0
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Likely this procedure has been repeatedly discussed here. But I want to put the pieces together in this certain domain. So let's say we need to get +/-12V at the outputs rated at 1A using classic voltage regulation circuit like this:

enter image description here

and we need to calculate transformer VA for it.

So we need to:

  1. Rectification

Our max current is 1A and output voltages are +/-12V so the 1A/50V should be fine, let's say 1N4001. To calculate diodes drop I have to multiply 2 conducting diodes (thanks to gbulmer) by forward voltage drop of 1N4001 (1.2V) getting 2.4V.

  1. Regulation

7812 and 7912 as mentioned in data sheet will drop 2V each.

So finally we should loose around 2.4+4V = 6.4V of total voltage drop.

  1. Transformer

Now, when we have our losses we may calculate the required VA of transformer. To get 12 volts out of regulators, we have to count our 6.4V drop, right? So then:

x = V(sOut)*1.4141-6.4v = 13V

6.4+13 = 19.4

19.4/1.4141

x = 13,71897319850081 (14V)

where V(sOut) - is transformer secondary voltage and 1.4141 is peak voltage factor.

So now when the V(sOut) is known, then transformer secondary VA is 1A*14V = 14VA and primary VA (mains - 220V) =

IpVp = IsVs

Ip*220v = 1*14

Ip = 14/220 = 0,0636363636363636A

The question: whether this calculation claims to be correct?

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  • 1
    \$\begingroup\$ In 1 'Rectification', why 'To calculate diodes drop I have to multiply 4 diodes'? There are only ever two diodes conducting in a full bridge rectifier, aren't there? \$\endgroup\$ – gbulmer Feb 10 '16 at 13:41
  • 2
    \$\begingroup\$ As an aside, quoting numbers like 15,41616575914009 and 0,0681818181818182A looks like you don't understand engineering, or errors. Much of the components are worse than 1% tolerance, so more than 1% significance looks wrong. Not crucial, but worth thinking about if this is for work, homework, exam or interview. \$\endgroup\$ – gbulmer Feb 10 '16 at 13:51
  • \$\begingroup\$ @gbulmer Thank you! I made a correction according to your "There are only ever two diodes". \$\endgroup\$ – Roman Feb 10 '16 at 14:07
  • \$\begingroup\$ As @gbulmer says, that level of precision will seem silly when you think that your mains voltage can vary by ±10% and your voltage regulators by several percent. Your transformer won't be ideal either. It's a bad world out there. \$\endgroup\$ – Transistor Feb 10 '16 at 14:38
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There are a number of things wrong with this. The transformer voltage must be chosen to be sufficient to provide 12V at the output. If we allow 2V for the regulator, 2V for the rectifier and, say, 2V for ripple in the filter capacitor, then we need 12.7VAC on each half of the winding. Better add 15% for line voltage brownout/dips, so 15VAC is about as low as I'd want to go. So we have 30VAC CT for the winding, not 18VAC CT. Big difference.

Now, let's calculate the minimum filter capacitors. The capacitors are charged only at the peaks of the waveform so they need to hold up for 1/2 cycle. Let's use 50Hz so it will work in most places. So each capacitor is:

\$ C_{min} = (1/100) \frac{I_{out}}{\Delta V}\$ or 0.005F (5,000uF) for 2V p-p ripple. Better to use the next size up so maybe 5600uF/35V (we have to account that the transformer output voltage will rise with a light load and mains voltage might be on the high side). At this point you should check ripple current rating of the capacitor, but I'll omit that step for the sake of brevity.

Now we're in a position to calculate the transformer RMS secondary current which will be 1.61 * the output current, so 1.6A (ignoring the regulator Iq, which is pretty much negligible). Now the transformer VA can be calculated as 48VA.

The heatsinks for those regulators much each dissipate (worst case with line voltage 10% high) about 9.5W, so about 20W for the pair (at this point we can see there's not a lot of extra there for VA vs. average current- the diodes will dissipate 4W, the output 12W and the regulators 19W, so only 1W for extra heating of the transformer winding- but that's at line 10% high).

Now to the diodes- you've made a 2A/50V bridge rectifier, barely enough for the output current. Each diode sees about 55V reverse voltage every cycle worst case with a light load (assuming line 10% high and 20% regulation in the transformer) which is somewhat more than the rating even without transients on the line- fortunately 1N4001 diodes are usually made the same as 1N4004 diodes so they will not fail.. usually. I would use 4x 1N5404 diodes for this application, or a packaged 3A or 6A bridge.

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  • \$\begingroup\$ You said "and, say, 2V for ripple in the filter capacitor" - may I use it as constant value in calculations ? \$\endgroup\$ – Roman Feb 11 '16 at 0:53
  • \$\begingroup\$ It's a design choice. If you make it too big the caps get too bulky and expensive too small and the input voltage has to be that much higher so the regulators get hotter and the transformer has to be bigger. It's also likely to run into the ripple current limit on the caps. Somewhere around 10% of the nominal unregulated voltage is probably reasonable. \$\endgroup\$ – Spehro Pefhany Feb 11 '16 at 1:31
  • \$\begingroup\$ You said "Now we're in a position to calculate the transformer RMS secondary current which will be 1.61 * the output current, so 1.6A"... But according to the pdf in the provided link, I use "Capacitor Input Load" and the I D.C. = 0.62 X Sec. I A.C. So where the 1.6A comes from? Thanks! \$\endgroup\$ – Roman Feb 15 '16 at 23:47
  • \$\begingroup\$ @Roman 1/0.62 = 1.61 \$\endgroup\$ – Spehro Pefhany Feb 15 '16 at 23:48
2
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There are several ways that presentation could be improved.

First, numbers. Quoting numbers like 15,41616575914009 and 0,0681818181818182A doesn't take into account actual tolerances and errors.

Much of the components are worse than 1% tolerance, so more than 1% significance looks wrong. This is not crucial to getting a number. However thinking carefully about errors and tolerance, and using appropriate levels of significance is important when this is for work, homework, exam or interview.

Secondly, presenting equations as
x = IpVp = IsVs
x = Ip*220v = 1*15
x = 15/220 = 0,068A (my fix)
is unnecessarily complex and confusing, "x" is not needed. Worse x doesn't represent the same value in each equation, so they are not a set of equations.

In the first step x = "IpVp" or "IsVs".
In the last step, "x = 15/220 = 0,068A", x represents "Ip".
Remove the "x =" and explicitly and consistently state what you mean:
IpVp = IsVs
Ip*220v = 1*15
Ip = 15/220V = 0,068A

Thirdly, in "1. Rectification" the diode drops are calculated as:

To calculate diodes drop I have to multiply 4 diodes by forward voltage drop

However, in a full-bridge rectifier, there are only two diodes conducting at any instant, so there are only two diode drops to consider.

I found the different uses of x unnecessary and confusing in the rest of the explanation, so I will stop here.

Edit: I can't see anything allowed for discharge and ripple on the capacitors. Drawing 1A from 2200uF (e.g. ~2.2ms at 1A) from a capacitor charged every 10ms, isn't going to work well. Include the calculation for ripple, and make the capacitors much bigger.

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2
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1. No. there is only one diode at a time doing the positive side rectification and one diode at at a time doing the negative side rectification, so at a maximum of 1 volt each, that's a 1 volt drop for the positive supply and a 1 volt drop for the negative supply.

2. No. if the regulators each need 2 volts of headroom and the diodes drop 1 volt, that's 3 volts for the positive supply and 3 volts for the negative supply.

3. In order to get 12 volts out of the regulators, then, you'll have to put at least 15 volts into them, which means that the BFC capacitor ripple can never dip below 15 volts on either the positive supply side or the negative supply side.

4. The capacitance required is determined by:

$$ C = \frac{I\ t}{\Delta V} $$

where

C is the capacitance, in farads,

I is the DC load current, in amperes,

t is the ripple period, in seconds, and

\$ \Delta V \$ is the allowable ripple, in volts.

5. The 15 volts required into the diodes is DC, so since \$ DC = RMS \times \sqrt{2} \$ (for a sine wave) the transformer must put out at least 10.6 volts, RMS.

6. If the DC loads require 1 ampere each, at the same time, that's 24 watts, and that has to come through the transformer. In addition, since the transformer will be supplying current to the load as well as charging the reservoir capacitors, a conservative estimate for both is 1.8 times the load current, so for 24 watts into the loads, the transformer must supply an additional 20 VA to charge the capacitors, which means the transformer should be rated for at least 44 VA.

The devil's in the details though, so here's an LTspice schematic and circuit list so you can simulate the circuit and play around with it if you want to get a better handle on what's actually going on in there. I don't have 7812 and 7912 models in my library, so I cheated a little and made the regulators out of discretes.

enter image description here

Version 4
SHEET 1 992 680
WIRE 208 -256 80 -256
WIRE 352 -256 208 -256
WIRE 432 -256 352 -256
WIRE 512 -256 432 -256
WIRE 848 -256 608 -256
WIRE 896 -256 848 -256
WIRE 80 -208 80 -256
WIRE 208 -208 208 -256
WIRE 352 -176 352 -256
WIRE 400 -176 352 -176
WIRE 560 -176 560 -192
WIRE 560 -176 480 -176
WIRE 848 -176 848 -256
WIRE 560 -160 560 -176
WIRE 640 -160 560 -160
WIRE 352 -128 352 -176
WIRE 560 -128 560 -160
WIRE 640 -128 640 -160
WIRE 208 -96 208 -144
WIRE 208 -96 -112 -96
WIRE -192 -80 -304 -80
WIRE -304 -48 -304 -80
WIRE -192 -48 -192 -80
WIRE 352 -16 352 -64
WIRE 560 -16 560 -64
WIRE 560 -16 352 -16
WIRE 576 -16 560 -16
WIRE 640 -16 640 -64
WIRE 640 -16 576 -16
WIRE 656 -16 640 -16
WIRE 848 -16 848 -96
WIRE 848 -16 656 -16
WIRE -112 0 -112 -16
WIRE -32 0 -112 0
WIRE 352 0 352 -16
WIRE 352 0 -32 0
WIRE -112 16 -112 0
WIRE 576 32 576 -16
WIRE 656 32 656 -16
WIRE 352 48 352 0
WIRE -304 64 -304 32
WIRE -192 64 -192 32
WIRE -192 64 -304 64
WIRE 848 64 848 -16
WIRE -160 96 -160 -96
WIRE -144 96 -144 -96
WIRE 80 96 80 -144
WIRE 80 96 -112 96
WIRE 80 144 80 96
WIRE 208 144 208 -96
WIRE 576 160 576 96
WIRE 656 160 656 96
WIRE 656 160 576 160
WIRE -192 176 -192 64
WIRE -160 176 -192 176
WIRE -32 176 -32 0
WIRE -32 176 -80 176
WIRE 352 176 352 112
WIRE 416 176 352 176
WIRE 576 176 576 160
WIRE 576 176 496 176
WIRE 576 192 576 176
WIRE 80 256 80 208
WIRE 208 256 208 208
WIRE 208 256 80 256
WIRE 352 256 352 176
WIRE 352 256 208 256
WIRE 432 256 352 256
WIRE 528 256 432 256
WIRE 848 256 848 144
WIRE 848 256 624 256
WIRE 896 256 848 256
WIRE -32 288 -32 176
FLAG -32 288 0
FLAG 896 -256 +12
FLAG 896 256 -12
FLAG 432 -256 RAW15
FLAG 432 256 RAW-15
SYMBOL ind2 -208 48 M180
WINDOW 0 -33 73 Left 2
WINDOW 3 -29 39 Left 2
SYMATTR InstName L1
SYMATTR Value 3
SYMATTR Type ind
SYMATTR SpiceLine Rser=1
SYMBOL ind2 -96 112 R180
WINDOW 0 -36 69 Left 2
WINDOW 3 -47 41 Left 2
SYMATTR InstName L3
SYMATTR Value 30m
SYMATTR Type ind
SYMBOL ind2 -96 0 R180
WINDOW 0 -42 65 Left 2
WINDOW 3 -51 40 Left 2
SYMATTR InstName L2
SYMATTR Value 30m
SYMATTR Type ind
SYMBOL schottky 96 -144 R180
WINDOW 0 48 33 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D1
SYMATTR Value B560C
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 96 208 R180
WINDOW 0 48 33 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D2
SYMATTR Value B560C
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL res -64 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName Rsim
SYMATTR Value 100k
SYMBOL schottky 224 -144 R180
WINDOW 0 48 33 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D3
SYMATTR Value B560C
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 224 208 R180
WINDOW 0 48 33 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D4
SYMATTR Value B560C
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL polcap 336 -128 R0
WINDOW 0 -42 33 Left 2
WINDOW 3 -60 67 Left 2
SYMATTR InstName C1
SYMATTR Value 4700µ
SYMBOL polcap 336 48 R0
WINDOW 0 -41 34 Left 2
WINDOW 3 -60 65 Left 2
SYMATTR InstName C2
SYMATTR Value 4700µ
SYMBOL res 496 -192 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 150
SYMBOL npn 512 -192 R270
WINDOW 0 54 31 VRight 2
WINDOW 3 89 -13 VRight 2
SYMATTR InstName Q1
SYMATTR Value 2SCR512P
SYMBOL pnp 528 192 M90
WINDOW 0 49 63 VLeft 2
WINDOW 3 85 113 VLeft 2
SYMATTR InstName Q2
SYMATTR Value 2SAR512P
SYMBOL res 512 160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R3
SYMATTR Value 150
SYMBOL res 832 -192 R0
SYMATTR InstName R4
SYMATTR Value 12
SYMBOL res 832 48 R0
SYMATTR InstName R5
SYMATTR Value 12
SYMBOL zener 576 -64 R180
WINDOW 0 49 30 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D5
SYMATTR Value PTZ12B
SYMBOL zener 592 96 R180
WINDOW 0 39 31 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D6
SYMATTR Value PTZ12B
SYMBOL polcap 624 -128 R0
SYMATTR InstName C3
SYMATTR Value 10µ
SYMBOL polcap 640 32 R0
SYMATTR InstName C4
SYMATTR Value 10µ
SYMBOL voltage -304 -64 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 170 50)
TEXT -224 -136 Left 2 !K L1 L2 L3 1
TEXT -168 264 Left 2 !.tran .2 uic
\$\endgroup\$
  • \$\begingroup\$ Check statement 1 again. There will be one diode conducting on each side (or none). Max voltage drop for diodes will be 0.7 V on each side. \$\endgroup\$ – Transistor Feb 10 '16 at 14:41
  • \$\begingroup\$ @transistor 1V is not unreasonable- datasheet says 1V maximum at 1A and the forward current will be much more than 1A. \$\endgroup\$ – Spehro Pefhany Feb 10 '16 at 14:49
  • \$\begingroup\$ Hmm. This datasheet confirms your assertion on the summary table. The graph on page 2 makes it look like about 0.92 V. I never assumed so high a figure. I have a reel of 1N4001 diodes I got about 30 years ago. I'll look at them all anew! \$\endgroup\$ – Transistor Feb 10 '16 at 15:03
  • \$\begingroup\$ @EM Fields Wow! Thank you for your contribution to my education! \$\endgroup\$ – Roman Feb 11 '16 at 17:46
  • 1
    \$\begingroup\$ @Roman: AHA!!! Copy the text and paste it into a file, then save the file with any name you like, but with a ".asc" filename extension. Then just left-click on the file and LTspice will find it and open it in its schematic editor. If, for some reason, that doesn't work, then launch LTspice, click on "open", navigate to the file and left-click on it. \$\endgroup\$ – EM Fields Feb 11 '16 at 18:51

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