Help Calculating Voltage after Rectification

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vdi_nenna

vdi_nenna

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#1
Hi,

I'm trying to calculate the required secondary voltage of a transformer that will produce 18-20v after diode bridge rectification. Need 16v after voltage regulator to power a circuit.

So, if the transformer secondary and bridge voltage is 15v * 1.414 = 21v.
Is the 1.414 figure for peak only?

I have a LM317 set to regulate at 16v. I need a few volts over 16v for the LM317 to function.

Is this correct?


Thanks,

Vince
 
#2
I think the general rule of thumb is to go with an ac voltage at least as high as the required dc regulated voltage. So for 16V regulated output 15V AC may not be high enough.

if a transformer is rated at 15V then I believe that is the RMS value, multiplying by 1.414 gives the peak to peak value (I think).

Note also if you use a CRC or CRCRC network before the regulator (this should dramatically reduce the ripple output from the LM317 if done properly I believe) It will drop the voltage so you need an even higher ac voltage if you plan to do this.

I'm going off my memory of simulating my LM317 based supply, but I think the dropout voltage was close to three volts, ie you would want 19V before the regulator for 16V output, 18V I think will be cutting it a bit fine, and you might see increased ripple.

I still haven't actually built it, but if you would like to do some simulation, the LTspice model of my YARPS powersupply would probably be a good starting point.

Tony.
 
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#4
I can't say from actual experience (I've only simmed), you might want a second opinion, but I have seen a lot of people posting with problems when they have tried to use a transformer with a lower ac voltage than the desired regulated DC voltage.

What I do know is my non regulated PS for my poweramp has 45V secondaries, and after rectification (no load) shows 63V dc. Once a decent load is present it drops quite a bit (2 - 3 V depending on the load).

Note I also corrected an error in my post, I had after rather than before the regulator for the crc....

Tony.
 
#5
LM317 would need at least 2 volts to function at 1A (slightly less at a lower load, see the datasheet), so you'd need at least 18 volts after the rectifier. Divide that by 1.414, then add voltage drop on the rectifier diodes (say 2 volts for a full bridge), add a little extra for the variations of the mains voltage (usually +/- 10% in the US, I believe) and you get roughly 16 volts AC on the secondary. That assumes you will have adequate capacitors after the rectifier, so that the difference between peak and trough in the unregulated voltage is small.
Also, you can get by with a 15V transformer if you choose an LDO regulator such as LT1084; it will generate less heat, too.
 
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#9
Also, you can get by with a 15V transformer if you choose an LDO regulator such as LT1084; it will generate less heat, too.
Click to expand...

I already have a bunch of LM317/337.

Higher voltage will lead to more dissipation on the regulator, although at 6mA it is nothing anyway.
Click to expand...

20v then. I have heat sinks, if needed.

Here is the link to the Hammond Mfg catalog.
Click to expand...

Thanks James, that did the trick! 20v sec. will get 18v avg. after rectification (full-bridge w/cap). That's enough for what I need.
 
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#10
Unregulated Power Supply Design

The link above has all of the basics. Then just be aware of your regulator's dropout voltage spec and make sure that worst case combination of AC mains under voltge, transformer regulation, and load current demand will not cause the regulator's input minus output voltage to be less than its dropout voltage spec, and then leave a little safety margin.
 
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#12
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