mauifan wrote:This isn't about crystal oscillators, but it does talk about op amps in a somewhat intuitive way. In particular, Pete mentions that the input impedance of real op amps is on the order of 100-1000K.
Hmm.. that seems a bit low. I generally figure op amp impedance based on the input bias current.
For bipolar op amps (like the LM358), bias currents are usually 50-100nA (10-20M). For JFET amps like the TL081, the typical bias current is 30pA (30 gigohms).
mauifan wrote:That suggests to me that if the impedance feeding into the inputs was on the order of 10k or less, the op amp would behave more or less like the ideal op amp.
Not exactly.. the 100k-1M value is what you'd see if you put a resistor in series with the input terminal. That doesn't account for feedback though, and you always use op amps with feedback.
The official formula for input impedance through an amplifier is Rin * (1 + AB) where 'A' is the open loop gain, 'B' is the amount of feedback, and 'AB' is the closed loop gain. The basic idea is that your feedback network supplies some of the current that goes through Rin. The higher your loop gain gets, the more current your feedback network supplies.
Rin -- the 100k-1M value mentioned above -- does appear in that formula. It just gets multiplied as your gain increases.
In addition, input currents don't actually change the linearity or gain of your amplifier. They show up as voltage offsets at the output.
Let's take a simple voltage buffer as a first example, and assume we know the impedance of the negative input is 100k. To compensate for that, we use a 100k resistor for negative feedback. With no other inputs, the circuit will balance when the current through the feedback resistor equals the current that flows into IN-. Let's call that 1uA (which is high compared to the values in most datasheets). Under those conditions, the voltage across the feedback resistor will be 100mV.
That's an output offset error. For an ideal op amp Vout would be exactly the same as Vin-.
Now let's turn that into an inverting amplifier by connecting another 100k resistor to IN-. If we set the signal voltage exactly the same as Vin-, no current will flow through the input resistor and the circuit will be identical to the buffer above. If we raise the signal voltage 100mV above Vin-, all the bias current comes from the input resistor and none can flow through the feedback resistor. That sets Vout = Vin-, 100mV below where it was before. Rasing the signal 100mV lowered the output 100mV, exactly what you'd expect from a 'gain of -1' amplifier.
As far as the signal source is concered, it sees 100k of impedance from the input resistor. It also sees a parallel combination of Rin and the feedback resistor, but the voltage at the far end of the feedback resistor changes. The output offset error supplies all the input bias current, and the only other path for current to leave IN- is through the feedback resistor. That means the signal sees a total impedance of 200k.
If we take the input and feedback resistors up to 1M, the 'gain of -1' behavior stays exactly the same but the output offset error rises to 1v. The error still supplies all the bias current, so the 'only way out is through the feedback resistor' rule still applies, and the input impedance is 2M.
If we drop the resistors to 10k, the output offset error drops to 10mV, but the input impedance drops to 20k.
Sometimes it's okay to accept a higher offset error in order to get higher input impedance, and sometimes it's okay to accept lower gain in exchange for less error.
mauifan wrote:So if the input impedance of the amplifier is about 100K, it would make sense to keep the impedance of the filter/feedback network on the order of 10k or less, no?
Not necessarily.. that value certainly isn't more important than your loop gain or the intrisic values of your crystal.
Like I said, the non-ideal properties of an op amp show up as output offset errors, and those have the very nice property of being more or less independent of the input signal or gain. There are three major ways to deal with them:
1) "Just ignore it" -- a very popular option if you only want to see how the input signal changes.
2) "Tweak the reference voltage" -- if you have 100mV of output offset error, change IN+ by 100mV so your reference voltage error cancels your output voltage error.
3) "Use a capacitor" -- caps make DC errors go away, as long as you can afford to live with the phase/frequency issues that caps impose.
For an oscillator, you get #3 for free thanks to the resonant elements in the feedback path, but you can also use #2 to center the output if the offset causes clipping.
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