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I'm trying to design a power supply circuit that can output four voltages from the same circuit using linear voltage regulators. I'm still learning a lot of this, and trying to stick to the LM7800 series for this application. The circuit currently uses the following regulators with a 15V input supply to produce four output voltages:

  1. LM7812 straight to output 12V DC
  2. LM7809 to output 9V DC
  3. LM7805 input connected to LM7809 output for 5V DC (to try and avoid excessive heat output from the LM7805)
  4. LM7810 on a 4kΩ+2kΩ voltage divider (if these values are too much, any ratio of R1/R2 = 2/1 will work) to step down to 3.3V DC

To that effect, I have a few questions about the design of the circuit:

  1. Is it a good idea to use the voltage output of one voltage regulator to power the input of another? I did read that the LM7800 series exhausts excess energy as heat, so I'm trying to avoid driving the LM7805 with the full 15V input.
  2. If #1 is not a good idea, would it be possible to use a voltage divider before the input of the LM7805 instead?
  3. Is there a voltage regulator in the LM7800 series that provides 3.3V directly instead of relying on a stepped-down LM7810? I did a quick search on Google for LM78033, but most of the results that I got were for defunct pages or for products that are no longer offered.

My primary circuit design constraint is simplicity. I'd like to make it compact enough, preferably using THT components, to fit on a PC board that can be turned into a DIP package for breadboard use.

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  • \$\begingroup\$ There was a very similar question asked just the other day: electronics.stackexchange.com/questions/95342/… \$\endgroup\$ – The Photon Jan 6 '14 at 5:14
  • \$\begingroup\$ Do you have a reason for not using an adjustable regulator? Buying 4 of the same part number is likely to be cheaper than buying onesy-twosy of 4 different part numbers. \$\endgroup\$ – The Photon Jan 6 '14 at 5:17
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    \$\begingroup\$ What current do you want to be able to draw from each output voltage? \$\endgroup\$ – Wouter van Ooijen Jan 6 '14 at 7:28
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    \$\begingroup\$ Look at LM317 datasheet. It can be used for all the voltages you are wanting. LM7810is rare and not advised. LM7809 similar. LM7805 is common but past its use by date. \$\endgroup\$ – Russell McMahon Jan 6 '14 at 8:16
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If you take the input for the 7805 from the 7809 output, you must include the current drawn from the 7805 into account when determining the power dissipation in the 7809.

You should not use a voltage divider to make 3.3 volts - the voltage on your 3.3 volt line will vary depending on the current drawn from the 3.3 volt line, as that current will have to flow through the "top" resistor of the divider, increasing the voltage drop across it. Look for a dedicated 3.3 volt regulator, or use an LM317 or similar adjustable regulator.

Depending on the currents you need from each supply, it may be best to use a switching regulator to get from the 15 volt supply to 5 volts.

Make sure you have adequate heatsinks on the linear regulators!!

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Is it a good idea to use the voltage output of one voltage regulator to power the input of another? I did read that the LM7800 series exhausts excess energy as heat, so I'm trying to avoid driving the LM7805 with the full 15V input.

If you are supplying (say) 1A at 5V and your input voltage is 15 volts then the 7805 acts like a series resistance of \$\dfrac{15v-5v}{1A}\$ = 10 ohms and, the 1A flows through this resistance and dissipates 10 watts.

It doesn't really matter how you string series voltage regulators together i.e. if you fed the 7805 input from a 7810's output, somewhere along the line 10 watts would be dissipated as heat.

If this is a concern then do consider using a switching regulator - there are plenty to choose from and you can get 0.5A switchers that are pin compatible with standard 78xx regulators: -

enter image description here

They are a little more expensive that the linear range but are much more efficient. Farnell stock quite a few of them too here

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