I'm looking at building a bench supply using a 24V 1.3A DC power supply and a ON Semiconductor MC34167. Along with an appropriate inductor and MOSFET as suggested in the datasheet. I want to make the bench supply adjustable between something around 5V and 30V.

Now the problem I'm having, being an uneducated hobbyist only starting out, is figuring out exactly how to make the circuit adjustable with a potentiometer. I'm also wondering how to determine what voltage the filter capacitor should be, I know the one at Vin should be at 40V to accommodate voltage peaks and the Vout one should be the same for a max output around 30V. I don't want the solution handed to me, I'm just looking to be pointed in the right direction. Perhaps something I should research. Any help would be very much appreciated, thank you. :)

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  • \$\begingroup\$ I have yet to decide on the filter capacitor voltage, I have however figured out how to adjust the output. The data sheet talks about R1 and R2 determining the output voltage, according this formula; Vref(R2/R1+1). Vref being 5.05v and R2 6800 Ohm while R1 I selected to be the potentiometer; which needs to be around 1376 Ohm for 30V. Now the problem I have is I can not find a 1.4k logarithmic pot, would a linear pot work? \$\endgroup\$ – rob j loranger Mar 15 '13 at 13:39

Don't underestimate the challenge of what you're attempting to do here. A variable-output power supply isn't trivial, as you now have to consider maximum and minimum load at multiple output voltages.

Also, your non-inverting buck-boost converter will be a negative impedance (constant-power) load on your source power supply - if the source supply output voltage decreases, the current being drawn by the buck-boost will increase. This can have implications for the stability of your source supply.

All that being said, as to your question, there are always two considerations for the output capacitor of a switching power supply:

  1. Voltage rating
  2. Ripple current rating

The expected maximum sustained DC voltage on the output should not exceed 70% of the capacitor's voltage rating, for reliability purposes.

Similar voltage derating should also apply to the input capacitor. Fixturing is going to play a role as to how much work your Vin cap will be doing, as well as the number and nature of caps on the output of your source supply.

You'll need to evaluate your power train and determine under what input and output combination you will see the highest peak-to-peak inductor current, which will dictate the peak-to-peak output ripple voltage and power dissipation of the capacitor via its ESR. You should also derate the ripple capacitor current to 70% of its maximum - more margin is better, as cooler capacitors last longer than hot ones.

(IPC-9592A is a good standard to get an idea of component derating guidelines, among other things, for high reliability power supplies.)

Further to your comment; what will your output voltage be when your pot is adjusted to zero? I don't think it'll be within your target range.

  • \$\begingroup\$ Thanks for the guideline about capacitor selection, I would otherwise have probably selected something a bit too close to my Vin/Vout's. Is there another converter I should consider? One that will draw less as the output drops? Referring to the pot; I had the wrong resistor initially. R1 is added when an output voltage other that Vref is required, so R2 should be the pot. In which case if I used a 7.5k pot; 5.05(7500/1500+1) = 30.3 and at 0, 5.05(0/1500+1) = 5.05. I have homework to do :) (IPC-9592A) \$\endgroup\$ – rob j loranger Mar 15 '13 at 23:37
  • \$\begingroup\$ All switching converters show negative impedance - it's the input power that dictates the output power, unlike a linear regulator (which is more like a constant current load, but is generally much less efficient). \$\endgroup\$ – Adam Lawrence Mar 16 '13 at 19:59

We start from the formula:

Vout = Vref (R1 + R2) / R1 

if we set R2=6.8K, then using the formula

R1 = Vref.R2 / (Vout-Vref)

we see that the minimum value of Vout is Vref because if Vout is less than Vref then R1 will give a negative value.

If we fix the value of R2 = 6.8K, then the value of R1 for Vout = 30V is:

  R1 = (5.05x6.8) / (30-5.05) = 1.376K

Moreover with R1 = 100K we obtain an output value:

Vout = 5.05 (6.8K + 100K) / 100K = 5.4V

If R1 = 250K we get Vout = 5.187V.

Ideally with R1 = infinity would get Vout = Vref.

  • \$\begingroup\$ the calculation of Vout is as follows; Vref(R2/R1+1), which is R2 Divided by R1, then add 1, then multiply by Vref \$\endgroup\$ – rob j loranger Mar 17 '13 at 0:44

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