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I am trying to get ±5V using voltage regulator L7805 and L7905.

The datasheet says to use AC source as input and transformer, but I am not really familiar with transformers, and I don't know if I can handle 110V AC as a source.

So I came across YouTube clip, creating ±5V with DC input.

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Figure 1: Circuit Diagram

Then I also learned that size of capacitors are enough as long as they are more than what datasheet requires, from searching through StackExchange How do I decide what capacitor to use in a circuit?

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Figure 2: Datasheet from L78xx and L79xx

I choose 220uF, 50V. This was the biggest I have right now.

enter image description here Figure 3: Test on breadboard and DC adaptor

I used DC 12V as an input.

I tested L7805 and L7905 individually, to check whether they are producing voltage +5V and -5V, respectively.

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Figure 4: Individual regulator test

From voltmeter reading, I get that individual operation is fine.

Then I tried the whole circuit diagram (Figure 1), and I don't get the voltage properly.

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Figure 5: Whole circuit test

As can be observed from Figure 5, L7905 produce voltage of -5, but L7805 doesn't. I replaced L7805 with other L7805 to see if that L7805 was damaged, but it wasn't.

I searched more on Stack Exchange, and found one, Incorrect Output From 7805 but it doesn't solve my problem.

Then I found an website, that suggests I need more input voltage.

I changed to DC adapter of 18.75V with 3.15A as an input, however, it doesn't solve my problem either.

Was it had to do with wrong(?) circuit diagram (Figure 1) from the very beginning?

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    \$\begingroup\$ The schematic you've drawn will work sort of okay, as long as no net current flows in the ground. Otherwise your neutral point will drift because of the capacitors you're using to clamp it. It's not really a design I'd recommend using. \$\endgroup\$ – Hearth Mar 12 '19 at 17:19
  • \$\begingroup\$ Large-value resistors across the neutral-point clamping capacitors will help, though. The smaller the resistors, the more power wasted but also the more current imbalance can be handled without losing regulation. \$\endgroup\$ – Hearth Mar 12 '19 at 17:21
  • \$\begingroup\$ @Hearth is there any recommendation on the circuit diagram using DC input? or do I have to figure out how to use AC input? \$\endgroup\$ – user65452 Mar 12 '19 at 17:21
  • \$\begingroup\$ If you want to use DC input and have a usable bipolar power supply, you want a switching converter, not linear regulators. Using AC input would be the easiest method for a beginner. That doesn't have to be 110V AC, though; you can get wall-wart transformers that output just 12V AC to your board, for instance. \$\endgroup\$ – Hearth Mar 12 '19 at 17:23
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    \$\begingroup\$ Not only does this design smelly fishy, the reality is that building your own power supplies is rarely worth the trouble - the reasons to do it would be if you had a very unique requirement (which you do not) or you were studying power supply design in which case you would be looking at classic architectures, or you were designing the power supply of a product, which this is far from. Buy something suitable (at the very least something with true positive and negative outputs, even if you then need to regulate more) and put your effort elsewhere. \$\endgroup\$ – Chris Stratton Mar 12 '19 at 17:36
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The circuit above has a floating ground and the supply feeds both. This is not the way these dual voltage regulator circuits are built, they usually use two supplies, a positive and negative supply, then regulate it down to the voltage needed.

The problem with regulators is they have dropout and need more voltage on the input terminal than what they regulate to on the output terminal. With a positive regulator, this is easy since most of the voltages we have are positive, you have a 12V source and regulate down to 5V. With the negative regulator, you'd need a -12V (or at least roughly -5.5V).

So either find a dual supply input (like ±12V) OR do this:

enter image description here

These regulators are 7805 drop in compatible and are DC to DC regulators, they have little loss and can generate a negative rail from a positive one.

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  • \$\begingroup\$ Of course, if the requirement isn't necessarily to have a drop-in replacement, there are also single devices that can provide both positive and negative supplies from a single positive input. \$\endgroup\$ – Hearth Mar 12 '19 at 17:45
  • \$\begingroup\$ @Hearth The 7805 drop in DC/DC's are compatible with breadboards \$\endgroup\$ – Voltage Spike Mar 12 '19 at 17:50
  • \$\begingroup\$ That doesn't mean other ones aren't. This one comes to mind. \$\endgroup\$ – Hearth Mar 12 '19 at 17:58

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