Welcome to the weekly What-If series, also available via the Café wiki.

This question came from figuring out how to drive an SPI chip over a few meters distance. Normally, you’re not supposed to do that: I2C and SPI in particular are designed for use on a PCB, i.e. distances in the order of dozens of centimeters at best. Fortunately, in this case speed was not an issue, so what if we simply take the clock signal way down, to say 10 KHz (50 µs on + 50 µs off). Can we create a solid SPI setup across a few meters of wiring?

To try this out, I decided to take a reel with some 70 to 80 meters of multi-core wire:

Being untwisted and unshielded, this is probably as bad as it gets for signalling purposes. Then I started with a clean 10 KHz square wave of 0..5V from the signal generator:

Here’s what happens to the signal on the signal generator side (i.e. same probe placement), when simply connecting the open wires of the above cable to signal and ground:

This is caused by the load and the reflection of the signal on the end of that long cable. Think of it as an echo of that voltage change, propagating back and forth and colliding with the signal – causing this fairly substantial change in signal rise time and shape..

But to really see the effects, we have to also look at the signal at the other end, where it’s going to be picked up and used. I made the following setup, with three different probes:

(sorry for the confusion: “B” and “M” were mixed up in this drawing)

• YELLOW probe on the incoming signal generator end
• MAGENTA probe on the outgoing signal
• BLUE probe on the outgoing ground connection
• RED is the MAGENTA minus the BLUE level (as math function)

All the vertical scales are set to the same 1V per division from now on:

Note that there are some really strange effects in there: the magenta output signal is “ringing”, as is to be expected with such a long coil of wire and lots of stray capacitance, inductance, and resistance. But the blue “ground” level is also ringing, almost as much in fact! And the yellow input is funny, too. Let’s zoom in a bit on that rising edge:

What does this mean for signal propagation? Well, as you can see, everything rattles and shakes in situations like these. It really makes a difference for example as to how the end is connected to power. If you use the (blue) output ground as reference, then the signal will appear as the difference (i.e. the red line), which has less extreme ringing than if you use the (magenta) output referenced to the incoming ground pin.

None of these effects are problematic at this 10 KHz signalling rate. If you need to pass signals like these, it definitely pays to A) keep the signalling rate low, B) reduce the steepness of the edges, and C) add a fairly substantial load resistance at the end of the wire (between signal and ground, say 330..1000 Ω). The reasons and details of all this will have to wait for some future posts.

What I do want to mention, is that you can actually see the speed of light in action here. Let’s zoom even further into that rising edge (sorry, blue and magenta are reversed now):

It takes over 400 ns for the yellow rising edge to reach the other end. In vacuum, light travels some 120 meter in that amount of time – not so very different from the 80 meter or so of cable the electricity has to traverse. Pretty amazing what you can do these days, with nothing more than a signal generator, a modern oscilloscope, and a coil of wire!

Welcome to the weekly What-If series, also available via the Café wiki.

There’s still much more to say about all this – as can be expected. One of the suggestions made in the comments was to use a few diodes in forward direction. Since these each have about 0.65V drop, three of them ought to bring down the voltage from 5.0V to 3.3V.

Time to hook up the signal generator and scope again. Be prepared for some surprises!

As I mentioned in my original post, the “official” way to handle this, is to use a “level converter” chip, which is based on some active circuitry, i.e. some transistors or MOSFETs. The resistor solutions described yesterday are less accurate, as you’ll see…

Here is a 1 MHz square wave @ 5V (yellow) feeding yesterday’s 4.7 kΩ + 10 kΩ resistive voltage divider to produce a 3.3V signal (purple):

That’s quite a different signal coming out! A typical capacitive charge / low-pass effect.

Note that the signal generator has a ≈ 10 nS rise-time, i.e. the edges are not completely vertical, and that the Hameg scope probe has a 14 pF loading capacitance, according to the specs. So some of these effects are artefacts of this measurement setup.

Let’s raise the frequency to 10 MHz (the horizontal scale is now a very fast 10 ns/div):

Hardly a square wave, and no well-defined 0-to-3.3V transitions either, as you can see.

Now let’s try this circuit:

The reasoning being that the diodes will “drop” the voltage from its high 5V level to 3.3V:

Quite an asymmetric effect, although lowering the resistor to 1 kΩ ought to improve it.

The above is again a 1 MHz square wave input, and here’s the same at 10 MHz:

This final setup is so far off the desired voltage levels that it probably won’t work.

Given these outcomes, I’m inclined to stick with a single 10 kΩ resistor in series. Or perhaps drop it to 1 kΩ to get better rise and fall times. Active MOSFET-based level-shifting circuits are starting to make a lot more sense now – they exist for a reason!

Welcome to the new weekly What-If series, also available via the Café wiki.

Yesterday’s post was about mixing 3.3V logic levels with 5V logic levels.

The real trouble, apart from things not working, is damaging some component, of course. And in electronics, almost all damage comes from overheating. At some point (well over 100 °C, usually), chips really do get “fried” and irrevocably damaged occasionally.

The trick is to avoid overheating. This in often easy to do by limiting the current that can flow. How do you limit the current? Simple: make sure there are some resistors in series!

But first, let’s examine what’s really going on when we connect a 5V logic “1” to an input which handles only 3.3V logic levels:

On the left, the typical circuit inside a digital chip at each I/O pin. The essential ingredients are two ESD protection diodes to VCC and GND, respectively, and the MOSFET input gate.

The diodes are placed in the “blocking” direction, i.e. they normally do nothing but block the current. The MOSFET gate is a very high impedance input, essentially it acts just like a capacitor, in fact. The rest of the input circuitry can be ignored for our purposes.

So normally, input signals on such a pin just “float” and follow whatever voltage is applied. Until the voltage is too high, or too low, that is.

On the right, I’ve redrawn the exact same circuit, but “lifting” the input well above VCC level. To make this clear, I’ve drawn the input pin above the VCC pin, as an intuitive way to represent voltage as height in the diagram.

Now you can see what happens above VCC + 0.65 or so: the top diode will start to conduct in this situation. So if the input is set to 5V, and VCC is 3.3V, then the diode will become a conductor and try to pull either 5V down or 3.3V up, to reach equilibrium again.

This is a danger sign. The amount of current flowing will rise as long as these voltages are more than about 0.65 V apart. If both power supplies are very strong, there could be several amps of current – the 5V supply could even start powering the 3.3V circuit!

And that’s where the damage-through-heat comes in: a 1 A current over a diode with a 0.65 V difference leads to 0.65 W of heat being dissipated. Far more than these tiny on-chip diodes can handle. In fact, they really are only designed to carry up to a milliamp or so. The result: a “blown” on-chip diode, and since the voltage difference continues to exist, the damage will probably propagate to other diodes and components on the chip.

There are a few ways to prevent this. One is to use special “level-converter” chips, designed to take one voltage on the input, and then generate a different voltage on the output.

But there are much simpler ways to deal with such small differences, especially when the pins are all close together and on the same board:

On the left, a traditional 4.7 kΩ + 10 kΩ resistive “voltage divider”, which takes one voltage and divides it down to a lower voltage. This works, but in this case we can even get by with a single resistor, as shown on the right.

The reason the single resistor works, is that current will start to flow through the diode as before, but now also through the 10 kΩ resistor. Since the circuit seeks equilibrium, in this case, there will be about 5.0 – 3.3 = 1.7 V across resistor + diode, i.e. ≈ 1 V across the resistor. With a 10 kΩ resistor, it only takes 0.1 mA of current to generate a corresponding voltage drop, so that’s when equilibrium will be reached. These current levels are completely harmless and can easily be sustained by the on-chip diode.

So there you have it: we’ve explained why direct connections might lead to overheating, and how a 1..10 kΩ resistor in series can prevent it, while still allowing the circuit to work.

All this required, was some theory and a basic understanding of the internal circuitry.

PS. These resistor solutions are sensitive to noise and capacitive loading (they act as low-pass filters), so this only works well when signal lines are short, a few cm or so. For reliable high-speed signaling over longer distances, a level-shifter chip would be a better way to go.

Welcome to the new weekly What-If series, also available via the Café wiki.

As you probably know, mixing 3.3V logic with 5V logic is usually a bad idea, but why and what if you need to do it anyway?

This topic is too complex for a single post, but let’s just start and see where it leads to:

• VCC in these examples is the supply voltage, i.e. either 3.3V or 5.0V, depending on which chip we’re looking at. The goal being to connect a mix of these.

• Digital I/O pins work by treating everything below a certain voltage as “0” and everything above another voltage as “1”. In between, the levels are undefined and could be interpreted either way, depending on temperature, stray capacitances, moon phases, who knows…

• For the ATmega328, for example, everything between -0.5V and 0.3 x VCC is treated as “0” and everything between 0.6 x VCC and VCC+0.5V as “1”.

• For 5V signals, that translates to: under 1.5V is “0” and over 3.0V is “1”, respectively.

• Also relevant, is that unloaded output pins tend to be very close to the 0V and VCC ground and power supply levels, respectively.

So mixing is in fact not a problem at all for the following scenario: 3.3V levels for output signals, tied to 5V levels for the input signals. If all you need, is to read logic levels on say a 5V Arduino Uno from a 3.3V JeeNode, then just tie the signals together and you’re all set.

Connecting a 3.3V level output pin to a 5V level input pin works fine.

The problem occurs in the other direction: a “1” output on a 5V logic level is about 5V, whereas the maximum allowed input level for a “1” on a chip powered by 3.3V is 3.3+0.5, i.e. at most 3.8V.

Hooking these together without taking care of the problem will lead to problems, such as overheating chips and even smoke or fire (although this is almost impossible with simple battery- or USB-powered hookups).

Tomorrow I’ll describe the cause of these problems, along with some simple solutions.