-5
\$\begingroup\$

What happens if we give some resistance directly to the input signal instead of giving some part of output resistance back (i.e., negative feedback) ?

\$\endgroup\$
2
  • 8
    \$\begingroup\$ Can you clarify this question some? Maybe a schematic of what sort of arrangement you mean? \$\endgroup\$ – Bitrex Dec 10 '11 at 5:34
  • \$\begingroup\$ What Bitrex says. And if you are already editing your question: Please rename the title to something more descriptive like "resistance on positive input of opamp". \$\endgroup\$ – 0x6d64 Dec 10 '11 at 7:54
1
\$\begingroup\$

There is much confusion here. Adding resistance to the signal going into a amplifier will generally attenuate that signal. Some amplifiers work such that the gain is inversely proportional to the input signal impedance. This is the case for the basic opamp inverting amplifier:

The gain is the ratio of R2 to R1 but also negative, assuming of course that the opamp open loop gain and bandwidth can support this. The input impedance is R1 because the negative input of the amplifier is actively held at ground.

However, the basic non-inverting opamp amplifier has high input impedance:

In this case the gain is (R2 + R1)/R1 (again assuming this is significantly less than the opamp open loop gain) and the input impedance is just the input impedance of the opamp. That is generally quite high, and is infinite for a ideal opamp. In this case adding additional impedance to the input signal will have little effect unless it is a significant fraction of the opamp input impedance. For example, adding 10 kΩ in series with IN will do little if the opamp input impedance is 1 MΩ, and that's low as most opamp input impedances go.

You also talk about "giving back" part of the output resistance. This makes no sense. You can feed back voltage and sometimes current, but you don't "give back" or feed back resistance.

\$\endgroup\$
0
\$\begingroup\$

If you do not provide a feedback resistance on an op-amp then the gain will saturate.

The gain for an inverting amplifier is defined as the ratio between input and feedback resistances, so a 1k input and a 100k feedback gives a 1:100 gain.

With no feedback resistor the feedback resistance is infinite, so you get a 1:infinity gain.

That is bad.

\$\endgroup\$
1
  • \$\begingroup\$ It's not bad, it's just called a comparator. Whether that's what you wanted or not is another matter altogether. \$\endgroup\$ – W5VO Dec 10 '11 at 23:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.