You can omit the resistor on the input side, since the source is push-pull.
However, you still would need a pull-up resistor on the output. The way the circuit works is that if one side is low, it pulls the other side low. On the other hand, if the input is high, it turns off the transistor, and the two sides are isolated from each other (hence the requirement for the pull-up resistor).
You still have to worry about RC delays when the device is transitioning from low to high. The R would be the pull-up resistor value and the C would be the input capacitance of your IC, capacitance of your MOSFET (C_oss? I forget exactly which one it is....), and board parasitics.
You should try simulating the circuit using spice (LT-Spice, for example) to see how quickly it responds. The flip-flop's input is around 4.5 pF, and you can check your microcontroller's datasheet for its input (likely in the 5 to 20 pF range).
If you need something faster, you could reduce the pull-up resistor value, though this will use more power.
You couldn't directly use a pair of diodes in series between the chips because that wouldn't let the source pull down the load. You could use two diodes with a pull-down resistor, though this also runs into RC delay issues.
One good (wastes a little power) solution would be to use a voltage divider to divide the input down to 3.3 V.
If you care about power consumption, you need a CMOS inverter. One example would be the 74LVT04, where you'd use two inverters in series (so that your signal isn't inverted). Don't forget to ground all the unused inputs. If it's OK to invert the signal, you could use the SN74LV1T04, which only contains a single inverter.
C
OSS = 7 pF @ 1MHz, V
DS = 10V and V
GS = 0V
C
F-F = 4.5 pF
microcontroller capacitance is not supplied, assume 20 pF
PCB capacitance ~3.24 pF based on existing design, but is unlikely to change significantly
C
tot = 34.74 pF
R = 10 kOhm
Tau = 3.4E-07 s
however, Tau is only 63% charge, which is below the threshold, 2*Tau is going to be well above the trigger threshold
f
MAX=1/(2*Tau) = 1/6.894e-7 = 1.45 MHz
I'll have to see if I can find a SPICE app to test it, but my calculations look like it's going to be ok, do you agree?
Without knowing the latch-up current for the MCU, if I selected a 100kOhm resistor I'd have a current of 17µA, which is greater than the combined input leakage currents and (presumably) less than the latch-up current.
Does that seem right?
I didn't check the math, but the numbers seem reasonable.
You still need to worry about RC delays. If you use 100k series, and have a load of 25pF, that's a f_3dB frequency of 80 kHz, and you wouldn't want to run any digital signal more than 1/5 of that... so you'd be limited to perhaps 20 kHz digital signals.
Ok, so maybe I have made a mistake in the above calculation, or there's a gap in my knowledge, where does the f_3dB come from?
EDIT:oh, ok, you're talking about 100k ohm resistor in series without MOSFET, which would make it:
C
F-F = 4.5 pF
microcontroller capacitance is not supplied, assume 20 pF
PCB capacitance ~3.24 pF based on existing design, but is unlikely to change significantly
C
tot = 27.74 pF
R = 100 kOhm
Tau = 2.774E-6 s = 2.77µs
f
MAX=1/(2*Tau) = 180kHz so too slow.
EDIT 2:Ok, so if I lower the series resistance to 1kOhm I'd have a frequency of 18MHz, divide by 5 and I can run up to 3.6MHz with a maximum current of 1.7 mA? I'm not so sure I've calculated this correctly.