0
\$\begingroup\$

So I've used a resistor and diode to level-shift from 5v to 3.3v, but that's been for general logic - the fastest thing I've done with that is 9600 bps serial.

I now have a need to feed a 3.3v powered microcontroller (an ATXMega32E5) from a 10 MHz clock signal, which I only presently have available as a 5v square wave (an output from an NB3N551 powered at 5v).

I propose to use a 10 kΩ series resistor followed by a 3.3v zener diode. Circuitlab simulation seems to show this would work, but I am not sure I trust it to be a complete enough simulation.

schematic

simulate this circuit – Schematic created using CircuitLab

Is that simple zener clipper likely to work acceptably, or is something more sophisticated called for?

\$\endgroup\$
4
  • 2
    \$\begingroup\$ No need for more sophistication. However, Zeners usually have substantial capacitance, so 10k-RC might filter quite a lot from 10 MHz clock. Use a simple resistor divider, 500/1000 Ohms or something. Or maybe nothing at all, the output voltage swing of oscillator is likely much less than 5 V. \$\endgroup\$ – Ale..chenski Feb 25 '18 at 19:55
  • \$\begingroup\$ Do you know the slew rate needed for 10MHz at say 10% ramp? Normally drivers are 25~75 Ohms not 10k. This demands knowing all component and stray capacitance. So not too good. \$\endgroup\$ – Tony Stewart EE75 Feb 25 '18 at 20:32
  • 1
    \$\begingroup\$ Why are you running NB3N551 on 5V when it runs on 3.3V which is what you need? Keep connections short or controlled impedance. \$\endgroup\$ – Tony Stewart EE75 Feb 25 '18 at 20:34
  • 1
    \$\begingroup\$ @TonyStewart.EEsince'75 The NB3N551 feeds other things that must get 5v output. \$\endgroup\$ – nsayer Feb 25 '18 at 21:15
0
\$\begingroup\$

If capacitance (track + diode+ load) is 30pF at 0V with 10k the rise time ~ 300ns in a 50 ns half cycle. Even a Rohm low capacitance zener=3pF + 3pF optimal load is 60 ns rise time. No good. Use an active level converter.

\$\endgroup\$
7
  • \$\begingroup\$ While it's true that the proposed 10K resistor will not work, it's not at all valid to leap from that to assuming that active conversion is needed. \$\endgroup\$ – Chris Stratton Feb 25 '18 at 22:01
  • \$\begingroup\$ maybe, but under 100 Ohms is needed with stray, trace and load capacitance with say 10pF and 2% slope, and power sequencing to ensure Vdd before input. An active translator gaurantees this. \$\endgroup\$ – Tony Stewart EE75 Feb 25 '18 at 22:11
  • \$\begingroup\$ I was afraid of that. I use a 74AHC1G04DBV elsewhere to perform a sine-to-square conversion (as a self-biased inverter). What do you think of using one here as a level shifter? The fact that the phase changes doesn't matter. \$\endgroup\$ – nsayer Feb 26 '18 at 1:12
  • \$\begingroup\$ I gather its close the clock and runs on 3.3V and the connection is farther away? and power sequence is guaranteed to not cause latchup on the target? You cant drive CMOS before power is applied. \$\endgroup\$ – Tony Stewart EE75 Feb 26 '18 at 5:47
  • \$\begingroup\$ All of this is inside of a ~1 sq in area. The 3.3v comes from an LDO powered from the 5v bus. The LDO spec is 50µs startup, and the clock startup is delayed way beyond that relative to power-up (which is ok because the controller ignores the external clock until way, way later). \$\endgroup\$ – nsayer Feb 26 '18 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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