basic questions hope everyone can help

[NicholasTok]

hello anyone just need to clarify some questions below is an example of the connections

The question is, what will I get from Vout when I have a AC input parallel to a DC input?

lets assume that my AC input has a frequency range from 20Hz to 200Hz with an amplitude of 0.4V, what will I get at Vout?

 

[Fom]

If yours AC source is ideal voltage source (zero output resistance) you will get Vout=Vac.

 

[NicholasTok]

If yours AC source is ideal voltage source (zero output resistance) you will get Vout=Vac.

you mean my Vout will be 0.4V?

 

[DrBob13]

Hi,
Adding to what has been correctly stated before:
Assuming your AC generatorhas a very low AC output impedance (near ideal) AND is AC coupled as stated previously you will get 0.4V AC but it will sit on a 5V DC offset voltage, do not forget this. You will be able to see this quite clearly if you use a DC coupled scope to view your signal.

 

[Fom]

According to your picture AC input and Vout are the same node. One node cannot have different voltages at the same time. So answer to your question "what's Vout voltage" is Vout = VACin.

 

[laktronics]

Hi Nicholas,

Fom is right for the case of DC coupling shown in the diagram, assuming an AC source resistance of zero ohms, output will be equal to AC voltage. For DC coupling case if AC voltage has internal resistance, then the output voltage will be a sum of DC and AC after application of Superposition Theorem. And if the AC voltage is AC coupled, and is of Zero internal resistance, the output will be full AC voltage riding over 5V DC. And in this case, if AC source has internal resistance, the amplitude of AC will get reduced to the extent of attennuation by 1KOhms.

Regards,
Laktronics

 

[NicholasTok]

Hello Fom , DrBob13 & laktronics, thanks for replying.

I'm so sorry as the electronic terms used is too technical, I am not really very sure about the AC ridding on DC theory.

I have make a picture using waveform to illustrate what everyone have said but not sure if the picture drawn was correct, please correct me if I am wrong.

Regards
Nicholas Tok

 

[DrBob13]

Hi,
Yes you have got it about right within the limitations already discussed regarding generator impedances, AC/DC coupling etc. You are adding he instantaneous values of the two signals which in this case like yuo have drawn, as one is DC the other is AC so you have no AC phase difference to worry about, so instead of having zero volts as the AC crossing point you have 5V. This is sometimes refered to a DC offset of +5V and is commonly used when you have a single supply rail op-amp which is biased so that its output sits at half the supply voltage. For example if the op-amp is connected to a 10V rail an offset of 5V is applied to the non inverting input so the output can swing from near zero volts to nea\rly +10V, in other words a 5V peak signal can be output, of the dc offset is not then required passing the signal through a suitable value capacitor will remove the DC offset. You can add two AC signals as well but in this case the resultant output is much more complicated to visualise, at least in the beginning of your electronics training, you can try adding two sinewaves in a spreadsheet like EXCEL and display the result graphically if you want to see what happens in the AC case, it will work for DC to incidentally.
I hope that helps.
Rgards,
Bob.