Problem: I haven't find yet an analog multiplexer fast enough with a small parasitic resistance. Maybe the discrete MOSFET solution would be enough
3. use the clock signal to drive a switching circuit, built from MOSFETs or an analog multiplexer. The switching circuit would "chop" a continuous voltage (which is easy to adjust) thus the output level would be adjustable. Problem: I haven't find yet an analog multiplexer fast enough with a small parasitic resistance. Maybe the discrete MOSFET solution would be enough (did some quick simulation tests with a 2N7002A with promising results ... but I need to have a P-channel complementary to have a true switch).
How do you implement variable level with diode clipping?
You could get the continuously variable output level with a logic gate (buffer or inverter) supplied from a variable voltage.
To drive transmission lines, you want a good match to 50R to make the square edges look clean.
This can be done by always having some minimal attenuation to improve match.
Instead of MOSFETs and voltage switching, use bipolar transistors and current switching, or fast diodes and current switching. The output level is then set by the easy to control DC current source driving the emitters of the differential pair, or driving the diodes.
What's that AQY2B25X part? Is there a typo? Google returns literally nothing when doing exact match search.
(it's clear that it's some optical switch, I wanted to find its datasheet)
3. use the clock signal to drive a switching circuit, built from MOSFETs or an analog multiplexer. The switching circuit would "chop" a continuous voltage (which is easy to adjust) thus the output level would be adjustable. Problem: I haven't find yet an analog multiplexer fast enough with a small parasitic resistance.
Finally, here are the specifications I try to achieve:
- output frequency: 10 kHz (if needed I'm ready to start at 100kHz or even 1MHz) to 100 MHz (I can accept to go a bit lower, such as 80 or even 60 MHz if the signal gets to distorted).
Instead of MOSFETs and voltage switching, use bipolar transistors and current switching, or fast diodes and current switching. The output level is then set by the easy to control DC current source driving the emitters of the differential pair, or driving the diodes.
That's an interesting approach, it also looks like it was used in some pulse generators such as the Philips PM5715, which I borrowed the idea of chained attenuators in the output circuit. I may try it too.
Tektronix used it on most of their pulse generators. They used the diode version for vertical channel switching in oscilloscopes which gives some idea of the potential performance.
Right now I'm still on the analog switch method (schematic attached) using SN74LVC1G3157 but I may look at the references suggested by PCB.Wiz.
By the way, will the switch mess up the op-amps output or something like that?
Right now I'm still on the analog switch method (schematic attached) using SN74LVC1G3157 but I may look at the references suggested by PCB.Wiz. By the way, will the switch mess up the op-amps output or something like that?
It doesn't look like it's worth using better op-amps to perform this function, so I'm gonna dig further the idea of the simple discrete BJT output circuit. I don't want my output level range ot be wide nor accurate; I'll use a more sophisticated pulse generator (I do have one already) driven by this circuit if I have special needs anyway. If I can achieve some control between ~1V up to 5V, mostly to cover most basic needs in the lab. I could maybe even ditch the DAC and use a more minimalistic approach, but this will be determined later.
I don't know much about LDOs and I'm kind of concerned about their output impedance/capacitance at high frequencies.
Since you want a square-wave output, is there any reason not to use the NL27WZ17, as I showed a few posts above, or the similar-performance SN74LVC2G04 Dual Inverter Gate? Feed the Vdd pin from a variable or selectable voltage (1.8V to 5V) and you get the square-wave amplitude you want. Add a series resistor to establish the 50 Ohm output impedance (I paralleled two buffers and output resistors to reduce the effect that any gate impedance variability had on output Z. I assume that the gate output Z will vary somewhat with Vdd, but unless you need high precision a compromise resistor value should work.
If you are driving the gate directly from the 3.3V Si5351 output you will want to add a buffer / translator to jack that signal level up to +5V before driving the NL27WZ17 (if that will be operating at +5V). The NL27WZ17 can accept +5V logic-level inputs even when running at Vdd = 1.6V.
Since you want a square-wave output, is there any reason not to use the NL27WZ17, as I showed a few posts above, or the similar-performance SN74LVC2G04 Dual Inverter Gate? Feed the Vdd pin from a variable or selectable voltage (1.8V to 5V) and you get the square-wave amplitude you want.
I'm not a fan of this method because I don't find it really proper. Output impedance and slew rates tend to vary quite drastically when changing the supply voltage of these chips as you said.