Rail splitter circuit question - Page 1
Forum section: electronics / beginners

author: CrazzzyEE
author_rank: Newbie
timestamp: August 17, 2021, 02:05:49 pm
content: Hello fellow humans!I have been working on a project that requires both negative and positive voltages. I have made this circuit to “split” the 11.4V Li-Po battery into +5V and -5V. Circuit -https://imgur.com/a/wdg9ZElSimulation -https://www.multisim.com/content/vy5dRfiYiVuewNDLmKGP6W/dual-rail-supply-from-li-po-battery/open/The circuit works fine with low power ICs I have in my main circuit (main circuit is ECG amplifier and filter) but as soon as I connect my microcontroller(Arduino NANO 33 IoT) to the positive rail of the splitter circuit the voltage on the rail drops from 6V to 4V and respectively the negative rail goes from -6V to -8V. I want to know if this behaviour is odd and if it is possible to fix it?I want to be able to draw more current on the positive rail of the splitter, while the negative rail will not have the same load.

author: iMo
author_rank: Super Contributor
timestamp: August 17, 2021, 02:31:28 pm
content: Lower the R18 and R19 values to something like 1k.

author: Manul
author_rank: Super Contributor
timestamp: August 17, 2021, 03:55:51 pm
content: In this circuit current from follower is limitted by what current 10K resistor provides and transistor hFE. So output can not provide a lot of current. Actually you could just connect both bases together and drive with op amp. Like a standard complementary follower. Maybe some current limitting resistor in series with opamp output. I do not see purpose for such arrangement like in your circuit.

author: pqass
author_rank: Super Contributor
timestamp: August 17, 2021, 04:16:05 pm
content: You can try the following rail-spliter design or adapt your op-amp driven one to use this output stage. But you'll need the emitter resistors otherwise there could be excessive current drawn between the rails. You can experiment with 1 ohm or 0.5 ohm instead of the 10 ohm shown.If you need more power, you can lower the base-to-rail resistors or use darlington transistors instead. In the latter case, you must double-up (in series) each diode (for a total of 4).See:https://electronics.stackexchange.com/questions/333896/a-question-about-splitting-a-regulated-power-supply-by-resistors-vs-discrete-com

author: pqass
author_rank: Super Contributor
timestamp: August 17, 2021, 05:08:02 pm
content: I've simulated one possible implementation and it looks like you'd be limited to at most 100mA per rail if you use ordinary 600mW small signal transistors.However, you can get 1A per rail if the 470ohm resistors are replaced with 330ohm and the transistors are 5W capable.Play with it here:http://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgoqoQFMBaMMAKABcKU9wNCLeefKhAYJoAFjRg8CLNmLYMxccSjRsKYnmY48eSNmxhx3MHBAATOgDMAhgFcANm3aduG2gI9QQTEOOgFZUIwDH1FMBDIJAYYDDBsEJRmNEhxSPEkMyorOycXACd+PjBiFGKQFAQhcBYi2T4qvgaQXHEfViLxDUF-HuIa8QxIOr7y727y8XMqIZGAJTHWqZ6qNf9hMGgkdZgEFgsl0L5JyurKyxsHZwOlpqXvctzrlwB3VsJuadme76gWd6JbgDJYgkaAz5nE49e4jADmSz+p3EJh84IqSN+MwBFRBpzBLAADiBPtx7qSNsI0USSXoQCCKfcIGscRSQS0Ce8KX8Wn8RopPHx2QJIlSWQKKoZmgI-sz-hKWuS6UzqUV9MJCFR1VCfNVRtq2iBtXzkHAccbzNzse9tUqyed0dqGXSCRKnXxtaLwNS3eZ7ga0t6WcSgTrQ7LqSHISDQyqWbZKtMw5D-eAtjtTfBaNAMJR0JkUNM9JoKDmsln0YWqFXKsr8P93lwyfXFQ7RqVyphyh36YNhmaij2u+AypSfHMzRLTr9+iUfcMHmg3MsxfKF6djsvN3L0T2Y9GhDie39w9bE9WMN3R8ORkAEDIT: Full disclosure... I forgot to lower the transistor hFE to 25 in the 1A example (was still at default of 100). That means that 330 ohm base resistors is too high. You'll need to lower them to 100 ohm to maintain equal output voltage split while maintaining a 1A asymmetric draw.

author: CrazzzyEE
author_rank: Newbie
timestamp: August 17, 2021, 05:53:09 pm
content: Thank you fellow human! This solution worked just fine for me. My whole circuit is powered now!

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2021, 08:27:43 pm
content: Reducing the resistor values will work, at the cost of greater power dissipation. Another option is to add another diode in reverse parallel, with each diode, so the op-amp can decide how much current is provided to the bases, rather than having to mop-up the excess.By the way, that web based app is terrible. The grey on white background makes it difficult to read and it has to be one of the worst editors I've used, in terms of moving parts and wires around. Had to import the image into Gimp to increase the contrast, so it's easier to read.

author: pqass
author_rank: Super Contributor
timestamp: August 18, 2021, 01:57:10 am
content: Quote from: Zero999 on August 17, 2021, 08:27:43 pmHmm... why have 4 diodes? I don't think the original forward-biased diodes would be needed at all.The base-to-rail resistors will, by default, turn on each transistor and also decide (via high-value) the minimum Ic drawn (1.2mA see below). The opamp will source or sink additional base current when there is an imbalance in split output current; up to 40mA base current for 1A asymmetric output.Reading the LM358 datasheet, it says that it can source 40mA but only sink 20 mA. So, you should be good for 500mA split output with it.Updated simulation here:http://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgoqoQFMBaMMAKABcQAWTvcFXnlTD8o4EAwTQwnQoU4JsCTihWEIMQpDClZKYnpTYVxECiQATOgDMAhgFcANm3ZcepvALxV+vKkxDY0MQKeIR4wZxYyrwMMFjYGMSQCV7ShJiEphbW9k4sAE5cXuDcRd74oqyFgqYVNZgolQUBaHy8RlRuQvBwzR3u7a3YYZU9kCwASi1+YJn9DF2inEJg0EhUG2ssAO6mGI3Dg1SHUDt7jW79XSwA5tMBI1ezouO7NScfI69lRQ+epwAHCj6Wq8BAgk7qQHAi5gkFoXhQ77g2E-a67FF-H4ncYJTrFNw1ZiNJEsPHYyCcbEjUnk+oVTEIsQbZphRGlNmg0a9Qr7cq8PnnUSSQhnDCUrGYnFnTlMxkVb7iqluTHXclK34a4nM07qiW4KkayEvFhAhpc83Gllm-a-c1M0k2cDCC0VJmI1brZBjCjQDCUSAoTjYeShULYEySf3e+BnbXaEku6RU77x0rBkmlcaFYiZFRU3O-TrinkgQv5stzL5cEu48X3JhzVpuWn17i8HxYzut440kTtto676Ftzm9GVru26W7bUV8tB3X1pnL20Ok3me5M-rCDsgSy2RzODf9Tsnt17nKHlgAZRAWpdWqzolsDgAznRRCgb3f9RKNeaNhAF930-GUPC5QUmWzEBZQqWVbW6UtBWTO9Wig31RV2e0KjPXwzhPcCTk7b5txESERHGIFyPaEY12tLscNo8DSV2IicLQhV8JGS5mzw1jHkeIYvlNLFLj7RETSopiaJkocRLYzx+Tkt4jBKZViiglggA

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2021, 07:58:11 am
content: Quote from: pqass on August 18, 2021, 01:57:10 amThat's true, but reducing the biasing resistors increases the quiescent current.The extra diodes enable the op-amp to supply the current, to the transistor bases, rather than simply diverting current away from them. The downside is poorer transient response, as the op-amp's output has to swing faster, which can be remedied by adding a capacitor across each diode. This circuit is prone to oscillation anyway, due to the huge output capacitance. It requires some frequency compensation to work properly.