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Design of powersupply

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toa
#1
Nov6-09, 05:13 AM
P: 6
Folks

I am about to build a 2 * 8A DC power supply from a transformer supplying 2 * 8A 18V AC (transformed from 230V 50Hz house current).

I have been dabbling a bit in electronics, but I am no expert. A formula that I found on the web, ( C = 0.7 * I / (dE * f) where C is the capacitance needed to provide less than dE volts ripple when the load is I ampères and f is the AC (and hence half the the DC rectified) frequency) lead me to believe that a capacitor of 1F on each output will be more than enough to stay within 1 volt ripple during maximum load.

My primary question is this: When the transformer's secondary voltage is rated at 18 volts, will a couple of 25V capacitors do the job, or will they be in danger of breaking? In other words: Does the 18V rating describe the maximum voltage output from the transformer at the top of the period, or is it some sort of average, so that the capacitors may be subjected to higher voltages than they can handle when there is no load on the power supply?

My secondary question is this: Do any of the distinguished members of the forum see other problems with a simple design using just the transformer, a couple of rectifiers (bridges), the aforementioned capacitors and a fuse or two to build such a powersupply?

Best regards,
Torgeir
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vk6kro
#2
Nov6-09, 06:08 AM
Sci Advisor
P: 4,032
The rating of a transformer always gives the RMS value of the voltages.

So, to get the peak value you have to multiply by 1.414 (ie Root 2). So, 18 volts from the transformer means 25.4 volts peak.
This (minus the diode voltage drops ) is what your capacitors will charge to.

So, a 25 volt capacitor would be in danger and you should look for the next higher one. Maybe 40 volts.

Now, whether this will be a good power supply depends on what you want to use it for. 25 volts is an unusual voltage so maybe you have a particular use for it.

You can get a much cleaner output and a more stable voltage if you use a regulator. This is a simple integrated circuit that gives you either a fixed output or a variable one depending on the IC chosen.

Did you mean 1 FARAD capacitance? That seems very high. Usually something up to 5000 uF is enough.
toa
#3
Nov6-09, 06:43 AM
P: 6
Thanks a lot, Sir, that cleared up my main issues.

The quality of the output voltage is not a big issue, I believe, because I am going to pass the output DC to several smaller "client" power supplies, which are based on regulators. For these, I aim at somthing around 12-15V maximum output voltage and a maximum load of around 1.5A. I trust the client power supplies' regulators to flatten the ripple considerably.

It now seems that I should use a 15V or perhaps even a 12V transformer, which, given the diode losses at different stages, will pan out at the 12-15V that I aim for at the client supplies.

The only thing that baffles me a bit is the capacitor value of 5000uF that you suggest. Is that really enough to get a reasonably small ripple at 8 amps of load?

vk6kro
#4
Nov6-09, 08:39 AM
Sci Advisor
P: 4,032
Design of powersupply

Regulators require an input of about 3 volts more than their output. Some are better than others at this.

So, as long as your ripple does not drop below this figure, then you will see no ripple in the output of the regulator.

Large electrolytics are expensive, so it is worth being careful about including excessive capacitance you don't need.

As well as the capacitance, you need to consider the ESR (equivalent series resistance) of the capacitors. This needs to be as low as possible.
toa
#5
Nov6-09, 09:06 AM
P: 6
OK, I think I am well on my way now. Thanks a lot for your help!
berkeman
#6
Nov6-09, 12:48 PM
Mentor
berkeman's Avatar
P: 41,098
Quote Quote by toa View Post
Folks

I am about to build a 2 * 8A DC power supply from a transformer supplying 2 * 8A 18V AC (transformed from 230V 50Hz house current).

I have been dabbling a bit in electronics, but I am no expert. A formula that I found on the web, ( C = 0.7 * I / (dE * f) where C is the capacitance needed to provide less than dE volts ripple when the load is I ampères and f is the AC (and hence half the the DC rectified) frequency) lead me to believe that a capacitor of 1F on each output will be more than enough to stay within 1 volt ripple during maximum load.

My primary question is this: When the transformer's secondary voltage is rated at 18 volts, will a couple of 25V capacitors do the job, or will they be in danger of breaking? In other words: Does the 18V rating describe the maximum voltage output from the transformer at the top of the period, or is it some sort of average, so that the capacitors may be subjected to higher voltages than they can handle when there is no load on the power supply?

My secondary question is this: Do any of the distinguished members of the forum see other problems with a simple design using just the transformer, a couple of rectifiers (bridges), the aforementioned capacitors and a fuse or two to build such a powersupply?

Best regards,
Torgeir
Is your transformer already in an enclosure that is UL approved (or the equivalent in whatever country you are in)? If not, are you using safety-agency-approved construction techniques, especially in grounding, switch, and fuse placement?
Bob S
#7
Nov6-09, 02:02 PM
P: 4,663
One problem with using large capacitors to minimize the ac ripple is the large surge current in the rectifying diodes to provide the average discharge current during a very small part of the ac cycle. The peak diode current is much higher than the average current, and the peak diode heating scales as I-squared times the diode drop, which increases logarithmically with the current. The high peak current also raises the heating of the transformer windings (I-squared R).

Bob S.
vk6kro
#8
Nov7-09, 08:43 AM
Sci Advisor
P: 4,032
I worked out a value of about 43000 uF for a capacitor that would give 1 volt peak to peak ripple on a 25 volt supply supplying 8 Amps.

The voltage has to drop 1 volt between cycles which are 120 Hz rectified 60 Hz.

To work out a time constant, this would have to drop to 37 % of the peak value or to 9 volts.

The voltage from the discharging capacitor drops 1 volt per cycle so it would drop 16 volts (25 - 9) in 16 cycles if it were not being recharged.
This would happen in 16 cycles of the 120 hz waveform or 16/120 sec = 0.133 seconds

So, 0.133 seconds = R * C = 25 V / 8 A * C
C= 43000 uF.

This is a slightly simplified method but it may give you a starting point. It would tend to give a higher value of capacitance than was really needed.
Bob S
#9
Nov7-09, 11:41 AM
P: 4,663
If the power supply has a 43,000 uF capacitor, and the discharge current is 8 amps, using I =C dV/dt, the droop is dV = I dt/C, where dt =~ 1/120 sec.

so the droop is dV = 8/(120 0.043) =~ 1.5 volts. This means that all of the discharge current during a 180 degree half cycle has to be recharged in about 20 degrees of phase. So the instantaneous charging current is about 10 times the 8 amps, or about 80 amps. Because the transformer winding resistance losses go as I-squared R, the winding losses will increase by about a factor of 10 over a circuit with no capacitor.

Bob S.
vk6kro
#10
Nov7-09, 05:30 PM
Sci Advisor
P: 4,032
Yes, pretty bad.
The original proposal was for a 1 FARAD capacitor.
Bob S
#11
Nov7-09, 06:40 PM
P: 4,663
Quote Quote by vk6kro View Post
Yes, pretty bad.
The original proposal was for a 1 FARAD capacitor.
Here is a thumbnail with (only) a 43,000 uF capacitor. Using 4 Schottky diodes. Input voltage = +/- 25.5 volts peak.

Left scale (blue) output voltage ripple ~ 1 volt, ave volts out about 23.5 volts.

Right scale (red) peak ac input current in V1 ~ +/-50 amps.

Bob S
Attached Thumbnails
Bridge_rect1.jpg  
toa
#12
Nov8-09, 01:44 AM
P: 6
Thank you so much for all the information. As usual, things turn out to be more complicated and need more attention than meets the eye of a little-experienced amateur. I will try to digest the advice, and do further research (and experimentation). Hopefully, I will end up with a workable solution.

Best regards
Torgeir
The Electrician
#13
Nov8-09, 07:12 AM
P: 771
Quote Quote by Bob S View Post
So the instantaneous charging current is about 10 times the 8 amps, or about 80 amps. Because the transformer winding resistance losses go as I-squared R, the winding losses will increase by about a factor of 10 over a circuit with no capacitor.

Bob S.
This would only be true if the charging current were a 100% duty cycle square wave, so that its RMS value was the same as its peak value.

What determines the winding losses is the RMS value of the charging current. The winding losses won't increase by a factor equal to the ratio of the peak charging current to the DC current in the load, but rather by a factor equal to the ratio of the RMS value of the charging current to the DC current.

In your simulation you didn't include an impedance in series with the AC source with a value typical for a transformer with a rated current suitable for the load.

I happen to have a transformer rated for 8 amps at 20 VAC. I wired up a bridge with some high current Schottky diodes, and captured some scope traces of the ripple voltage and charging current with 10,000uF and with 40,000uF capacitances.

The transformer has a series impedance of (130 + j30)milliohms, referred to the secondary with the primary shorted.

The charging current in the two cases is almost the same, because the series impedance of the transformer dominates, but the ripple voltage is much lower with more capacitance, as we would expect.

The first image shows the ripple voltage and the charging current with 10,000uF and the second image shows the same, but with 40,000uF. The DC load current in both cases was 8 amps.

The charging current is much less peaky, with a greater conduction angle, than it would be if the transformer had no series impedance; its peak value is about 25 amps. The RMS value of the charging current is 13.5 amps. The value of I to be used in calculating the I^2*R losses in the windings is the RMS value (13.5 amps), not the peak current (25 amps).

The ratio of RMS winding current to DC load current in this case is 1.6875. A rule of thumb often seen is that for a capacitor input power supply, the transformer VA rating should be about 1.6 times the DC output power.

The upshot of all this is that after the capacitance has reached such a value that the capacitor impedance (at the dv/dt of the rising edge of the ripple waveform) is less than the transformer impedance, adding more capacitance will reduce the ripple voltage with no additional penalty in transformer or diode losses.
Attached Images
File Type: bmp With10kuF.bmp (16.0 KB, 9 views)
File Type: bmp With40kuF.bmp (15.1 KB, 7 views)
Bob S
#14
Nov9-09, 01:08 PM
P: 4,663
Quote Quote by The Electrician View Post
This would only be true if the charging current were a 100% duty cycle square wave, so that its RMS value was the same as its peak value....

In your simulation you didn't include an impedance in series with the AC source with a value typical for a transformer with a rated current suitable for the load.....

I happen to have a transformer rated for 8 amps at 20 VAC. I wired up a bridge with some high current Schottky diodes, and captured some scope traces of the ripple voltage and charging current with 10,000uF and with 40,000uF capacitances.

The transformer has a series impedance of (130 + j30)milliohms, referred to the secondary with the primary shorted.

The charging current in the two cases is almost the same, because the series impedance of the transformer dominates, but the ripple voltage is much lower with more capacitance, as we would expect.

The first image shows the ripple voltage and the charging current with 10,000uF and the second image shows the same, but with 40,000uF. The DC load current in both cases was 8 amps..
Electrician-
Thanks for the measurements of the output impedance and for the bench tests. This is very informative.
Bob S


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