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I am designing a mixed signal circuit and I need +5/-5 dual supply for my opamps. Could the following configuration work, else what are my alternatives?

 Li battery 3V--> Boost regulator--> 14V-->14 to 7---> Linear reg(e.g.,7805)--->+5V    
 (coin or LiPo)                             0 to 7---->Linear reg(e.g.,7905)--->-5V
 The 7V would be the actual ground throughout the circuit for both analog and digital

Given that, I have a microcontroller and other digital components in the circuit, I will be drawing more current from the positive supply than I would be sinking from the negative supply, so a net positive current is drawn from the boost regulator (OR is there something wrong with this logic?)

Would this scheme work? Any other suggestions for achieving the same results? Any specific battery specs or otherwise I need to worry about?

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    \$\begingroup\$ No. You can't invert the voltage with an LDO. Why are you boosting to +14? \$\endgroup\$ – DrFriedParts Sep 12 '14 at 10:56
  • \$\begingroup\$ @DrFriedParts: I am boosting to 14 to get a +7 for 7805 and a -7 for 7905 if we create a ground for the circuit in middle of the 14V output. These need minimum 7V to give a regulated 5V output. 7905 is a negative output LDO, so if we use the middle ground and lower end of the 14V supply, we are giving it a negative voltage in effect as input. \$\endgroup\$ – Samyuktha Sep 12 '14 at 11:18
  • \$\begingroup\$ @SpehroPefhany Sorry but what is a ground referenced dh? \$\endgroup\$ – Samyuktha Sep 12 '14 at 12:28
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    \$\begingroup\$ Even if this could work (and you'd need more circuitry to create a virtual ground - and the virtual ground might cause problems if you're planning on using a ground referenced charger) .. It wastes a lot of battery power. It's feasible, but I think a bit more voltage or better regulators would be a good idea. If the current is small from the negative rail there are better ways. \$\endgroup\$ – Spehro Pefhany Sep 12 '14 at 12:36
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    \$\begingroup\$ Why not just boost to 10V and be done with it? \$\endgroup\$ – Matt Young Sep 12 '14 at 13:43
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I would use a boost regulator form 3v to 5v, and then a voltage inverter like the Analog Devices ADM8829 or the Maxim MAX1697, about $2.30 in single quantities, to get the -5v.

Both take an input up to 5.5v, and produce a negative voltage of the same magnitude, with an output of either 25 mA or 60 mA which should be enough for op-amps. They only need a minimum of external components (two caps). Quiescent current is 600 uA for the AD part, and a few hundred µA for the Maxim part. Very simple, you don't have to deal with virtual grounds or the like.

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  • \$\begingroup\$ Thanks, that helps! So, essentially, the inverting IC will be sinking current so, the boost regulator should also be able to sink current,right? What else should I look for in such a boost regulator? \$\endgroup\$ – Samyuktha Sep 12 '14 at 15:58
  • \$\begingroup\$ @Samyuktha BTW, I want to mention for prototyping there is a voltage inverter available in a DIP package, MAX660. A lot more expensive ($8) but good for breadboarding. Similar specs to the ones in my answer. \$\endgroup\$ – tcrosley Sep 12 '14 at 16:43
  • \$\begingroup\$ @Samyuktha There literally thousands of boost regulators. You haven't mentioned what your current requirements are. Assuming it is less than 600 mA, you might want to look at the TPS61202DSC, $2.87 in singles at Digi-Key for 2.87 ). All boost regulators require several external components; if you are laying out a board, follow the recommendations in the datasheet. For prototyping, there is also a 5v boost regulator in a DIP package, LT1302-5 for $6.16. \$\endgroup\$ – tcrosley Sep 12 '14 at 16:59
  • \$\begingroup\$ Consider this part, as well: cds.linear.com/docs/en/datasheet/3260fa.pdf You will still need a boost converter, but you get the inversion plus nice LDO functionality on both rails. \$\endgroup\$ – user49628 Sep 12 '14 at 22:37
  • \$\begingroup\$ @user49628 Frankly I'm not sure what the advantage is to have to boost the voltage past +5v and then drop it back down using an LDO with only an output of 50 mA on +5v (compared to the 600 mA you can get from the boost regulator by itself), and also 50 ma on -5v. And the combo is twice as expensive: $7.26 in singles + $2.87 for the boost reg = $10.23 (compared to $2.87 + $2.30 = $5.17). \$\endgroup\$ – tcrosley Sep 12 '14 at 23:31
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  1. The 78XX and 79XX series are (to my knowledge and a quick google check) NOT Low Drop Out (LDO) regulators. They require (as you note) some 2 Volts above the regulated voltage on the input.
  2. You would do better to use a switching power supply to generate your negative voltage - this way, you can have one ground level for all of your circuit rather than having a virtual audio ground 7 Volts above your digital and other ground planes.
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  • \$\begingroup\$ There are just a couple of switched negative regulators and they are really costly, is there another way? I plan to use the 7V as the only ground for both digital and analog. Digikey lists 78xx, under LDO so I was confused, thanks@JRE \$\endgroup\$ – Samyuktha Sep 12 '14 at 12:24
  • \$\begingroup\$ I found switched negative converters in several places. Does $2 for a chip sound like too much? (TPS65130) \$\endgroup\$ – JRE Sep 12 '14 at 12:36
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Yes, it will work. Or at least it can be made to work with some extra effort. But it's probably not the best way to do it; the best solution would probably be an inverting switching regulator for the negative rail. But I will discuss your idea here.

As @JRE said, the 7805 and 7905 aren't LDO regulators; they're linear regulators (and LDO is another type of linear regulator, "low drop-out").

But that's beside the point. The 78xx can only source current and the 79xx can only sink current. But to create the virtual ground at 7V, you'd need to do it with a regulator that can both sink and source current, because at different times either the high side or the low side might be using more current. An op amp wired as a voltage follower would be one example (following the 7V from a resistive voltage divider), but it needs to be beefy enough for the difference between the high side and low side currents.

At least that's the theory if you want a robust ±5V regulator for generic use. If your circuit always uses more current on the high side, the virtual-ground regulator could be a sink-only type.

Since your power source is a battery, the virtual ground is perfectly okay, but if you connect a grounded battery charger and at the same time your circuit's output is connected to some grounded equipment, suddenly your virtual ground could appear like +7V and blue smoke may be released from somewhere.

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  • \$\begingroup\$ Thanks @Rennex, for the detailed reply. I understand switched regulators better now. Could you suggest any resource where I can learn more about them? \$\endgroup\$ – Samyuktha Sep 12 '14 at 16:03
  • \$\begingroup\$ +1 This is a nice summary of the issues in play. Just a small suggestion, I would remove the first sentence and begin with the second. \$\endgroup\$ – DrFriedParts Sep 13 '14 at 3:35

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