# Need some help in designing an single supply op-amp

#### yaantey

Joined Oct 7, 2011
48
Hey,

Attached is a circuit and some explanations.
1) Can someone explain to me the highlighted parts in simple english what it means and why it means that.

2) Also if I am designing a non-inverting single supply op-amp what is the best resistor values for the feedback loop to have a gain of 100x and why is it? (The input signal is 20m Vrms)

3) What values should I choose for the voltage divider biasing? (The input signal is 20m Vrms)

#### PaulEE

Joined Dec 23, 2011
423
-Guitar and cap: guitar = signal, cap = AC coupling to input pin (+)
-first op-amp stage: two 1 M resistors; voltage divider. brings (+) to 1/2 of supply voltage
-feedback loop on first opamp: provides gain for AC due to cap to ground (47uF)
-second opamp: no voltage divider because output of first opamp is already at 1/2 supply voltage
-second opamp: another feedback loop with AC gain due to capacitor
-10K ohm output resistor for current limit into ADC port, protection diodes,...output

GAIN of non-inverting opamp stage: 1+R2/R1, use moderate values, 10k's worth of ohms
voltage-divider biasing: design such that output impedance of equivalent voltage source that the two resistors creates is sufficiently high as to not swamp the input signal, but low enough to drain input bias currents out of opamp input pin. otherwise, input pin bias currents will charge up the input capacitor and cause no signal at the output of your circuit.

OVERALL gain: input volts * (1+R2/R1) of first stage * (1+R2/R1) of second stage;

If you want to take ADC into account, we'd need to know the input impedance of the ADC; 10K output resistor in series with ADC will form divider out of the two impedances; much like another voltage divider.

I think those diodes are simply protection diodes and also lower the noise on the output. Not 100% on that, but pretty sure.

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• yaantey

#### yaantey

Joined Oct 7, 2011
48
design such that output impedance of equivalent voltage source that the two resistors creates is sufficiently high as to not swamp the input signal, but low enough to drain input bias currents out of opamp input pin. otherwise, input pin bias currents will charge up the input capacitor and cause no signal at the output of your circuit.
Can you show an example for the statement with some values, because I couldn't quite understand it.

Also in the explanation it talk about 3DB frequency which I quite don't know why its relevant. So, could you tell me the importance of 3DB frequency and how the filter created by the 1uF capacitor and 1M ohm resistor has the 3DB frequency.

Thanks.

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#### PaulEE

Joined Dec 23, 2011
423
Can you show an example for the statement with some values, because I couldn't quite understand it.

Also in the explanation it talk about 3DB frequency which I quite don't know why its relevant. So, could you tell me the importance of 3DB frequency and how the filter created by the 1uF capacitor and 1M ohm resistor has the 3DB frequency.

Thanks.
Sure.

You know you want some particular gain, right?

That means that my gain expression should approximately equal (enter your gain here).

Suppose the gain of the overall circuit depends upon the two opamp stages, and also suppose for ease of calculation, that the stages are the same gain.

The equation would then be overall gain = A^2, where "A" is the gain of one stage; --> 1+R2/R1

so, for a gain of, call it 100 V/V, you want one stage to have a gain of the square root of 100, or 10 V/V.

If you choose 99.9 Kohm for R2, R1 calculates out as:

10 = 1 + 99.9/R1; 9 = 99.9/R1; 9/99.9 = 1/R1; R1 = 11.1 Kohm.

Therefore, each stage would have a R2 = 99.9 Kohm and R1 = 11.1 Kohm;

each stage now has a gain of ten, and ten^2 = overall gain of 100.

The 3db frequency is the frequency at which a filter (in your case, low pass) attenuates the input signal in such a way as to yield [sqrt(2)/2]*input (or 0.707 * input) at the output.

the 3db frequency of RC low pass filter is approximately f = 1/2(pi)RC

the equivalent resistance of a voltage source equal to 1/2 the supply voltage at the "+" input pin with two 1M resistors is 1M in parallel w/ 1M = .5 M = 500 kohm. C = 10^-6 farad or 1 uF.

f = 1/2*pi*500,000*.000001 = 0.3 Hz, approximately. Note that I am ignoring the source impedance (the real, or resistive part) and the input impedance of the opamp, which is usually >>1 Mohm

Cool?

• yaantey

#### PaulEE

Joined Dec 23, 2011
423
meanwhile, the feedback resistors at each stage, along with the 47 uF caps, have similar filtering effects, which you can do rough calculations on if you'd like.

• yaantey

#### yaantey

Joined Oct 7, 2011
48
Thanks alot. I understand the calculations. So basically, the 3DB frequency refers to each frequency the filters filter out. For example the low pass filter made up of the two 1M ohm parallel resistors and 1uF capacitor passes low frequencies than 0.3 Hz through. Is that the meaning of the 3DB referring to?

#### PaulEE

Joined Dec 23, 2011
423
Thanks alot. I understand the calculations. So basically, the 3DB frequency refers to each frequency the filters filter out. For example the low pass filter made up of the two 1M ohm parallel resistors and 1uF capacitor passes low frequencies than 0.3 Hz through. Is that the meaning of the 3DB referring to?
In a nutshell, yes.

The 3dB point of a filter/etc. is a figure-of-merit of filters that describes the outer bound at which they are considered effective. It usually is termed the "half-power" point also. Check out this wikipedia link. Note the mention of 0.707 there, also.

http://en.wikipedia.org/wiki/3dB-point

Enjoy!

• yaantey

#### yaantey

Joined Oct 7, 2011
48
Thanks alot. You have greatly helped me. Cheers !!

Joined Dec 26, 2010
2,148
The 3dB frequencies are those frequencies at which the filter or coupling concerned introduces a power loss of very nearly two times: [ 10*log$$_{10}$$(2) = 3.0103 ]. In voltage terms this is a gain of 1/√2, or 0.7071...

For an AC coupling, this happens when the capacitor reactance equals the sum of the driving and terminating (equivalent) resistances. Do you know why this does not result in a gain of 0.5 (-6dB, more or less)?

Others were faster. Oh well...

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#### yaantey

Joined Oct 7, 2011
48
I need to put a bandpass filter of 700Hz to 2000Hz. If so, what should my biasing resistors be? How exactly do I know to choose this?

Thanks

#### PaulEE

Joined Dec 23, 2011
423
In this particular case, you'd need to tweak the values of the first filter and add another filter in series whose f3db frequency covers the second frequency...so:

using the equation 1/2piRC you can tweak the 1M resistors such that the cutoff frequency is 700 Hz; you then need to look into what a high pass RC filter looks like and use a similar equation to add another filter stage with another corner frequency (f3db) equal to 2000 Hz.

It's tricky because you want set it up such that the input impedance, as seen by the guitar signal, is large enough that it doesn't pose a burden to the signal, while, tweaking each of the filter stages to achieve your 700 and 2000 Hz boundaries.

Start by developing a simple circuit model on paper, and carefully work through the algebra.

Good luck #### Audioguru

Joined Dec 20, 2007
11,248
Never use a lousy old LM358 for audio:
1) It is too noisy (hiss).
2) It has crossover distortion.
3) It has trouble above only 2kHz due to its very low slew rate.

Why does the circuit have a dual-polarity supply when the input is properly biased at +2.5V so that the output never goes negative? Then the negative supply is not needed.

Many people say an electric guitar pickup sounds best when its load is 1M or higher. This circuit is less than half that.

Here is a guitar preamp with a high impedance input provided by a Jfet. An opamp with J-fet inputs could have been used instead.

#### yaantey

Joined Oct 7, 2011
48
@Audioguru

Im just referring to that circuit. I am not making a guitar amplify. I'm trying to capture baby's voice (using bandpass filter) and send the signal to micro-p for processing. I don't need more than 2k, so guess it will be alright then. It's just my own little project to learn some stuff #### Audioguru

Joined Dec 20, 2007
11,248
The gain is way too high at about 1620. If the baby cries then the 2nd opamp will be badly clipping.

The hiss all the time might cause trouble with the ADC.

I don't know why you are filtering away most audio frequencies.

#### yaantey

Joined Oct 7, 2011
48
I am just referring the circuit. Im not using the exact same circuit. Because when the baby cry it has a certain frequency range called hyperphonation above 700 Hz and below 2000 Hz.

#### PaulEE

Joined Dec 23, 2011
423
The gain is way too high at about 1620. If the baby cries then the 2nd opamp will be badly clipping.

The hiss all the time might cause trouble with the ADC.

I don't know why you are filtering away most audio frequencies.
I don't think he's trying to achieve perfection; I think he's at the the trial and error, learning stages....

#### yaantey

Joined Oct 7, 2011
48
I check bandpass filter. So, first I need a low pass filter to pass all the frequencies lower than 2000Hz and then a high pass filter to pass frequencies higher than 700Hz. Hence having a selecting a frequency range of 700Hz to 2000Hz.

Attached is the arrangement of the bandpass circuit I could think of, is it correct (if not can I get a correction)? Also when calculating the frequency using (1/2*pi*R*C*) do I need to consider resistors 1, 2 n 3 for both the filters. Example to get low pass total resistance would be (R1//R2 + R3).

Thanks

#### Attachments

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Joined Dec 26, 2010
2,148
The amount of filtering achieved by a simple RC low pass or high-pass will be quite limited, as the maximum gain slope is only 6dB per octave. This can be improved somewhat by having more than one stage of filtering, preferably separated by amplifying stages.

RC networks will also interact if directly connected together. Separating them by a stage of amplification will isolate them so that the response is purer and easier to predict. For more effective filtering, purpose designed filter circuits are required. This is a subject in itself, and not something I am particularly expert at. There is however a great deal of information available on the internet: try a search based on the term "active filter".

http://en.wikipedia.org/wiki/Active_filter

#### PaulEE

Joined Dec 23, 2011
423
I check bandpass filter. So, first I need a low pass filter to pass all the frequencies lower than 2000Hz and then a high pass filter to pass frequencies higher than 700Hz. Hence having a selecting a frequency range of 700Hz to 2000Hz.

Attached is the arrangement of the bandpass circuit I could think of, is it correct (if not can I get a correction)? Also when calculating the frequency using (1/2*pi*R*C*) do I need to consider resistors 1, 2 n 3 for both the filters. Example to get low pass total resistance would be (R1//R2 + R3).

Thanks
I'm actually out to dinner on my phone, I can look at the circuit when I get back to a computer in about 20 minutes...