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I have a multi-tap transformer and I need to get two different voltage from the transformer, variable and steady.

Is the below diagram correct, will it work this way?

schematic

simulate this circuit – Schematic created using CircuitLab

edit:

schematic

simulate this circuit

edit 2 (rewritten):

I designed a digital linear CV/CC mini PSU 20V/1A based on LM317 series regulator. I'm well aware that linear regulators specially LM317 are not efficient, but since I'm not going to draw more than few hundred milliampere the heat and efficiency isn't really an issue. According to LM317 datasheet at 1 A voltage drop is going to be ~2.5 V. I am planning to use Schottky diodes for the bridge rectifier so I get half of normal diodes' voltage drop on bridge rectifier.

I need a transformer at least 4 V higher than the maximum output (20 V) with four taps so I can use two relays and switch between taps to reduce the overall voltage drop on the regulator. Also LM317 needs negative voltage to be able to output 0 V therefore a -5 V would be enough.

For powering MCU and fan and op amps supply I need another 24V which should be steady.

The voltages that I need are:

  1. 8, 16, 24 V
  2. -5 V
  3. 24 V
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    \$\begingroup\$ The problem with this has already been explained in answers to your previous question electronics.stackexchange.com/questions/472297/…. You can't have common connections on both sides of the bridge rectifiers. \$\endgroup\$
    – Transistor
    Commented Dec 25, 2019 at 8:42
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    \$\begingroup\$ Why not make a further edit to your question, write the specification of the transformer that you've got and a specification for the finished product you're trying to build: output voltages required, their tolerances, current required, allowable ripple, etc. and we'll help you along. \$\endgroup\$
    – Transistor
    Commented Dec 25, 2019 at 9:48
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    \$\begingroup\$ @ElectronSurf, (1) You are still approaching this problem the wrong way. You are specifying the transformer first when it should be the last thing you do. (2) Write down what DC voltage ranges you want on the output and whether or not they are to be adjustable or fixed. (3) You seem to be insistent on using a linear regulator despite their inefficiency and the heatsinking problems it will cause. That's OK. So you need to say which regulator you are using and look up the minimum voltage drop across the regulator. \$\endgroup\$
    – Transistor
    Commented Dec 25, 2019 at 13:04
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    \$\begingroup\$ (4) From 3 you can work out the smoothing capacitor values to reduce the ripple. (5) Then work out what AC voltages will give you the required DC after the rectifiers. (6) If you are getting a transformer wound specially then add an isolated winding for the - 5 V supply. Please edit again and write a better specification. \$\endgroup\$
    – Transistor
    Commented Dec 25, 2019 at 13:07
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    \$\begingroup\$ (1) I showed how to do it with one secondary on your previous question. (2) How did you work out that you need an 8 V AC tap to generate 8 V DC, etc.? (3) To get an LM317 down to 0 V you only need -1.25 V. (4) Why do you say "For powering MCU and fan and op amps supply I need another 24V"? Any MCU is going to be 3.3 V or 5 V. \$\endgroup\$
    – Transistor
    Commented Dec 25, 2019 at 15:00

2 Answers 2

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. This is OP's edited schematic simplified to show only two taps and the bridge rectifiers exploded into their individual diodes. An earth has been added to the bottom of the transformer as a voltage reference point.

With the earth shown in the second circuit and the AC voltage frozen at V = -5 V and -10 V we can see a problem. D11 will be trying to pull the DC common line to -4.3 V (0.7 V diode drop) but D16 will pull it to -9.3 V.

To repeat: You can't have a common on both sides of the bridge rectifiers.


schematic

simulate this circuit

Figure 2. A simple solution. SW1 represents the multi-tap selector.

  • This solution uses half-wave rectifiers so the capacitors need to be a bit larger. Use one of the online calculators to check values.
  • The permanent 24 V is taken from the topmost tap on the transformer.
  • The negative rail is generated from the lowest tap on the transformer or from an independent secondary.
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no that's not going to work. you have parts of the transoformer short-circuited by diode paths.

the common negative must be directly connected to one of the secondary taps.

then you use half-wave or push-pull rectiifiers to get your output voltages.

the switches in the second curcuit don't limit the voltage on the switched output.

enter image description here

here I get 24V on the blue and green path and no current flows in the red path.

one possible solution is to use half-wav rectifiers, but that makes filtering the ripple harder.

enter image description here

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  • \$\begingroup\$ See the edit please, is it correct? \$\endgroup\$ Commented Dec 25, 2019 at 8:50
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    \$\begingroup\$ no. same problem. \$\endgroup\$ Commented Dec 25, 2019 at 9:02
  • \$\begingroup\$ Appreciate it if you draw a diagram of a transformer with more than 3 taps to show me how, I don't get the concept... \$\endgroup\$ Commented Dec 25, 2019 at 9:03
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    \$\begingroup\$ to do it right either use half-wave rectifiers, or use separate secondaries \$\endgroup\$ Commented Dec 25, 2019 at 9:23
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    \$\begingroup\$ not without using isolated transfomer secondaries. \$\endgroup\$ Commented Dec 25, 2019 at 20:41

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