Simple, Passive Interface from 1.8V Output to 3.3 volt circuit

=20

I suppose the chip maker's comment was based on your using an actual open-drain, unprotected by ESD clamps (at least to the

• rail, anyway.)

I forget all of which PICs have this feature (never heard of any of Atmel's that do), but one or two you might look at are the datasheets for the PIC16F84A and PIC18F2220, both of which limit the maximum pin voltage of RA4 to 8.5V and the PIC16F628A (for example), which limits maximum pin voltage on RA4 to 14V. There are some with even higher limits (PIC16HV540.) On those, you can just add a pull-up.

But you are using Atmel at 1.8V, so that is that. Aren't there some Atmel chips with one voltage for the cpu and another for I/O? I know they did that on some of their AT91 stuff.

=46ailing that, I would imagine a small mosfet like a BSS138 and two 10-22kOhm resistors (S and D sides.)

I take it you intend using an LCD without a controller like the HD44780.

Jon

You didn't say much about what style of LCD module that you have. If the module is a character mode device consider running it in 4-bit mode to cut down a lot on the actual number of signals that you have to interface the module. In similar manner many graphics LCD controllers support a variation of the SPI interface (Clk, Data & ChipSel) which can reduce the numner of interface lines quite a bit.

As far as the voltage level of the interface I'd suggest to stop keeping yourself in a rut with the parts that you have now. EIther bite the bullet and add the necessary buffers or get on with a display that is more system compatible with your requirements.

Do be aware that with character mode LCDs the timing needs are relaxed enough that you can often get away with building the necessary level translators out of a few resistors and some BJTs. These parts can be very cost effective but may very well take up more space on your board than a logic chip capable of performing the voltage level translation,

--

Michael Karas
Carousel Design Solutions
http://www.carousel-design.com

You MIGHT be able to get away with it (just barely) if all of the following are true:

- you can drive your output pin as either LOW (output) or HI-Z (input)

- the rated continuous current on your clamping diodes is non-zero

- the clamping diodes have a forward voltage of around 0.7V

- the VIH(min) of your LCD is 0.7 Vcc

- your 3.3V and 1.8V regulators are accurate enough If all of this is true then a simple pullup will give you 1.8+0.7=2.5 and

0.7*3.3 = 2.31 which gives you a 0.2V margin If there's nothing else using the 3.3V supply and you have an adjustable regulator and the LCD can tolerate it, you can even drop the 3.3 down by a few mV to give yourself more margin.

This is an ugly solution though, and personally I'd just add a level shifter unless I was REALLY cramped for board space.

Ask 'em what they mean. If it means some horror of a poor man's buck converter... I just wouldn't.

I didn't see a pin count - SPI or I2C kinda sounds like "two or three".

I would think an multichannel ( usually 2, 4 or 8 ) level shifter would be doable. They're like $3 for 8 channels. Is $4 "several"? They're very simple and can be deadbugged.

-- Les Cargill

Depending on the speed you need, you might be able to use common-base amps. NPN transistor, Drive the emitter, output at collector. Base resistor to 1.8V. Resistor limits the current in saturation. Maybe fets with gate at a higher voltage. Back when I was experimenting, they didn't make fets with low threshold voltage.

Place a diode in series with every data/control line to the LCD. Cathode to the driving chip, Anode to the LCD. Add a pull-up resistor (4k7) from the Anode to VCC of the LCD.

Adding a series diode will turn every push-pull output into an open drain/open collector equivalent.

Meindert

On the PIC32 they have VDD at 3v3, and 5V compatible IO. Dunno how they do it, but it you run OC and have pullups then it gets to 5V which is 1.7V above Vdd. Maybe your chip can do this too - that is run with the outputs at 1.5v above Vdd.

I assume you mean a character display not a graphics display.

At this point I am confused is the display SPI/I2C or are you talking about putting an LCD driver chip with I2C/SPI interface to drive an LCD display. So adding another chip to the design.

If the display has parallel interface like HD44780 type character LCD interface I would use a level translater chip as for that many lines (even in four bit mode) is simpler to do with less board space and components than by discretes.

If you only have two lines possibly four you could put on discretes, as others have suggested, I often use FDV303N FETs for this as these are bidirectional.

Fo more than four lines and often simpler assembly, less parts in BOM I use chips with bidirectional FET translaters which is easier and often have disable functions if both power rails not ready, which reduces risks of leakage powereing.

Most recent devices I have used are the TI TXS010x and TXB010x series. TXS for I2C/SMSbus and TXB for other types of lines.

They are CHEAP, in UK I bought 10 off of 4 and 8 bit versions for

1.51GBP (UK Pounds) or less each, these are small packages for me easier than bunches of discretes for PCB space and assembly time.

--
Paul Carpenter          | paul@pcserviceselectronics.co.uk
    PC Services
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I believe the OP stated that there are diode clamps to the micro's 1.8V supply rail -- and the usual specs state that the absolute max at the pin is to be about 0.3V above Vcc or below gnd. (There is often a max current specified to avoid latchup.)

Wouldn't the +3.3V pullup at the external diode's anode conduct via the external diode to the pin and from there through the high side protection diode to the 1.8V rail at the micro, dropping a few tenths of a volt in the external diode and a few more tenths with the protection diode? (The resistor will then limit the current.) (My LTspice simulation confirms this, but I can make mistakes.)

Were this going the other way, from 3.3V at the micro down to

1.8V at the LCD, yeah.

Jon

Are you running LCD glass directly from MCU outputs, without controller? If the LCD has to run from 3V, why can't be MCU powered from the same source?

Bootstrapping either upper or lower MCU rail; i.e. making 3V from 1.8V ?

Vladimir Vassilevsky DSP and Mixed Signal Consultant

I guess my original post was not clear. The LCD is a static or multiplexed display with as many as thirty couple segments. A multiplexed display has fewer connections, but needs four levels on the signals.

I understand how to use resistors and transistors. But there is no need. Before I would do that I would just switch to an MCU with an LCD controller built in.

It seems like I haven't forgotten anything obvious. There is no way to use a 1.8 volt clamped output to drive a 3.3 volt static LCD directly without adding at least transistors to each pin.

Rick

to

You're absolutely right.

It is however harmless to pull the micro's output pins up to 3.3V with a 4k7 ohm resistor. The output of the micro would then be at 2.1V, limited by the internal diode. The the resulting voltage at the LCD would then be 1.8 + 0.3

• 0.7 = 2.8V. Enough to be seen as a '1'. When the output drives a '0', the voltage would roughly be 0.1 +0.7 = 0.8V. Enough to be seen as a '0' by the LCD.

Meindert

Meindert Jon

No, you're not missing anything. If they were true open-drain pins then it'd be trivial, but the clamp to VCC kills it.

What you don't need to do though is spend several dollars on level translation. How about something like an LVC07 open-drain hex buffer? It'll run from 1.8V, switch over 5V happily, and Digikey's got them at

24 cents a pop at 100p quantities.

--
Rob Gaddi, Highland Technology -- www.highlandtechnology.com
Email address domain is currently out of order.  See above to fix.

You are thinking in terms of TTL inputs. This is directly driving the LCD. What matters is not even the exact voltage at the pin since there is no ground connection (or power), just the backplane and segments which are all switching and need a voltage range of close to 3 volts.

I think the guy with the MCU company was talking about how easy it is to buffer the outputs with driver chips... Not a very effective marketing tool when I can buy an MCU with integral LCD driver for around the same cost as a few buffer chips.

There are some companies that design their products for markets which have not yet evolved... and may never evolve.

Rick

Thanks for the reply. Even at $0.24 for a six pack, that's over a buck for enough to drive 32 segments and the backplane. For $2 I can just get a different MCU that has an integral LCD driver with proper I/Os. I think they even support higher voltages and multiplexed LCDs. I know the $3 one does and then I just don't need the original MCU with 1.8 volt I/Os.

Rick

e

what are the required Vh and Vl?

I'm thinking float the cpu and everything but the lcd a diode off gnd

that more or less splits the difference between 1.8 and 3.3, so you get Vl =3D ~0.7 Vh =3D~2.6

-Lasse

Chips that have the type of 5V tolerant I/O as you describe omit the diode to the + power rail in their ESD protection circuit. The lower diode in the clamp circuit to GND is designed to act as a zener in the reverse bias direction and starts to turn on at some voltage level above ~5.5 -> 6 volts. Thus the GND clamp diode provides ESD protection for either polarity of the ESD pulse.

--

Michael Karas
Carousel Design Solutions
http://www.carousel-design.com

Thanks - I sort of thought the diodes were parasitic. I guess these are pretty nifty zeners - or possibly they still use the same diodes that would have gone to Vdd, but instead take them to a common zener.