Op Amp Circuit Noise

[AdamZ]

See attached circuit diagram. I've designed a fairly simple circuit for amplifying voltages from a DAQ card (op amp A), measuring the applied voltage (op amp B), and measuring the current through a load (op amp C). The circuit works with little noise when all the op amps are directly connected to the voltage supply. I would like to supply op amp B and op amp C with a lower voltage than op amp A by using a voltage divider (with 1.69 kOhm and 1.78 kOhm resistors). For some reason, this resistor setup results in considerable noise (x10 noise for op Amp B and x2 noise for op Amp C).

I am using a constant voltage supply. Resistors are metal film, 0.4 W, 1%.

Suggestions are very much appreciated.

 

[shrtrnd]

What's the physical composition of your resistors? Many types create their own noise in some circuits.

 

[AdamZ]

What's the physical composition of your resistors? Many types create their own noise in some circuits.

Resistors are metal film, 0.4 W, 1%

 

[Alec_t]

I am using a constant voltage supply.

The supply for opamps B and C won't be constant if the current they draw from it varies with input signal amplitude.

 

[Anon_LG]

Have you tried decoupling capacitors?

 

[BobK]

You cannot use a voltage divider that way because the load is part of the voltage divider.

You need a regulator to drop the voltage.

Also, what is the reason for running them at lower voltage?

Bob

 

[davenn]

You cannot use a voltage divider that way because the load is part of the voltage divider.

You need a regulator to drop the voltage.

agreed !

Also, what is the reason for running them at lower voltage?

more to the point ... why do you need such a high supply voltage to the first Op-amp, 70V


Dave

 

[AdamZ]

You cannot use a voltage divider that way because the load is part of the voltage divider.

You need a regulator to drop the voltage.

Also, what is the reason for running them at lower voltage?

Bob

I don't really understand how the load is part of the voltage divider. To answer your question, op amps B and C can only accept +/- 20 V whereas op amp A can accept +/-35 V, therefore if I want to use 1 power supply, I need to reduce the voltages for B and C. I want the full swing for all op amps.

 

[AdamZ]

agreed !



more to the point ... why do you need such a high supply voltage to the first Op-amp, 70V


Dave

Since I want to swing to high values +/- 35 V at the output, the op amp needs a +/- 35 V supply.

 

[davenn]

Since I want to swing to high values +/- 35 V at the output, the op amp needs a +/- 35 V supply.

no probs

 

[BobK]

I don't really understand how the load is part of the voltage divider. To answer your question, op amps B and C can only accept +/- 20 V whereas op amp A can accept +/-35 V, therefore if I want to use 1 power supply, I need to reduce the voltages for B and C. I want the full swing for all op amps.

Okay,

What is the supply voltage to the op amp (U1) in this circuit?

 

[dorke]

So many issues with that circuit

1. If you will have a full swing of ±35V output from op-A and feed it(like you do) to the input of op-B with ±18V P.S rails,you will destroy op-B !
Since you will go above it's absolute max input voltage rating.

2.The use of high value resistors (M-ohms) isn't a good idea for low noise applications...you do know this right?
Same goes for high input impedance amplifiers...

3. As was suggested to you by others:
You do need to feed ops power pins with a power supply,
not a voltage resistor divider.
So you should use Voltage Regulators like the LM7815 and LM7915 to create a stable ±15V PS for opB and opC. from the ±35V you already have.
and do use decoupling capacitors...

4. How is op-C measuring the load current is beyond me

 

[hevans1944]

This is an ill-conceived circuit.

Clearly the load is high impedance, hence the attempt to "measure" the load voltage without "loading" by using op-amp B with an inverting input configuration and a 1 MΩ input resistor. This in conjunction with a 10 kΩ feedback resistor, yields a gain of 10,000 / 1,000,000 = 1 / 100 = 0.01. So even with a 35 V output from op-amp A, the output of op-amp B will be limited to 0.35 volts. So far, so good. But why is the 1 MΩ resistor even necessary?

The load is being driven by a low-impedance source, the output of op-amp A, so a simple voltage divider consisting of a 100 kΩ and 1 kΩ resistor in series between the load and common will produce a voltage across the 1 kΩ resistor approximately 0.01 times the input voltage. This can be applied to the non-inverting input of op-amp B, and the output of op-amp B would be connected to the inverting input to implement a unity-gain buffer with high input impedance. The reduction in resistance value will lower the Johnson noise injected into this buffer circuit by the 5 MΩ resistor. The AS820 is a good choice for this part of the circuit.

However, although the load may be high impedance, it is driven from the low-impedance output of op-amp A configured as a G = 2 non-inverting buffer. The 5 MΩ resistors are much too large for this task. Lowering their values will reduce the noise contributed by op-amp A. The ADA4700-1 is capable of 30 mA output, so resistors more on the order of 10 kΩ passing 35 / 20,000 = 1.75 mA would be appropriate. The high-voltage capability of the ADA4700-1 is a good match for this part of the circuit.

Op-amp C has its inverting input connected to the "ground" side of the load and its non-inverting input is grounded. That means the "ground" side of the load is connected to a "virtual ground" by means of the 5 MΩ feedback resistor. The voltage output of op-amp C will be the load current multiplied by 5,000,000. So a two microampere load current will result in a 10 V output from op-amp C. The low bias current presented by the AD549 makes it a good fit for this application because of the large feedback resistor and presumably large source impedance of the load.

The voltage divider resistors used at the positive and negative rails of op-amps B and C are not by-passed. If this circuit configuration is to be used, 10 μF to 100 μF capacitors should be installed between both rails and common. In addition, 0.1 μF capacitors should be installed between the op-amp power rail terminals and common, as close as possible to those terminals. This is a "fix a bad design" band-aid solution. You should install three-terminal voltage regulators to bring the ±35 V supplies down to ±15 V.

 

[AdamZ]

This is an ill-conceived circuit.

Clearly the load is high impedance, hence the attempt to "measure" the load voltage without "loading" by using op-amp B with an inverting input configuration and a 1 MΩ input resistor. This in conjunction with a 10 kΩ feedback resistor, yields a gain of 10,000 / 1,000,000 = 1 / 100 = 0.1. So even with a 35 V output from op-amp A, the output of op-amp B will be limited to 3.5 volts. So far, so good. But why is the 1 MΩ resistor even necessary?

The load is being driven by a low-impedance source, the output of op-amp A, so a simple voltage divider consisting of a 100 kΩ and 1 kΩ resistor in series between the load and common will produce a voltage across the 1 kΩ resistor approximately 0.01 times the input voltage. This can be applied to the non-inverting input of op-amp B, and the output of op-amp B would be connected to the inverting input to implement a unity-gain buffer with high input impedance. The reduction in resistance value will lower the Johnson noise injected into this buffer circuit by the 5 MΩ resistor. The AS820 is a good choice for this part of the circuit.

However, although the load may be high impedance, it is driven from the low-impedance output of op-amp A configured as a G = 2 non-inverting buffer. The 5 MΩ resistors are much too large for this task. Lowering their values will reduce the noise contributed by op-amp A. The ADA4700-1 is capable of 30 mA output, so resistors more on the order of 10 kΩ passing 35 / 20,000 = 1.75 mA would be appropriate. The high-voltage capability of the ADA4700-1 is a good match for this part of the circuit.

Op-amp C has its inverting input connected to the "ground" side of the load and its inverting input is grounded. That means the "ground" side of the load is connected to a "virtual ground" by means of the 5 MΩ feedback resistor. The voltage output of op-amp C will be the load current multiplied by 5,000,000. So a two microampere load current will result in a 10 V output from op-amp C. The low bias current presented by the AD549 makes it a good fit for this application because of the large feedback resistor and presumably large source impedance of the load.

The voltage divider resistors used at the positive and negative rails of op-amps B and C are not by-passed. If this circuit configuration is to be used, 10 μF to 100 μF capacitors should be installed between both rails and common. In addition, 0.1 μF capacitors should be installed between the op-amp power rail terminals and common, as close as possible to those terminals. This is a "fix a bad design" band-aid solution. You should install three-terminal voltage regulators to bring the ±35 V supplies down to ±15 V.

Thank you. Very helpful!

 

[AdamZ]

This is an ill-conceived circuit.

Clearly the load is high impedance, hence the attempt to "measure" the load voltage without "loading" by using op-amp B with an inverting input configuration and a 1 MΩ input resistor. This in conjunction with a 10 kΩ feedback resistor, yields a gain of 10,000 / 1,000,000 = 1 / 100 = 0.1. So even with a 35 V output from op-amp A, the output of op-amp B will be limited to 3.5 volts. So far, so good. But why is the 1 MΩ resistor even necessary?

The load is being driven by a low-impedance source, the output of op-amp A, so a simple voltage divider consisting of a 100 kΩ and 1 kΩ resistor in series between the load and common will produce a voltage across the 1 kΩ resistor approximately 0.01 times the input voltage. This can be applied to the non-inverting input of op-amp B, and the output of op-amp B would be connected to the inverting input to implement a unity-gain buffer with high input impedance. The reduction in resistance value will lower the Johnson noise injected into this buffer circuit by the 5 MΩ resistor. The AS820 is a good choice for this part of the circuit.

However, although the load may be high impedance, it is driven from the low-impedance output of op-amp A configured as a G = 2 non-inverting buffer. The 5 MΩ resistors are much too large for this task. Lowering their values will reduce the noise contributed by op-amp A. The ADA4700-1 is capable of 30 mA output, so resistors more on the order of 10 kΩ passing 35 / 20,000 = 1.75 mA would be appropriate. The high-voltage capability of the ADA4700-1 is a good match for this part of the circuit.

Op-amp C has its inverting input connected to the "ground" side of the load and its inverting input is grounded. That means the "ground" side of the load is connected to a "virtual ground" by means of the 5 MΩ feedback resistor. The voltage output of op-amp C will be the load current multiplied by 5,000,000. So a two microampere load current will result in a 10 V output from op-amp C. The low bias current presented by the AD549 makes it a good fit for this application because of the large feedback resistor and presumably large source impedance of the load.

The voltage divider resistors used at the positive and negative rails of op-amps B and C are not by-passed. If this circuit configuration is to be used, 10 μF to 100 μF capacitors should be installed between both rails and common. In addition, 0.1 μF capacitors should be installed between the op-amp power rail terminals and common, as close as possible to those terminals. This is a "fix a bad design" band-aid solution. You should install three-terminal voltage regulators to bring the ±35 V supplies down to ±15 V.

I implemented all changes today except for the voltage regulator (since I need to wait for delivery) and witnessed a 10+ fold improvement to the noise level. Thanks you are a pro!

 

[hevans1944]

Glad you got it working! I edited my original post to correct some arithmetical errors in the calculations, so you might want to go back and re-read it. The gist of it hasn't changed however. Lower the resistor values to lower Johnson noise, by-pass the power supply rails.

Can you give us some idea of what the load is, i.e., what you are trying to DO with this circuit without violating any confidentiality?

 

[AdamZ]

Glad you got it working! I edited my original post to correct some arithmetical errors in the calculations, so you might want to go back and re-read it. The gist of it hasn't changed however. Lower the resistor values to lower Johnson noise, by-pass the power supply rails.

Can you give us some idea of what the load is, i.e., what you are trying to DO with this circuit without violating any confidentiality?

It's for generating defects in silicon material using an electrochemical setup.

 

[hevans1944]

Would it also generate defects in or near the surface of an un-doped, semi-insulating, gallium arsenide wafer, selectively, through a photo lithographic mask? There is a device, called a GaAs PCSS (Photo-Conductive Semiconductor Switch) that can be fabricated by implanting oxygen ions in narrow, parallel, conduction channels to direct its photon-initiated avalanche conduction along parallel, non-overlapping, paths. I wonder if the same results could be obtained without the use of an expensive high-energy particle accelerator? Oxygen ion implantation to several micrometers depth apparently inhibits the photon-initiated avalanche conduction, preventing tiny branching, lightning-like filaments from forming, said filaments leading to rapid overheating and destruction after just a few discharges. A selective electrochemical process, applied through a removable mask, might achieve similar results to ion implantation at much lower cost.

There are other problems associated with fabrication of the GaAs PCSS, but removing the requirement for ion implantation would go a long way toward producing a practical, inexpensive, high-voltage (kilo-volts) high-current (kilo-amperes) optically-triggered switch with sub-nanosecond switching times. One practical application is the construction of a small high-energy, high-current, dielectric-wall proton accelerator for the treatment of surgically inaccessible cancer tumors.

 

[AdamZ]

Would it also generate defects in or near the surface of an un-doped, semi-insulating, gallium arsenide wafer, selectively, through a photo lithographic mask? There is a device, called a GaAs PCSS (Photo-Conductive Semiconductor Switch) that can be fabricated by implanting oxygen ions in narrow, parallel, conduction channels to direct its photon-initiated avalanche conduction along parallel, non-overlapping, paths. I wonder if the same results could be obtained without the use of an expensive high-energy particle accelerator? Oxygen ion implantation to several micrometers depth apparently inhibits the photon-initiated avalanche conduction, preventing tiny branching, lightning-like filaments from forming, said filaments leading to rapid overheating and destruction after just a few discharges. A selective electrochemical process, applied through a removable mask, might achieve similar results to ion implantation at much lower cost.

There are other problems associated with fabrication of the GaAs PCSS, but removing the requirement for ion implantation would go a long way toward producing a practical, inexpensive, high-voltage (kilo-volts) high-current (kilo-amperes) optically-triggered switch with sub-nanosecond switching times. One practical application is the construction of a small high-energy, high-current, dielectric-wall proton accelerator for the treatment of surgically inaccessible cancer tumors.

Perhaps take a look at the more comprehensive paper: http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0092880

 

[hevans1944]

Perhaps take a look at the more comprehensive paper: http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0092880

Very interesting way to create a nanometer sized fluid channel in a silicon nitride thin membrane. I assume you will use the Axopatch 200B to track the DNA strands as they make their way through the hole. Is it your intent to make just one hole per membrane, or is there a goal to make an array of holes?

Unfortunately, I don't think this particular technology is applicable to PCSS fabrication. <sigh>

Thanks for the link to the paper. I've always been fascinated with the electronics aspect of bio-engineering but (so far) have been unable to break into that field, even at the company where I was previously employed, where I thought I might make a contribution. Instead of hiring me, they farmed the work out to a third party to manufacture bio-sensors. Probably a good idea at the time, but I was disappointed. Now I am retired without access to all the fancy toys. I can appreciate the problem of trying to find a nanotube with an electron microscope in such a vast field. Narrowing down the search area with a mask, prior to creating the hole was clever.

Best of luck on getting your instrumentation working. The 1/f noise contribution to the "leakage" current through these holes appears to have an inverse relation to hole diameter. Any ideas on why that occurs? Does it have anything to do with the mean free path of ions in solution that find their way through the nanopore hole?