled in terms of impedance).
 

Thread Starter

jpanhalt

Joined Jan 18, 2008
10,222
Thanks for the prompt reply. Here's a little more detail. I have labeled the pertinent parts in the attached partial schematic the same as in the datasheet.

R1/R2 are the voltage divider and their values set the output voltage by establishing a value of 5.05V at the feedback pin. I am trying to make a 12 to 24V supply, and with R2 at 6.8K, the maximum current through the divider is 2.8 mA (i.e., 19 V across 6.8K). I am trying to use a digital pot for R1, which is limited to 1mA. Therefore, I need R2 to be at least 19K.

On-semi is silent in its datasheet about the effects of R2 on the filter consisting of RF and CF. It does say that the filter values are non-critical in the step-down configuration. It also implies that R2 can be reduced as necessary to attain the maximum output voltage without needing to change the filter.

For sake of argument, assume R1 was 1.8K , which will give the 24V output. The paralleled resistance of that with 6.8 K is 1.42K. Now, if I change R2 to 20K, my current need goes to 0.95 mA, and R1 is 5.3K. The paralleled resistance of those two is 4.2K. Since the filter values work with no R1 and with an R1 of 1.8K (as implied), it would seem that raising R2 to 20K with R1 of 5.3K and a parallel R of 4.2K is within the two extremes that the datasheet implies will work without changing the filter.

Now one reason for all of this is that I am virtually ignorant of filters. I calculate the 3dB point for the 68K/0.1 uF combination (i.e., RF and CF) as about 20Hz. That is without considering R2. I don't know how to do it with R2 also included in the calculation.

Thanks. John
 

mrmeval

Joined Jun 30, 2006
833
You should be able to use the digital pot to control a circuit that would give you a higher current variable resistance. I'm not quite good enough to know how to do that yet.
 

rwmoekoe

Joined Mar 1, 2007
172
hi john,

i'm sorry to get back to you so long, i slept over yesterday. it was way past my bedtime here in my country.

there are a few points we should know.

1.) to suppress the variability of the paralelled r, it is better to vary r2 instead, and put a fixed value for r1. it comes by nature too, since the current will stay the same all the way, and r1 still is dominant with the low resistance value.

2.) the cf and rf ( + (r1 // r2) ) actually forms a negative feedback at high frequency. it is a zero pole compensation for the error amplifier. the ratio of rf to r1 (paralleled with r2) should be about the same as the ratio of the values they've already tested (in fig 18, it is 10 times, ie. 68k to 6.8k). unless (but it is better) you would want to re-adjust them for critical damping specifically for your application. (It is not very critical for step down application, though). cf should follow with the same ratio as it is in fig 18 too.
 

Thread Starter

jpanhalt

Joined Jan 18, 2008
10,222
That is just what I wanted to know. When I saw the 10-fold ratio of RF to R2, I figured it was not just by accident.

My application is not too demanding, I just don't want it to become an oscillator. The datasheet shows a nice chart of the compensation damping (Figure 14).

Thanks for the help. John
 
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