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In the circuit diagram below, why is there a 100KΩ resistor (NOT R2) connected to the capacitor? To my understanding the capacitor-resistor act as a high pass filter to block the DC offset of the microphone, but since only the capacitor blocks DC why is the 100k resistor used? According to the author of the video (link below) he said, the 100k is used "not to overload the microphone's un-amplified output". I don't get this part.

Also, can only a capacitor be used in this circuit or any other circuit without the 100k resistor?

Passive RC high pass filter tutorial! Simple microphone-speaker circuit

Ricardo
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putu06
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  • Basically the same question but for an inverting amplifier is at http://electronics.stackexchange.com/questions/93496/what-is-the-use-for-non-inverting-input-resistance-in-a-negative-feedback-opamp – Fizz Jan 13 '15 at 22:22

2 Answers2

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The resistor is there to provide a DC path for the input bias current of the opamp.

It is normally selected to be the same as the DC resistance connected to the other input, so that the bias current does not produce a voltage offset at the output of the opamp. But in this case, the effective DC resistance on the inverting input is only 1k||100k = 990Ω, so that benefit is not realized here.

It is also selected to be high enough that it doesn't affect the frequency response of the circuit overall (in conjunction with the DC blocking capacitor). In this case, 0.1 µF and 100 kΩ have a corner frequency of

$$\frac{1}{2\pi R C} = 15.9 Hz$$

This means that for frequencies above this value, the resistor will have no effect in the AC signal, but there will be a rolloff (loss of amplitude) below this frequency. This "loading" effect is probably what the author of the video was referring to.

Dave Tweed
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  • May I ask how affect the frequency response of the circuit? In what way is "high enough"? – thexeno Dec 30 '14 at 15:36
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    See edit above. You can see that as the R value goes up, the corner frequency is reduced. You just have to decide what corner frequency is "low enough". – Dave Tweed Dec 30 '14 at 15:42
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    It's worth noting that, in terms of being `selected to be the same as the DC resistance connected to the other input`, it fails miserably, since the DC resistance to the inverting input is 990Ω. In this case, I can only assume it was chosen simply to avoid loading the mic output too much, or because the circuit already had some 100KΩ parts. – Connor Wolf Dec 30 '14 at 20:23
  • I think the answer would be more complete if you addressed the effects on the circuit of removing that resistor. – Nicolas Holthaus Dec 30 '14 at 21:22
  • @NicolasHolthaus: I think that would be too much detail for this question, and it depends on the specific opamp being used anyway. But feel free to have a go at it... – Dave Tweed Dec 30 '14 at 21:28
  • @DaveTweed I actually don't know and was curious :) – Nicolas Holthaus Dec 30 '14 at 21:29
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    @NicolasHolthaus: Ah, well, in that case, the LM324 shown in the schematic uses a Darlington PNP input stage, which means that the bias current is coming *out* of the input pins. Without a DC path for it, the right end of the capacitor will get charged to nearly +9V, and the output of the opamp will saturate as far as it can go in the positive direction. – Dave Tweed Dec 30 '14 at 21:34
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Dave Tweed's answer is excellent on the facts (and so I upvoted it). Since this is basically a newbie question that is covered/answered in most intro electronics textbooks, there is one addendum perhaps worth making: how to figure it out (or convince yourself)... using SPICE!

I'm using a different opamp, the NE5532, which probably has higher bias currents, but which is commonly used in audio. The circuit otherwise is basically the same, except that I've wisely added an output cap as well... which is not a bad idea as you shall below why: enter image description here

There's about -5V of DC bias on output (before the cap). And these originate from the amplfication of the input bias voltage (about -50mV) caused at the input by the current flowing through the positive-input biasing resistor R10. Now watch what happens when we increase this R10 resistor to 100Mohm (or remove it altogether).enter image description here

The output goes into saturation; we have a hint why it happened because of the input offset voltage is also much higher than before (about -200mV instead of -50mV).

You can also do a parametric sweep of some values for R10, in this case 50K, 100K, 150, 200K, which turns out to be enough to cause output saturation with the NE5532.enter image description here

And if you're curious about eliminating (as much as possible, in practice it won't be perfect) the offset voltage, then you need to add another resistor (R3=R10) to roughly match the input currents. This is only relevant if you want to live without the output cap as the circuit from the question tries to do. But that's basically another topic, which is the subject of a different question here.)enter image description here

Finally, I've uploaded the source code for one of the above (very similar) circuits, namely the 3rd/parametric one, so you (newbies) can experiment yourselves. You need the NE5532 opamp macromodel for the code to work as is (although practically any opamp will work the same way but will cause saturation at different R10 values) and of course the LTSpice IV simulator.

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Fizz
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