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I'm trying to power a home-made load cell transmitter using a 24V single SMPS. I need to make +12, 0 and -12 Volts that are capable of 50mA. I wish to power multiple channels of opamps and bridges.

I don't have much budget and availability of components in India.

I have an idea to use 1 LM7812 an 1 LM7912(negative) linear voltage regulators and a voltage divider setup to do this as per the circuit below.

schematic

simulate this circuit – Schematic created using CircuitLab

Would this work? I've modified it from the suggestions and articles elsewhere.

Somebody suggested me one other circuit but I am concerned about the current capabilities of the opamp.

schematic

simulate this circuit

Would this work? If yes, please suggest suitable op-amp.

Are there any other techniques that would do the job economically?

mastermind
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Ohbhatt
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  • Think about what will happen if you have mismatched loads between the rails. – winny Nov 19 '18 at 14:23
  • @winny That's what I am worried about. – Ohbhatt Nov 19 '18 at 14:38
  • How much current do you need? I have made such a contrapment for an audio circuit to prevent uneven clipping using an opamp and even divider like yours, but transistor+resisor buffered on the output. Wasted a lot of power and there are easier solutions. In your case, I would go for two switchmode converters or one isolated one. – winny Nov 19 '18 at 14:42
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    You mention you need to be able to supply 50mA of current. But I guess this is mainly through the +12 and -12 rails (e.g. powering dual-supply opamps). What exactly do you have to supply through the 0V rail? If the 0V rail simply serves as a reference and only goes to some opamp inputs or high-valued resistors, it means your current needs for the 0V rail are much much lower than 50mA, and the solution #2 is perfectly valid. – dim Nov 19 '18 at 14:55
  • Btw, both farnell.in and mouser.in ship in India. You would find pretty much any component on these(albeit a bit pricey ). Another option is ebay.com , but these ship from China and have quite long delivery times. – Tejas Kale Nov 20 '18 at 21:54
  • And another option is to explore the electronics supply market in your city. For example in Mumbai, you can find pretty much any component at Lamington Road. There's usually some shop selling components, with contacts to source unavailable ones in every city. – Tejas Kale Nov 20 '18 at 21:55

7 Answers7

19

You first idea will not work at all.

Your second idea will work, but many OP-Amps aren't going to deliver more than a few mA on their output, which limits the current your circuit may draw from the virtual ground. There are Power-OP-Amps available which may deliver up to a few ampere, but if you cannot get your hands on one, you can use a PNP/NPN transistor pair to increase the output current:

schematic

simulate this circuit – Schematic created using CircuitLab

The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though.

Janka
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  • can you please suggest a Transistor suitable for 50 or 100mA. – Ohbhatt Nov 19 '18 at 16:02
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    You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application. – Janka Nov 19 '18 at 16:35
  • Okay. I will take a look at that. Thank you for the time. – Ohbhatt Nov 19 '18 at 17:04
  • Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails. – Ohbhatt Nov 20 '18 at 11:48
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    The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.) – Janka Nov 20 '18 at 12:59
  • If you tell which distributor you want to buy from, we could help. – Janka Nov 20 '18 at 13:04
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    @Ohbhatt You could use multiple 2n2222 and 2n2907 in parallel with small emitter resistors (2r2 or so) to share the load if you're unable to source bigger parts. – Colin Nov 21 '18 at 09:11
  • I found the BD139/BD140 PNP/NPN pair with my vendor at a good price. Could they work? – Ohbhatt Nov 22 '18 at 06:55
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    Yes. But be careful, without a heatsink, the BD139 and BD140 may dissipate only up to 1.25W. A tiny fin on each transistor improves that drastically. – Janka Nov 22 '18 at 12:56
  • @Janka Thank you so much. You not only solved this problem for me but also taught me a lot. – Ohbhatt Nov 23 '18 at 05:27
7

You'd be better off using two 12V supplies, but if you insist...

#1 won't work.

#2 (given the very limited information you have supplied) might require the op-amp to dissipate as much as 600mW and stability would likely be an issue with capacitive loads. There are dedicated rail splitter chips which take stability seriously but they are not jellybean parts and, for example, the TLE2426 cannot handle the dissipation or current involved.

I suggest something more like this (assuming you have power to spare on your 12V supply:

enter image description here

This uses a ubiquitous TL431 shunt regulator and boosts it with a generic PNP power transistor.

The combination is like a precision power zener. Or just use a zener as below. Set Vo = 12V.

Then use this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that if you excessively load the GND to -V the +V to GND voltage will increase to as much as 24V. Usually that's acceptable but take care about capacitor voltage rating and so on. You can add a higher voltage zener (say 14V) across R1 as a preventative measure. R1 will dissipate less than 1W, under normal conditions, but the zener could dissipate as much as 1.3W if 50mA flows from +V to GND and there is no corresponding current from GND to -V.

You could use two 6.2V 1W zeners in series, for example. Keep the leads short, attach them to some PCB area and keep them apart so they run cooler.

Spehro Pefhany
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  • I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though. – Ohbhatt Nov 19 '18 at 15:43
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    I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents. – ThreePhaseEel Nov 20 '18 at 03:26
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Given your desire for as low power as possible, and my realization that this common problem is seldom approached this way. I came up with a self-oscillating switching solution just for the fun of it.

As with any switcher, single-tone emissions/ripple have to be considered (around 20kHz with these values). But if there is any significant ground current, I doubt you can be much more efficient (a more formal switcher with a separate oscillator can be made more efficient and could use a single inductor, but it would require more parts).

schematic

simulate this circuit – Schematic created using CircuitLab

It is basically a relaxation oscillator that modulates the average current through L1 so that it oscillates around the required ground current. M1 and M2 are switched on and off relatively quickly (some acceleration capacitors would help with efficiency) and C12 provides positive feedback so that the opamp/comparator saturates on crossing the threshold (otherwise the load would damp the oscillator and it would become a linear regulator instead).

L3, C10, and C11 are there to filter the ripple and to isolate the oscillation from the load, so as to avoid dampening it too much. C10, and C11 also do double-duty as the regulator input capacitance. Excess energy in L1 and L2 would be returned to the required rail and stored in them. M1 and M2 source-drain diodes are conducting in this design.

R3,R4,R5, and R6 are chosen so as to keep M1 and M2 below threshold when there is no ground current. Unfortunately this also reduces the overall gain of the oscillator loop.

I have not done a very careful analysis of all of the implications of this design (particularly because of it being self-oscillating), so overall stability considerations on load changes might be an issue.

I don't think there are ICs for this type of configuration, which unnecessarily increases the part count and the design constraints. The only ones I know of are the DDR memory termination voltage regulators, but those are intended to work at very low voltages.

Edgar Brown
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  • +1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime. – Janka Nov 19 '18 at 20:22
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    Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components. – Edgar Brown Nov 19 '18 at 20:27
  • Yes, exactly the latter was my concern. *What happens when the ground current changes direction.* – Janka Nov 19 '18 at 20:30
  • @Janka granted, in most applications it's overkill. Particularly in audio you do not need large un-balanced ground currents. In DDR memory, where these architectures are used, the controllers could handle up to 10A of imbalance. – Edgar Brown Nov 19 '18 at 20:31
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    @Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power. – Edgar Brown Nov 19 '18 at 20:34
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    @Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that.. – Edgar Brown Nov 19 '18 at 20:46
  • Thank you so much for your time but I might not be able to implement this circuit in my design. – Ohbhatt Nov 20 '18 at 06:08
  • @Ohbhatt I was aware of that. But it was hard to pass this fun exercise. My goal was minimizing the number of components to an acceptable level while maintaining efficiency and improving reliability, it is still too big for my taste. – Edgar Brown Nov 20 '18 at 06:16
  • +1 This is nice. Output will gain a firm midpoint reference. Supply fluctuations will appear at half magnitude and opposite polarity on the two rails. No loss of total supply voltage. – KalleMP Nov 20 '18 at 11:36
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The regulators won’t work. You have zero dropout allocated to them and your ground impedance is excessive.

The op amp is a better option, but it all depends on how much current you have going through ground. If the current is low enough you can just use a resistor divider with a couple of capacitors in parallel, if it is high you would need a hefty op amp.

You have a couple more options:

  1. You could use two zeners with series resistors to reduce ground impedance
  2. You could put together a class AB source follower with a few resistors and two transistors (basically what the op-amp is doing but with higher impedance)
  3. If your ground current has a well-defined and consistent direction, you can use a positive or negative 12V regulator or even a transistor off one of the rails (making sure to put a bypass diode).

But regardless of what you do, any ground current will result in wasted power (unless you figure out how to design a ground-switcher regulator of course).

Edgar Brown
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4

If your 24 V is well regulated you could just use a 7812 to create a mid point and call that your 0 volt rail.

schematic

simulate this circuit – Schematic created using CircuitLab

This will only work if the 24 V is independent of whatever you're powering, and as per Edgar Brown's comment, positive linear regulators like the 7812 can't sink current.

Colin
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  • That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify. – Ohbhatt Nov 19 '18 at 16:04
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    This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current. – Edgar Brown Nov 19 '18 at 16:15
  • That's a very valid point, thanks, @EdgarBrown, I've edited the answer. – Colin Nov 19 '18 at 16:22
  • I'm not having voltages any lower than -12V in my system. That would eliminate any chances of negative ground current. would that? and please refer the modification in the first solution. That could turn out to be even better. – Ohbhatt Nov 19 '18 at 17:27
  • What would happen if you also added a 7912 with its INP tied to -12V rail, its GND pin tied to the +12V rail, and its OUT tied to the 7812 OUT (inside-out version of OP circuit #1)? – amI Nov 19 '18 at 19:32
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    @EdgarBrown Instead of a trusty `7812` you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive. – crasic Nov 19 '18 at 20:16
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    @Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design. – user253751 Nov 19 '18 at 22:46
  • One more thing to keep in mind: having `U1` to drop 50% of the source voltage is a red flag in terms of overheating. As soon as you eat 50 mA, that's OK to dissipate less than 1 Watt by the chip package. Any increase of the current would bump that dissipation volume up rapidly. – Yury Schkatula Nov 20 '18 at 12:56
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I think NJM4556A would work

you can draw current from negative and positive rails but there difference not to exceed the OP amp output current.

Note: I'm not experienced, i suggest you to read the following post

EEVBLOG - my-negative-voltage-rail-doesnt-work

Hasan alattar
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There are many low-cost methods. But switching method may help you with a minimum component that available everywhere.

you can use a flyback converter with a minimum circuit:

Edited: The main circuit: mix circuit Ref: a mix of two links (http://uzzors2k.4hv.org/index.phppage=flybacktransformerdrivers, https://wiki.analog.com/university/courses/electronics/text/chapter-6)

Component list:

  • Zener diode

  • 555 IC

  • Mosfet

  • A toroid, the transformer can be made with wire and a toroid core

  • Diode in output

  • some capacitor

  • some resistor

  • some wire

Benefits:

  • you can generate any voltage in output even bigger than your first voltage

  • these components are available everywhere

  • you can generate any voltage even isolated voltage

  • you can increase your power by changing the Mosfet and selecting a bigger toroid.

The main references:

http://uzzors2k.4hv.org/index.php?page=flybacktransformerdrivers enter image description here

Additionally, you need a Zener diode for 12-15volt and a 555 IC.( your coil feed with 24Volt but for 555, you should generate a 12-volt power with a Zener diode).

in the output, you need a diode with a capacitor. link: https://wiki.analog.com/university/courses/electronics/text/chapter-6 enter image description here

It is dual polarity Full-wave rectifier using a center tapped transformer and 4 diodes

M KS
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  • is your edited version removes the connection between the zener voltage and supply? – Hasan alattar Nov 21 '18 at 14:42
  • @Hasanalattar No the main circuit(Eirik's flyback)works with 12-16 volt. I added a Zener as a regulator for this case, to convert 24V to 12V. Just I mix 3 circuits. a regulator and flyback and output coil for double voltage in output. – M KS Nov 21 '18 at 15:07
  • what i meant is that main circuit shorting 12-16 to 15-30 volts of the transformer!, and ne555 exceeds its rated vcc – Hasan alattar Nov 22 '18 at 05:10