I must confess I never tried this. But the series resistor has an important role: it is there to limit the current through the LED. If there's no resistor, the current may be limited in the end to a value which is too high for the LED or for the driver transistor. Theoretically you should graphically add the U-I characteristics of the diodes and the LED, and see on the the resulting characteristic what's the current for your Vcc value. But the main problem is that this current cannot be reliable predicted, since the U-I catalog characteristics of the diodes and LED give you an typical curve, and this curve will also shift with the temperature.
So while it may work, I wouldn't count on it working in every instance. But you may have some help from an unexpected place: the IC driving your LEDs. Sometimes the digital outputs have internal resistors or other ways to limit the output current, in order to avoid overloading them. So check the spec sheet for your attiny2313.
What is your application? From what you've mentioned so far, I'm going to go out on a limb and say that it sounds like you're trying to build a 14MHz local oscillator for a quadrature mixer/detector in the ham radio 20 meter band, so I'll address some issues specific to that application (though these issues are not exclusive to that application).
First, to answer your question, you'll want to get the 14MHz signal generated first, then apply the 90 degree phase shift. There are a couple of reasons for this, including, but not limited to:
- Once you have your in phase ("I", aka, not-phase-shifted) and quadrature ("Q", aka, 90 degree shifted) signals, you now have two separate signals, so if you want to take a 500kHz signal and use mixers to multiply it up to 14MHz, you now have to have two mixer chains, one for each signal.
- Your two mixer chains now have to be identical in every respect in order to ensure that no additional phase changes occur between your I and Q signals. Physics/murphy make this difficult.
So, it's easier and more practical to do the phase shift last. Since it sounds like you just need a fixed-frequency oscillator, this can be done pretty simply with passive components; see figure 6 (and the corresponding text) from this ARRL article.
You could also use a pair of D-type flip flops to generate your I and Q signals, which would work over a wide range of frequencies, but it requires your oscillator to be running at 14*4 = 56MHz; there's an example of this in the "Generating I/Q Quadrature Local Oscillator Signals" section of this page.
Now, even if you did one of the above and got your 14MHz I and Q signals generated, I think you would not be very happy with the result. Although you didn't mention it, from your description I'm guessing that your 500kHz op-amp oscillator is some form of RC relaxation oscillator? Those have terrible frequency stability (with respect to, say, a crystal oscillator), and to make matters worse, multiplying the signal up also multiplies the frequency instability!
Likewise, specially WRT local oscillators, phase noise is very important; multiplication of a signal will increase (worsen) the phase noise by a factor of 20*log(N), where N is the multiplication factor. Without going into too much detail on phase noise, as with most forms of noise, less is generally better.
So, given the fact that it sounds like you need a fixed-frequency oscillator on/around 14MHz, I would strongly recommend building a crystal oscillator. The Pierce oscillator design is particularly popular, as it is hard to beat in terms of simplicity and performance. You can use a jellybean transistor (e.g., 2N3904) for the amplifier, or use an inverting logic gate IC; there is a related question here which may help you get started; you can then do either a passive/active phase shift circuit, as mentioned above, to get your I and Q signals.
BTW, if my hunch about your application is correct, I suspect your comment "I'm afraid it has to be a sinusoid" is probably incorrect. Most mixers actually work better with square wave LO input, as it switches the mixer's diodes/transistors on faster.
Best Answer
I gave this picture in an answer yesterday and it seems appropriate today: -
The two graphs show the volt drop across a 1N4148 diode for different currents.
At 1 uA the diode voltage is about 275 mV.
At 10 uA it is about 380 mV.
At 100 uA it is about 500 mV.
To say a 1N4148 diode has a threshold voltage of about 0.62 is also to imply that this is at a current of 1 mA. We say this sort of thing a lot in EE and it's implied that we have in mind some sort of diode current, normally 1mA.
If you plotted these graphs on linear graph paper it would become more clear that a lot seems to be happening between 0.6 volts and 0.7 volts but, to be accurate there isn't really a threshold. Here's what I mean: -
This graph isn't related to diodes but some yearly events but, can you see that the linear graph "has a lot going on" at about 2008 to 2010.
Altogether this means that mixers using diodes can use quite small signals and not suffer excessive distortion.