I'm trying to understand this schematic of this differential probe: enter image description here

I marked some blocks. I have some questions about the block I marked "Long-tailed pair" (LTP). I have no experience with LTPs so excuse ma noob questions.

  1. I have simulated this LTP (just the LTP without all the other blocks in the schematic) in LTspice and it's bandwidth is around 26MHz. How to increase it's bandwidth (beyond 100MHz)? What are the main limitations here in this LTP?
  2. What is the purpose of BJTs (U2B, U2C) there? The only thing I could think about it is that it offloads the JFETs from the current. Is that right? Or is there another reason for that? When I remove BJTs (U2B, U2C, and their resistors R8, R9) then bandwidth increases to 52MHz. So I'm curious what is the purpose of them, because they alone halves the bandwidth.
  3. I understand that R12 and R13 are emitter degeneration resistors (or better say "source degeneration"?). Anyway, how to determine overall gain of such long-tailed pair?
  4. What is the output resistance/impedance of this LTP? Is it R14, R16 or is it R12, R13?

These are the parts I have used in simulation (LTspice does not have the exact transistors used in that schematic):
BJT 2N2222

enter image description here enter image description here

Now as I see these charts, it is weird. I considered it is a 26MHz bandwidth, but it looks like a high-pass filter. Because lower frequencies are at -13dB, and higher frequencies are at 0dB.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ What is the design aim of the circuit? What problem is it meant to solve? \$\endgroup\$
    – Andy aka
    Commented Oct 25, 2019 at 13:44
  • \$\begingroup\$ It's a diff probe \$\endgroup\$
    – Voltage Spike
    Commented Oct 25, 2019 at 15:18
  • \$\begingroup\$ @Andyaka it's a differential probe. \$\endgroup\$ Commented Oct 25, 2019 at 17:13
  • \$\begingroup\$ No, what is its intended specific use, I didn’t ask what it is, clearly and obviously it’s a diff probe but, it has some specific impedances at the input that suggest it has been designed for a specific purpose. \$\endgroup\$
    – Andy aka
    Commented Oct 25, 2019 at 18:41
  • \$\begingroup\$ @Andyaka I'm planning to make HV diff probe (like 100MHz bandwidth). I found several such projects online. This schematic belongs to one of them. My main goal is to learn. So I would like to fully understand this schematic. And understand it's limits and possibly how to increase bandwidth of it. \$\endgroup\$ Commented Oct 27, 2019 at 8:11

2 Answers 2


The bandwith of an input stage as differential pair is strongly dependent from: 1. DEVICE: each transistor has a cutoff frequency fT=gm/2/pi/Cpi/Cmu enter image description here 2. CONFIGURATION: simple differential stage has BW dependent also from impedance of current mirror you use enter image description here enter image description here

Check this book for more details https://www.amazon.com/Analog-Circuit-Design-Discrete-Integrated/dp/0078028191

For question 2 the BJT is a Vbe multiplier in my opinion is used to keep the JFET in the linear zone, becuase if it saturates then the BW collapse (you are not anymore in the linear zone)

For question 3, check the book the gain is frequancy dependend in low freq is -gm*(Rc//ro)

For question 4, I don't see where R14, R16, R12, R13 are. the output in small signal model is the same of common emitter stage normally quite large enter image description here


It is a side topic, but you might want the 555 to run a little cooler by adding a base resistor on Q4 or maybe change Q4 to a MOSFET and do something about the 'snubber' capacitor. It needs a series resistance to avoid nasty current spikes in Q4. A properly tuned snubber will help keep the switching noise down as well. Without the resistor, the switching noise is likely worse or maybe more concentrated in one spike but either way not as quiet as it could be.

The 555 has a push-pull output stage, so it will not appreciate the base-emitter diode direct to GND, and the BJT base may not be rated for the s/c output current of the 555 either.

Better yet, the power supply is way too complicated for what it is; have a look at offerings from TI among others or even just roll your own if you are confident enough doing such.

This is probably old news by now but back on topic: the JFETs offer high input impedance and the BJTs are low output impedance. The funky current sink in the tail is using a couple of current mirrors to get a cascode of sinks and keeping both in their linear range. And as long as that is the case, the red LED is off. Curious bit of circuitry there. Not elegant but nicely done anyway.

I also noticed the input divider I think has the cap values the wrong way around. The division ratio of the capacitors should match the resistor divider. So the 6n8 / 3 = approx 2n2 and for that to have the bulk of the voltage across it the shunt capacitance must have approx 4M/26k times the series cap value which is more like 300n (or thereabouts). Caveat here is stray capacitance etc. Have a look at http://electronics-diy.com/electronic_schematic.php?id=967

  • 1
    \$\begingroup\$ Welcome to EE.SE. When your answer has a part which is on-topic I recommend putting it at the beginning and add an header for your additional hints. So all readers can get what they expect first. This is not to discourage you from adding further information but to increase readability. \$\endgroup\$
    – Ariser
    Commented Mar 12, 2022 at 14:05

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