DC power supply 12V 2A

Discussion in 'Homework Help' started by xxxyyyba, Feb 26, 2015.

Aug 7, 2012
249
2
Hi!
I want to design DC power supply 12V 2A. Here is what I have done.
Firstly, we need to transform AC main voltage. I will use transformer with ratio Np:Ns=16:1 :

We must choose diodes with maximum forward current more than 2 amperes. I will choose UG8JTHE3/45.. This diode has maximum forward current 8A and maximum reverse voltage 600V, Is=1.0025^(-12)A, Emission Coefficient n=1.74999 (according to Multisim model).

Let's analyse circuit in case e(t)>0. I will not analyse case e(t)<0 since everything is same as in this case (only difference is of course that two other diodes conducts). In this case D1 and D2 are on, while D3 and D4 are off:

Diodes D1 and D2 are same, same current flow through them so voltages across them are also same.
Since we want to calculate maximum (peak) value of voltage on diodes and peak value of power on diodes, peak value of power on transformer's secondary and lowest value of passive load we can connect on our power supply, we solve this circuit:
VD1max=VD2max=VDmax

$I_m_a_x=I_s(e^{\frac{V_D_m_a_x}{nV_t}}-1)...(1)$
$V_D_m_a_x=\frac{1}{2}(19.4454-2R_l_o_a_d)...(2)$
If we put (2) in (1) we get:
$2=1.0025*10^{-12}(e^{\frac{19.4454-2R_l_o_a_d}{2*1.74999*0.02589}}-1)$ (Vt=0.0258 at temperature 27 degrees Celsius).
If we solve this nonlinear equation in Matlab, we get Rload=8.4395Ohms. This would be lowest passive load value we can connect to our supply. So when Rload=8.4395Ohms, peak value of current in circuit is Imax=2A. In that case, peak value of voltage across diodes will be VDmax=(1/2)*(19.4454-2*8.4395)=1.2832V. When voltage at transformer's output is at peak value 19.4454V and current is 2A, power that is dissipated is Ptransf.max=19.44542*2=38.8908W. For everything to work fine, we should choose some transformer with higher power rating than this. PIV (peak inverse voltage) on diodes is: PIV=Rload*Imax+VDmax=18.1622V<<600V.
Peak value of power on diodes is PD1max=PD2max=PD3max=PD4max=VDmax*Imax=1.2832V*2A=2.5664W.
Output voltage will pulsate from 0v to Imax*Rload=2A*8.4395Ohms=16.8790V.
My calculations match with Multisim. What do you think? What power rating should transformer have exactly? Maybe 50W?
It is necessary now to smooth pulsating voltage. What is best way to do this? Would it be enough to use just one capacitor and one voltage regulator?

Last edited: Feb 26, 2015
2. WBahn Moderator

Mar 31, 2012
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4,917
I've only skimmed your work, but you seem to be doing okay. A few things to keep in mind.

You have a 2A fuse in the primary, but the current in the primary will normally be 1/16 of the current in the secondary, so if you are wanting a 2A supply out of the secondary, I'd recommend using a smaller fuse in the primary, subject to the caveat of the next paragraph. But 2A wouldn't be a horrible choice at the end of the day.

Your current output (and input, for that matter) will not be sinusoidal, particularly after you put filter capacitors in there. What you will see will be virtually no current during most of the period with a sharp spike in current twice each cycle as the input voltage rises above the capacitor voltage and replenishes the charge stored in it. So your peak current might be very large. However, most fuses will not respond to transient peak currents even if they are many times the fuse's rating as long as the RMS current over some short time interval is low enough to keep the heating to an acceptable level. The same is generally true of diodes, though you might look to see if your diodes have a peak forward current rating. The reverse voltage spec, on the other hand, is a peak spec; however, the reverse voltage isn't subject to the kind of transient concentration that the current is. None-the-less, you need to realize that, with a filter capacitor on the output, that your bridge diodes will see a peak inverse voltage that is basically equal to the peak-to-peak voltage of the input waveform.

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3. Dodgydave AAC Fanatic!

Jun 22, 2012
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220V primary @16:1 reduction is 13.75V secondary,less volt drop across diodes 1.4v= 12.35V open circuit.

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Aug 7, 2012
249
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But I used peak value, 220*sqrt(2). All calculations (power dissipated on diodes, power dissipated on transformer etc.) I got are related with peak value of AC main voltage and peak value of voltage on output of transformer.

I used peak values for my calculations because I want to know what are absolute maximum values of voltages across components (and powers) I can expect so I can properly choose components for practical realisation of circuit.

Last edited by a moderator: Feb 28, 2015
5. WBahn Moderator

Mar 31, 2012
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Once you put filter capacitors in there, what will happen is that during each half cycle the load current will be supplied by the capacitor during most of the time and then, during a small fraction of the half-cycle, the diodes will be forward biased and a large current will flow from the transformer to replenish the charge on the capacitor. As you make the capacitors larger to reduce the ripple voltage, the time during which the diodes are forward biased will be shorter, resulting in even higher peak currents.

For some calculations, particularly anything in which bulk heat is the issue, it is the average power that will matter. But for others, such as sensitivity to peak electric or magnetic stresses or, to a somewhat lesser degree, where transient heat is the issue, then peak values will count.

xxxyyyba likes this.

Aug 7, 2012
249
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And how can I know what is peak of these currents? I thought that with calculations I showed in my fist message I came up to all informations I need to properly choose transformer and diodes.

I know that I'm annoying with all these questions but I want to be 100% sure I'm doing it right. I don't want do burn my house

Last edited by a moderator: Feb 28, 2015
7. WBahn Moderator

Mar 31, 2012
18,085
4,917
There are actually manufacturer references for how to size transformers when the output will be highly non-sinusoidal. There are several factors -- the very high peak currents are not, in and of themselves, the big problem because they are transient and the total power is what matters most. But because they are sharp transients that means that they have very high spectral content -- meaning that a lot of the energy is at much higher frequencies -- and since a transformer IS an inductor, that will cause a degradation in the performance.

8. MrChips Moderator

Oct 2, 2009
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xxxyyyba likes this.

Aug 7, 2012
249
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WBahn,
So can we make summary like this: my calculations in first message are correct but for proper realisation of this circuit in practice aditional things should be also considered (things you mentioned) ?