I have an application where I'm amplifying a small AF signal. It comes from the sensor as an AC signal at a DC offset of a couple of volts. I run the signal through a capacitor to remove the DC component and then pass this signal to an op amp.
In my application the measured signal varies from larger-amplitude, low-frequency signals up to tiny-amplitude, high-frequency signals (up to about 5 kHz).
The circuit works great until I get up to the higher frequencies/lower amplitudes and then I lose it. I'm currently using a series capacitor of 0.1 uF. Would it help to decrease the capacitance, to get a faster time constant? Is there something else I should try? Thanks for any help.
The ordinary limitation for capacitive coupling occurs at the lowest frequency to be passed, especially if the impedance receiving the signal falls as frequency falls, as with an input transformer.
Decreasing the capacitor will probably not help, since a coupling capacitor is not acting as a time constant filter element, but as a nearly short circuit at the signal frequency (no intentional attenuation). I suspect that some other factor is involved in your problem, like a receiving impedance that is falling as frequency rises, while a source impedance is rising. So we need to see more of the circuit to offer useful suggestions.
Drawing circuits in the forum never seems to work very well, but I can explain it since it is simple.
I have a reflective photosensor that is looking at a vibrating object. It consists of an infrared LED that shines on the object and an NPN phototransistor that picks up the reflection. The signal I get from the transistor is an approximate analog of the vibration. Here is the sensor:
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The collector is connected to +5V; the emitter is connected to a 220K resistor to ground. I split my signal from the connection between this resistor and the emitter and connect it to the capacitor (0.1 uF). After the capacitor I connect it to the op. amp.
As I mentioned before, I get a great signal up to a few thousand Hz, then I lose it. The signal gets higher in freqency, but also smaller in amplitude. I'm not sure if it's the frequency or the amplitude (or both) that is the culprit.
I wonder if it would help to just use a higher voltage across the phototransistor (like 24 V).
I think the problem is with the sensing transistor, not the coupling capacitor. The transistor has internal capacitance that keeps it from following the higher speed light variations. I think you need to think about how you can measure the variations in transistor current while you hold the voltage across the transistor constant, so that as many of those internal capacitances as possible, do not have to charge up and down as the transistor varies its conductance in response to light variations.
One way to do this would be to capacitively couple the emitter directly to the - input of your opamp (no input resistor) so that the negative feedback around the opamp holds that node in spite of input current changes. Effectively, you are amplifying the transistor's current output, rather than its voltage swing. This may require an increase in the size of the coupling capacitor, to handle the low frequency end of the range.
Thanks John. I think I'll try your suggestion. If I have no input resistance, then what should I use for a feedback resistance...fairly low? Can I go too high on the coupling capacitor value?
The capacitance in the transistor is about 30 pF. Do you think the higher operating voltage I proposed would help? The output is rated up to 30 volts. Thanks again!
The feedback resistor is forced by the opamp to carry all the transistor current variations, since there is no other place for it to go. If you expect 10 uA of current variation, then a 100k feedback resistor will convert that (by ohm's law) to a voltage swing of 1 volt. The only limit on how big the capacitor can be is the stabilization time when power is first applied and the stray capacitance from it to the surroundings. Another practical limit may be the leakage current. But, at the frequency of interest, you essentially want the coupling capacitor to have zero impedance, so larger is generally better. You may not need it or the bias resistor at all, if the feedback resistor on the opamp and the output swing can soak up not only the variations, but the average current, as well. It may be better to do it that way, with a low enough gain that the opamp doesn't saturate, and then capacitively couple that stage to a second gain stage, since the opamp output is a low impedance source that can drive the second stage much more robustly that the transistor could.
It will help some, sine raising the voltage decreases the internal capacitance some, by increasing the thickness of the reverse biased base to collector junction. You might expect the capacitance to fall by half as the bias voltage goes from a few volts to 20 volts.
But with the current to voltage converter amplifier, that capacitance has little effect.
Well, I think I got it figured out. For one thing, I found that I was barely driving the infrared LED. It is rated for 50 mA max. and I was driving it at 6 mA. So it was way too dim to get a good reflection. So I changed my current limiting resistor to make the LED run a lot hotter.
The "op amp" that I originally refered to is actually a 339 comparator and the signal was too weak to detect enough to "square it up". I ran the sensor output directly to the capacitor like you suggested, then to a pre-amp, then on to my final comparator. It appears to work now.
The keys were to boost my LED output, eliminate the resistor between the sensor and the coupling cap, and adding a pre-amp stage.
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