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Stable DC power supply point for Ultra Low Noise Amplifier (ULNA)

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Hello good people,

I need help. I'm posting a schematic of an ultra low noise amplifier. It is to be powered by batteries. For that reason, I need a stable DC solution in the place of R4 resistor. I was proposed to achieve that by using another JFET and an integrator circuit, but I am not quite sure how to apply that. Any ideas?

Best wishes


Rich D.

Active Member
Maybe I don't understand the question, but I believe batteries are low-noise already. Adding a voltage regulator with a good filter should achieve the goal of a "stable DC voltage".
Can I assume the "integrator" you speak of is just a fancy word for a capacitor-based filter? If so, there must be a 10,000 examples out there.

I would head first to an LM317 which is pretty low noise. If run from a 12 volt battery I would choose a low-dropout regulator and sprinkle it with a few good quality capacitors in the filter -er- integrator.
It can also be used for many of the other voltage sources shown in the schematic.

If you need more design details than that I'm sure we can recommend some decent regulator chips. Likely the best schematics for these are included in the chip specifications.

schmitt trigger

Well-Known Member
Use another LT1028A as an unity gain follower, which would be buffering a resistor divider which you will calculate to obtain 7.4 volts from the 11.1 supply.

And you will tell me: "the 7.4 volt supply will drop with the battery voltage". Sure, but you will be applying that same voltage as bias to the non-inverting supply, so the net offset will be zero.

In a real world circuit, Ids will vary with both individual transistor transconductance and to a lesser degree temperature changes. But I see that your signal input is AC-coupled, you can also AC-couple your output.
Dear Rich, thank you for your detailed answer, and I apologise for my scarce layout of an issue.

The purpose of the system is the low-noise acquisiton of low-frequency signals and this schematic is only the first part, ultra low noise amplifier. After that the signal is to be lead to ADC, and then through a transmitter and optical cable to a PC where the data is displayed and stored.

Batteries are indeed low noise. What I meant is that I need a stable DC voltage, so the varying voltage of the batteries doesn't affect the output. And by an integrator, I meant an op amp with a capacitor in the feedback loop (what I have been suggested to use). Thank you very much for your suggestion, perhaps it is a better option, since I don't know how to apply the 'JFET + integrator' solution anyways. Seems like a simpler solution as well. And please excuse my lack of experience and knowledge.

Rich D.

Active Member
An op-amp is indeed a good idea for use as a stable, low-current voltage source, and in fact may be a bit less expensive and flexible.

If I may split hairs though: As far as I know a voltage regulator includes an error amplifier (an op-amp), the filter capacitors may or likely may not be in the feedback loop, but as long as they do the job of reducing AC variations it shouldn't make a difference. The main difference is that a voltage regulator is generally used for a higher current output. Your circuit does not need a high current, so an op-amp is a good choice.

I think the smart design would use an op-amp.

The performance would be affected by the reference voltage of the op-amp's input - make sure the voltage divider is supplied with a stable source and filtered properly. Temperature could be an issue if measurements are taken over a long time. If so, consider using a zener diode or a 3-pin reference IC such as a very cheap AN431 can really help with the voltage reference and thermal performance. The op-amp as a voltage follower would not really degrade an accurate output. You'll get a stable DC with any temperature and almost any battery power remaining.
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