# Computing stuff tied to the physical world

## What if I mix 3.3/5.0V – part 2

In Hardware on Apr 25, 2013 at 00:01

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.

1. In the second case (resistor in series), how does the circuit “know” to reach equilibrium at 3.3V if 5V is applied?

• Ohm’s law: U = I x R, i.e. there will be U / R current through the resistor, with U the voltage difference between two ends. There’s 5V on one side and 3.3V + a diode drop on the other, i.e. roughly 4V. That determines the amount of current.

A fancy way of saying this is to also mention Kirchhoff’s voltage law: if one point is at 5V and the other at 3.3V (with a shared ground), then connecting them will by definition cause 1.7V to end up across the new connection. If the voltages change (due to a change in current load, for example), then the difference will adjust as well.