Designing a linear DC power supply

[Whiley]

Homework Statement

Design, giving the necessary ratings and characteristics necessary for the transformer, bridge rectifier, reservoir capacitor and regulator, a 12 V, 2 A regulated power supply. Also identify any heat sink requirements. The available mains supply is 230 V ±10% @ 50 Hz.

Homework Equations

The Attempt at a Solution

So far I have:
Desired output + diode drop + allowance for regulator +10%variation in supply
(12V+1.4V+2V)+10%=16.94V RMS
minimum Vs_peak=16.94(sqr(2))=24V

 

[Whiley]

I have so far gone as far as working out the minimum secondary winding peak voltage.
I have 2 answers working differently so I would like to clear something up before I can proceed.

I have used in the method posted above 1.4V for diode drops and 2V for regulator headroom. My calculations are done in RMS before finding the peak. Should I change the 1.4V and 2V to RMS or are they assumed to be already?

 

[CWatters]

I have used in the method posted above 1.4V for diode drops and 2V for regulator headroom. My calculations are done in RMS before finding the peak. Should I change the 1.4V and 2V to RMS or are they assumed to be already?

No see below.

So far I have:
Desired output + diode drop + allowance for regulator +10%variation in supply

The output of a bridge rectifier and capacitor looks like this...

https://en.wikipedia.org/wiki/Ripple_(electrical)

At the bottom of the ripple the voltage must still be above the minimum needed for the regulator so you need something like..

Peak voltage = Desired output + headroom for regulator + ripple voltage + diode drops

If you use a big enough capacitor to make the ripple small you can approximate the ripple voltage using..
I = CdV/dt (which comes from Q = CV)
so
dV = Idt/C

You know I is 2A, you know approximate dt from the period of the waveform. Pick a value of C to give you a modest ripple voltage.

 

[CWatters]

Also.. Power diodes typically have a higher forward voltage than signal diodes. Could be 1.2-1.5V per diode rather than 0.7V. Perhaps have a look at a few data sheets for diodes in the 2-10A range.

 

[Whiley]

Also.. Power diodes typically have a higher forward voltage than signal diodes. Could be 1.2-1.5V per diode rather than 0.7V. Perhaps have a look at a few data sheets for diodes in the 2-10A range.

Thanks for your reply, I didnt consider the ripple before, thanks I will work on that. As for the diodes, the lessons state approx 0.7V drop in reference to power supply applications. I will have another look, perhaps its mentioned and I missed it.

 

[Whiley]

Also.. Power diodes typically have a higher forward voltage than signal diodes. Could be 1.2-1.5V per diode rather than 0.7V. Perhaps have a look at a few data sheets for diodes in the 2-10A range.

Also... the reason I hadn't considered ripple is because i am following the following procedure:

1. Define the desired d.c. output voltage, VO, and the maximum current, IO, that the supply is to provide.

2. Select a regulator for the desired output voltage that is capable of delivering the required current.

3. Note the minimum allowable input voltage to the regulator (or allow at least 2 V above the regulated output voltage, i.e , VO + 2 V).

4. Determine the minimum peak secondary voltage of the transformer, allowing for 2 V dropped across the bridge and a possible 10% supply voltage variation. The rms secondary voltage will be of the peak value.

5. Select a suitable transformer to give at least this secondary voltage and suitable VA rating.*

6. Calculate the maximum rectified voltage presented to the input of the regulator. This will be on no load, i.e. an added percentage for poor transformer regulation and + 10% for voltage supply variation. 1 2 28 Teesside University Open Learning (Engineering) © Teesside University 2011

7. Calculate the maximum ripple voltage.

8. Calculate the minimum capacitance of the smoothing capacitor.

9. Calculate the capacitor ripple current.

10. Select a suitable capacitor.

etc..

 

[CWatters]

That procedure isn't bad but its normally more of an iterative process..

The voltage you calculate at step 3 is a lower limit below which the linear regulator stops working or "drops out". So you cannot specify that as the peak voltage. You must shoot for a higher peak voltage to allow for the diodes, the issues mentioned in step 6 (10%) and a budgetary estimate for the ripple voltage (perhaps 1V).

Steps 7 and 8 are usually done together. For example you normally calculate the minimum capacitance required to reduce the ripple voltage below the value you budgeted for earlier. Then you round up the calculated capacitance to the nearest standard value and finally recalculate the actual ripple current that will occur with the chosen capacitor.