0%
0%

# Low voltage differential probe

200 MHz differential probe

Similar projects worth following
74 views
0 followers
In this project, I am aiming to create a 200 MHz, x10 differential probe. I am basically a newbie in electronics so I do not have much experience in developing such circuits.

My objective is to mainly learn a lot during the development of this project and if finally a decent differential probe is built, even better!

I'll write like “lecture” logs of each building block of the probe: their purpose and how they work, so I can understand better what I'm doing.

These are the other projects (thanks!!) that I'll use to start:

— Craig D's project: https://hackaday.io/project/191837-pd150 (awesome project!)
— Bud Bennett's project: https://hackaday.io/project/169390-a-10x-100mhz-differential-probe
— Christoph's project (based on Bud's): https://hackaday.io/project/175351-a-1x-100-mhz-35-v-common-mode-diff-probe
— Petteri Aimonen's project: https://hackaday.io/project/181065-modular-differential-probe

• ### 1st building block: input attenuation

Miquel Tutusaus06/29/2024 at 11:36 0 comments

# Passive probes input attenuation

Typical passive probes have a switch that lets you change between x1 or x10 attenuation. Dave from EEVblog has an amazing YouTube video explaining how these probes are built and how the attenuation is obtained:

My oscilloscope has a 1 MΩ resistor with an 18 pF capacitor in parallel as input impedance, assuming that the passive probe has a 9 MΩ resistor in series (as shown in Dave's video), the equivalent circuit would be:

C5 is a variable capacitor (the flat screw seen on the probe) which is used to compensate this voltage divider, why?

Before answering this question, let's take a look at the circuit above. Basically is a resistive voltage divider with two capacitors in parallel. We know how a resistive voltage divider works, so why do we need these capacitors?

A resistive voltage divider works perfectly well at DC, but as we increase the frequency, the parasitic capacitance of these resistors will affect the response of the circuit, let's see an example:

This is the frequency response of an ideal resistive voltage divider (with no parasitic capacitance). With R5 equal to 9 MΩ and R6 to 1 MΩ, the attenuation is 0.1 V/V (or -20 dB). Now let's see the difference when we add a 5 pF parasitic capacitance on both resistors:

As seen, the frequency response has changed drastically to a perfectly constant attenuation for all the frequency range to a change of the attenuation from -20 dB to -6 dB. This is not desired for an oscilloscope probe, as we do not want to modify the signal that we are measuring, so how can we solve this problem? With a frequency compensated voltage divider!

This Analog wiki article explains really well this type of voltage dividers (Analog Wiki link). Basically, a frequency compensated voltage divider is the circuit we have presented before, a resistive voltage divider with the capacitors in parallel, so how can we achieve this frequency compensation?

"The compensated divider employs pole-zero cancellation to suppress undesired frequency dependence caused by any stray capacitance on the output side of the network. If the resistor and capacitor values are adjusted so that the pole and the zero of H(s) are superimposed, |H(jω)| becomes independent of frequency." (from Analog Wiki)

Basically, we have to fulfill that C6/C5 = R5/R6, easy! Let's try it, with R5 = 9 MΩ, R6 = 1 MΩ and C6 = 18 pF, C5 should be equal to 2 pF:

By using a frequency compensated voltage divider, we can obtain a constant attenuation regardless of the frequency of the input signal, nice!

"C5 is a variable capacitor (the flat screw seen on the probe) which is used to compensate this voltage divider, why?"

Basically, its function is to tune the value of the capacitor so the ratio C6/C5 = R5/R6 is fulfilled.

# Differential probes input attenuation

Now, let's apply this concept to a differential probe. We know that we have to use a frequency compensated voltage divider to attenuate the differential input voltage of the probe, but why do we need to attenuate the signal? Can we just subtract both inputs and feed this voltage to the oscilloscope? Well, yes and no.

• Let's say that we want to measure the differential voltage of a shunt resistor used to measure current. The typical differential voltage swing across the shunt would be less than 1 V, which is inside the (typical) allowed voltage range of broadband difference amplifiers. Therefore, no attenuation is needed!
• Or let's say that we want to measure the Drain-Source voltage of a power MOSFET switching inside a power converter, with a typical differential voltage swing of 10 V. This voltage swing is much higher than the (typical) allowed voltage range of broadband difference amplifiers. Therefore, attenuation is needed!

So, to be able to use this differential probe in both cases, we will add a -20 dB (x10) attenuation input stage. The basic circuit...

Read more »

Share

## Similar Projects

Project Owner Contributor

### Vacuum Tube Synth Filter

256byteram

Project Owner Contributor

### Open Source Bench Top Power Supply

acp.calin

Project Owner Contributor

### DIY Smartphone Charger

Naman Chauhan

Project Owner Contributor

Craig D

# Does this project spark your interest?

Become a member to follow this project and never miss any updates