2
\$\begingroup\$

I found a similar question here and understood the following: input signal must be within the input common-mode voltage range of the OpAmp. Take for example AD8055 which can be powered by dual or single voltage supply. The specsheet describes IO characteristics for only dual power supply:

TA = 25°C, VS = ±5 V, RF = 402 Ω, RL = 100 Ω, Gain = +2, unless otherwise noted. enter image description here

However, I wish to power this OpAmp with a single power supply. Let's say I will provide -Vs=0V and +Vs=5V. How do I determine IO characteristics than?

Here's another example: LT1818 OpAmp. They actually provide IO characteristics for dual and single power supplies:

TA = 25°C. (Note 9) VS = ±5V, VCM = 0V, unless otherwise noted enter image description here

TA = 25°C. (Note 9) VS = 5V, 0V; VCM = 2.5V, RL to 2.5V unless otherwise noted enter image description here

Now they introduced positive/negative notion for IO voltage. How to understand it? Is it the maximum/minimum voltages, voltages that used for Vin+ and Vin- respectively, or they omit the '-' sign for the negative voltages? It does implicitly say that on a single 5V supply, the output swings from 1V to 4V with a 100Ω load connected to 2.5V.

My video signal will will range +0.5V to +2.5V. I want to use any of these two OpAmps as buffers G=+1. I know that if I use dual +-5V power supply, they will work. However, I wish to stick to single +5V supply. Would any of them still do the buffer job for my video signal range?

\$\endgroup\$
  • 1
    \$\begingroup\$ The second one will not swing below 0.8V (at Rl = 500 ohms) so ... probably no. (If Rl = several kilohms, maybe, but you don't specify that). The first will apparently not get closer than 1.8V of the negative rail, so, again ... no. \$\endgroup\$ – Brian Drummond Jun 13 '14 at 17:04
1
\$\begingroup\$

You cant use AD8055 amplifier with single 5V supply because:

enter image description here

It needs at least 8V between + and - supply.

For LT1818 - im not sure where my red vertical line should be, please somebody confirm if I'm right or wrong.

In my opinion - this is "within the input common-mode voltage range":

enter image description here

Even if I'm wrong about "red vertical line" - this amplifier can't work with +0.5V signal at 0V at Vs- pin.

\$\endgroup\$
  • \$\begingroup\$ Thank you. Clear enough. Let's see if anyone else has to add something. \$\endgroup\$ – Nazar Jun 13 '14 at 20:43
1
\$\begingroup\$

The front page of the data sheet tries to paint the best picture it can and on the front page for the AD8055 it says that the supply is +/- 5 volts. Look no further, this device is not suitable for +/- 2.5 volt supplies. End of story.

Look a little further if you must but on page 4 it shows this: -

enter image description here

You should also look at absolute max ratings too. One that can trip out is the maximum differential voltage between input pins - for this device it's +/-2.5 volts so don't go using as a comparator!!

\$\endgroup\$
1
\$\begingroup\$

"... However, I wish to power this OpAmp with a single power supply. Let's say I will provide -Vs= 0 V and +Vs = 5 V ..."

That is not sufficient to ensure the proper operation of your amplifier in single-supply mode. By holding the common-mode input voltage constant, the inverting-amplifier configuration eliminates common-mode-rejection errors. This is achieved by creating a positive floating voltage (typically Vcc/2) - called reference voltage applied to the non-inverting input, as shown in figure below:

Single-supply

Maintain that offset voltage away from ground and Vcc.

By not holding the common-mode input voltage constant, an example of circuit that is susceptible to errors due to violation of CMRR is the voltage follower configuration (unity-gain or buffer). Depending on the input voltage, the output may contain clipping or reach saturation.

Voltage Follower

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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