0
\$\begingroup\$

I want to know the current spectral density of a Keithley 6221 current source.

Source specification Keithley 6221 from datasheet.

The datasheet above provides values in RMS for a given Bandwidth from \$10~\text{Hz} - \text{BW}\$, not clearly specifying the bandwidth \$\text {BW}\$. I am not exactly sure what Output Response Bandwidth (BW) into Short means, but figuring from the \$\text {BW}\$ in brackets I used it as bandwidth. In a first approximation the RMS value can be calculated from the spectral density \$\text {SD}\$ by $$X_{RMS}=SD\cdot\sqrt{\text{BW}}\cdot\sqrt{\text{ENBW}}~~~~~~~~~~~(1)$$ with the equivalent noise bandwidth \$\text {ENBW}\$ (rectangular approximation valid for high bandwidths). Assuming the above values for the 100 mA range this yields $$\frac{20e^{-6}\text{A}_{\text{RMS}}}{\sqrt{1e^6~\text{Hz}}\sqrt{\text{ENBW}}}=20\frac{\text{nA}}{\sqrt{\text{Hz}}},~~~~~~~~~~~(2)$$ which seems a lot to me. Note that a \$\text{ENBW}\$ anywhere from 1 (brick wall filter) to 1.57 (1 Pole) does not change the result significantly and can thus be neglected.

Is this the correct way to determine the spectral density?


I read through the thread Determining noise spectral density, which states that the spectral density cannot be calculated due to lack of information. If the bandwidth I used is correct, this is not the case here.

Sideinformation: I feed a 3 Ohm inductance with the current source and I can measure the voltage density over the inductance with a spectrum analyzer. I would like to get the calculation and the measurement into accordance. The voltage noise can be calculated from the current noise and the resistance, once the current noise is known.

\$\endgroup\$
0
\$\begingroup\$

Spectral density is a function. You have a total integrated over a bandwidth (stated as 'typical' so it is not dependable anyway).

There is not enough information to determine the PSD function from a single number. For example, here is the typical noise of a real op-amp showing the PSD (the top one showing the nasty peak is hidden deep within the datasheet, the bottom one is what they show on the front page).

enter image description here enter image description here

You can see the 1/f noise in the lower graph, the white noise and the excess noise caused by the circuit design that causes the peak at ~400kHz.

In the case of your current source there may also be significant shot noise at low currents.

\$\endgroup\$
  • \$\begingroup\$ The datasheet above provides to RMS values, one for 0.1 Hz to 10 Hz, and one for 10 Hz to \$BW\$. I guess the reason is that the lower values shall somehow display an 1/f component, however highly undefined. I conclude, that the datasheet values are quite useless in terms of noise, my calculation is only valid for white noise and that one has to measure the device oneself, as usual. \$\endgroup\$ – Irenaius Oct 17 '18 at 11:21
0
\$\begingroup\$

What current noise will you get from a 10Meg Ohm resistor in 1MHz bandwidth?

1KOhm is 4 nanoVolts/rtHz

100Kohm is 40 nanoVolts/rtHz

10Meg Ohm is 400 nanoVolts/rthz

Current noise density is 400nanoVolts/10MegOhm = 400 e-9/1e+7 = 400 e-16 or 40 e-15 or 40 femtoAmps/rtHz.

Scale up by sqrt(1 MHz), and find 40fA * 1000 = 40 picoAmps.

But............this is the current noise for a single resistor. That Keithley has numerous circuits involved in implementing a programmable current source.

\$\endgroup\$
  • \$\begingroup\$ It can be seen here that the current noise from resistors is a lot lower, which makes sense, because the current noise in the keithley includes a lot more sources, which also depends on the output signal. However, this does not answer my question. \$\endgroup\$ – Irenaius Apr 24 '17 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.