# driving analog frequency through a digital component

Discussion in 'General Electronics Discussion' started by ami85t, May 25, 2014.

1. ### ami85t

71
0
Feb 19, 2014
Hi to all,
I have a technical question which is actually a theoretical one:
What's the result of driving an analog frequency through a digital component, For example - trying to feed a sinusoidal wave through a shift register and receiving a digital frequency at the output?
I guess the answer depends on the characteristics of the analog signal and the digital component - accuracy, amplitude rate and the amount of time that the analog wave exists at a logic state (There's a term for that I think...).
Thinking about it theoretically my answer would be, at least for the specific case, why not? If the amplitude rate of the
analog wave is included in the '1' logic rate of the shift register so the frequency of the digital output will be the same as the

Thanks,
Amitai

Last edited by a moderator: May 25, 2014
2. ### duke37

5,364
772
Jan 9, 2011
Most digital circuits use edge triggering and need a sharp change of voltage on their input.
If you use an analog signal, then nothing may happen or the circuit may go bananas. To avoid this the inputs of a slowly changing signal are passed to a schmitt trigger which rapidly changes from one level to the other, so generating a digital signal.

KrisBlueNZ and ami85t like this.
3. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Apart from the problem of feeding an analogue signal into a clock input, the general answer is that digital inputs have a threshold voltage. When the applied voltage is above the threshold, it is regarded as high; when it's below the threshold voltage, it's regarded as low.

The threshold is often somewhere between 1/3 VCC and 2/3 VCC but it is not tightly controlled, and when the input voltage is near the threshold, the IC may not interpret it in a clean and consistent way. This can also cause extra current consumption in the device, depending on how the input stage is implemented.

Digital devices are supposed to be driven with signals that swing quickly and cleanly between low and high; that is, voltages cross through the no-man's-land between 1/3 VCC and 2/3 VCC (or whatever the thresholds are for the particular logic family you're talking about) very quickly and cleanly, minimising the uncertainty that occurs during the transitions.

In applications where you have a slowly changing voltage and you want to convert it into a digital signal, you can use a Schmitt trigger. This circuit has a characteristic called hysteresis which eliminates the uncertainty around the switching point. Look it up on Google or Wikipedia if you're interested.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
Or more importantly, in this application, by defining the behaviour between VInput HIGH(min) and VInput LOW(max)

5. ### LvW

604
146
Apr 12, 2014
An interesting exception is the CMOS inverter which allows analog (linear) amplification.
The reason is: If biased at Vcc/2 the output is at app. Vcc/2. For this purpose, a (high-valued) resistor beween output and input is sufficient.
This property allows to use the quasi-linear part of the transfer characteristic (between "low" and "high") for amplifying an analog input.
This application illustrates that - in principle - it is problematic to speak about "digital components".
For my opinion, it is more correct to say: For digital applications, there are components which are operated exclusively between two fixed states only.

Arouse1973 likes this.

5,164
1,087
Dec 18, 2013
All signals are analog. As Bob Dobkin (CTO Linear Tech) said digital is just fast analog. Apart from certain micro processor inputs, is there a minimum time for a digital gate to switch?

7. ### BobK

7,682
1,688
Jan 5, 2010
Analog and digital are not properties of a signal. They are two different means of encoding information into a signal. A signal might be a voltage, a current, a frequency, a phase difference, a duty cycle and probably others I cannot think of right now. Any of them can carry either analog or digital information. The difference between analog encoded information and digital encoded information is that the former is continuous, i.e. any change in the signal changes the information it carries, the latter is discrete, all signals within a specified range carry the same value and there are a finite, usually small number of valid ranges (often 2.)

Note that by my definition a PWM signal is naturally analog since it can vary continuously, though in most implementations it is digital because it is restricted to discrete steps.

Bob

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,839
Jan 21, 2010
Whilst the above is all good and true, "digital" circuits are often specced in such a way as to minimise the amount of time
spent in the (designer's) no-man's land between logic levels.

We're all probably aware of the increase in current draw of logic with increasing frequency. [For the rest of this I'm talking about CMOS.] For the most part that gets described in terms of the current flowing into or out of gate capacitances (or at least that's the initial explanation given to students). However, in fact, a significant portion is caused by crowbar currents as both mosfets pass through their linear regions.

LvW's example of a CMOS inverter biased for linear operation is a great example of this. I recall seeing this in a National Semiconductor datasheet at about the time when most of the 4000 series datasheets were stamped "preliminary". I'm not sure how much earlier this circuit was devised, or if it was an independent thought or the result of testing the devices to characterise them. Nevertheless, it's a great example, and for a whole lot of reasons.

If we step away from this for a moment, and look at the design goal of minimizing power consumption is a microcontroller circuit, we find a number of mechanisms. If we ignore the load switched by the device, the next most important item is possibly clock speed. However, as you work through the items, one is perhaps surprising. That one is that you should keep the inputs close to the supply rails.

It is at this point we return to LvW's example. When configured as an amplifier, a CMOS inverter is operating almost as a class A amplifier (perhaps one could argue AB with lots of bias. The current consumption is many times higher than for the inverter operating with an input that is in spec for the logic family.

So that's one thing. It also might be worth looking up meta-stable states, because you can get yourself there too.

Arouse1973 likes this.
9. ### ami85t

71
0
Feb 19, 2014
Thanks to all,
I didn't understand deeply all that you have all brought but the mean idea
I have understood.
Amitai