6
\$\begingroup\$

Let me start out saying I'm not trying to build anything specific or useful here, just trying to learn the behavior of circuits.

Is it possible to turn a 0 to 5V square wave pulse into a -500mV to +500mV pulse using just passive components and where the only voltage source available is 5V?

What I'm aiming for is using some resistors and capacitors (and inductors, if necessary) where there's a capacitor divider network where half the square wave ends up below ground potential (presumably due to a discharging capacitor or two).

It's easy enough for me to use two capacitors to create a divider network to get a 0 to 500mv or 0 to 1V square wave, but what I've failed at is being able to bring the square wave below ground potential in LTSpice.

I know it's possible to simply change the voltage source parameters in LTSpice to pulse from -500mV to +500mV. That is NOT what I'm looking for. I want to know how to do it by adding components to a 0V to 5V pulsing voltage source.

\$\endgroup\$
2
  • \$\begingroup\$ Is the duty cycle 50%? If it's not 50%, is it at least constant? \$\endgroup\$
    – The Photon
    Commented Jan 16, 2018 at 4:13
  • \$\begingroup\$ Yes, the pulse generator duty cycle is 50%. \$\endgroup\$
    – acker9
    Commented Jan 16, 2018 at 4:29

3 Answers 3

Reset to default
10
\$\begingroup\$

You can use a 5:1 resistive divider followed by a DC blocking capacitor. You are going to have to use a large enough capacitor to get the RC time constant well below the fundamental frequency of the square wave.

schematic

simulate this circuit – Schematic created using CircuitLab

This can be simulated to see the various waveforms and experiment with the values and frequencies.

\$\endgroup\$
6
  • \$\begingroup\$ I wasn't successful in LTSpice with this circuit. Resulting waveform is 0v to approx 1V. Waveform image. \$\endgroup\$
    – acker9
    Commented Jan 16, 2018 at 5:58
  • 3
    \$\begingroup\$ Let the simulation run for much longer. \$\endgroup\$
    – vofa
    Commented Jan 16, 2018 at 6:26
  • 1
    \$\begingroup\$ I figured out what I was did wrong in my first attempt to duplicate your circuit in LTSpice. I didn't run the simulation long enough. It looks like this circuit doesn't "stabilize" until approximately 6 seconds out! After that, it stabilizes to -500mv to +500mv. I'd appreciate you or any others adding what the math is to figure out the "stable" time (or "settling time", if that's a better term) for the above circuit. Successful waveform image \$\endgroup\$
    – acker9
    Commented Jan 16, 2018 at 6:29
  • 1
    \$\begingroup\$ @acker9 To understand the behaviour of that circuit mathematically you need either to know linear differential equations theory or Laplace transformation techniques. If you can understand either one, post a new question, linking to this answer for reference in order not to be mistaken as an homework question (which will be shut down quickly), you might get an answer. Note: on this site we try to keep the threads well focused on one single question, that's why I told you to post another question. \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Commented Jan 16, 2018 at 6:35
  • 1
    \$\begingroup\$ You can also set the initial voltage of the capacitor to -0.5V, to avoid waiting for it to settle at that voltage in the simulation \$\endgroup\$
    – Pelle
    Commented Jan 16, 2018 at 8:22
0
\$\begingroup\$

Transistor divider

The primary is connected between your signal and 2.5V, seeing a signal of +/-2.5V. Through transformation to the secondary, this signal is divided by 5, resulting in a signal of +/-500mV.

How do you get the 2.5V DC? If you have 5V DC available, you can just use a resistive divider and buffer it with a capacitor (top circuit). Alternatively, if your input signal is exactly 50% duty cycle, you can just average it (with a resistive divider and smoothing capacitor).

There are two limitations: 1: both circuits require some settling time dictated by R and C 2: the transformer has to be suited for the pulse frequency.

\$\endgroup\$
1
  • \$\begingroup\$ While it does answer OP's question, I doubt he means to complicate with a ferrite transformer, or with the fairly large currents. It wasn't me who downvoted, though. \$\endgroup\$
    – a concerned citizen
    Commented Jan 17, 2018 at 8:45
0
\$\begingroup\$

It simply is not possible (but with one small exception, described later).

The reason for that is the type of signal you want to transform. A square wave consists of very sharp edges which cause the spectrum to spread to infinite frequencies.

The passive networks posted by @Dean Franks and @bla add low-pass behaviour. This is unavoidable because of the dependencies of capacitors and inductors.

To be less theoretical: To have a (almost) perfect square pulse with (almost) perfect edges you need to charge the capacitors or inductors with very high currents or high voltages, which are in fact limited on your passive network.

If your requirements does not force you to have an (almost) perfect square pulse on your output, the provided solutions are quite fine.

To come back to exception: You can get rid of your storages (capacitors and inductors) if your application allows you to have different voltage definitions and electrical potentials. In that case, your output ground (0V) must not be same as the input ground and you can stick by an simple voltage divider. (Taking the schematic from @Dean Franks and remove C1)

PS: That has been my first answer on stack exchange. I hope to give you and all the readers a meaning full reply to the op's question and i am happy to be a part of that greate knowledge-community.

\$\endgroup\$
3
  • \$\begingroup\$ Since it's your first try, I'll go easy on you and not downvote. See Dean Franks answer. Capacitive coupling will do the job just fine. \$\endgroup\$
    – WhatRoughBeast
    Commented Jan 16, 2018 at 18:11
  • \$\begingroup\$ Capacitive coupling cannot produce a square wave! (As you can see in the ltpice simulation) That is why I expllicity try to distinguish between “a not-perfect square wave” and an “(almost) perfect square wave” in dependency to what your requirements needs. \$\endgroup\$
    – Finkman
    Commented Jan 16, 2018 at 19:33
  • \$\begingroup\$ @Finkman I think your reply adds value to the knowledge base and I appreciate that you have decided to participate. For my purposes, perfection of the square wave isn't a requirement. \$\endgroup\$
    – acker9
    Commented Jan 17, 2018 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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