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I'm using NE5534 OpAmp in a voltage follower configuration. No other parts, just the opamp.

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Power supply is 12VDC and GND. Vin is square pulse with 5V amplitude.

However, after making prototype, I can see that Vout can't go to 0V. At 0V input, output is 1.2V or so.

OSCILLOSCOPE SHOT. Vin is YELLOW while Vout is GREEN.

enter image description here

Why is this?

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    \$\begingroup\$ Not a rail-to-rail opamp. If you are lucky, there are rail-to-rail (IN and OUT) ones that can simply replace yours. See if you can order free samples. \$\endgroup\$
    – TQQQ
    Commented Jun 14 at 2:42

6 Answers 6

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The NE5534 datasheet indicates that, for a +/-15 V supply (30 V total), the output voltage swing will be 24 - 26 V, implying that the output cannot get closer than 2 volts to the supply rails, so your 1.2 V above the negative supply (which happens to be Ground) is very good.

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There are two problems with what you are trying to do and, both are indicated in the NE5534 data sheet. Here's an extract from page 5: -

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What the green highlight is telling you is that the guaranteed usable input range is from 3 volts above the negative rail to 3 volts below the positive rail. In your case that's +3 volts to +9 volts.

In blue is a similar story for the output swing.

In addition you should take note of the absolute maximum rating for the inputs: -

enter image description here

If your input can be 0 volts (GND) then you are right at the absolute limit and this is a concern. The NE5534 is really more suited to a dual supply (+ and -) where the signal reference is at the mid-point.

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That is normal for real-world op-amps.

This specific op amp type cannot both properly detect input down to the negative supply voltage and cannot properly drive the output down to the negative supply voltage.

If you have 0V as the negative supply, this op-amp circuit works properly down to only about 3V or 2V but not below that.

There are op-amps that can detect the input voltage beyond the supplies. For the output voltage, even the best rail-to-rail output op-amps can only drive the output within millivolts of the supplies.

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You are violating the common mode range. If you want to use this opamp, then you need to use dual supply.

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NE5534 is designed to work without surprises with symmetric ±12V or ±15V supplies. At +5V it suffers from very limited input and output voltage range, and is not suitable for your application.

To buffer a 0V-5V signal, the minimum supplies would need to be -3V, +8V.

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The problem

It consists in the fact that the op-amp output voltage cannot reach the limit values (0 V and 10 V) determined by the supply voltage. But why?

The solution

The obvious solution (if you are going to use this op-amp) is to supply it with two voltage sources of opposite polarity (V+ and V-). But why?

To answer this question, let's see, step by step, how this idea was arrived at.

Conceptual circuit

For this purpose, we can use a conceptual equivalent circuit where the op-amp complementary output stage is modeled by a potentiometer (think of the two pot halves as the two complementary transistors forming the stage). Imagine that when the input voltage varies from zero to maximum, the pot wiper moves up and its transfer ratio K changes from 0 to 1.

Single-supplied potentiometer

K = 0.5: When we put the wiper in the middle position, the output voltage is equal to half of the supply voltage (Vout = 5 V). This is the high OP's level.

schematic

simulate this circuit – Schematic created using CircuitLab

K = 1: When we move the wiper to the highest position, the output voltage is maximum but does not reach the supply voltage. The reason for this is that there is a residual resistance R1 between the potentiometer and V+. That is not the OP's problem though.

schematic

simulate this circuit

K = 0: The problem of OP appears when we move the wiper to the lowest position but the output voltage (Vout = 833 mV) does not become zero. Now the reason for this is that there is a residual resistance R2 between the potentiometer and ground.

schematic

simulate this circuit

We can see this graphically if we sweep (change linearly) K from 0 to 1; Vout changes linearly from 833 mV to 9.167 V.

STEP 1.3

"Shifted down" potentiometer

The problem in Schematic 1.3 above is that the output voltage is raised by 833 mV when it should be 0 V. Then let's lower it by this much by "pulling down" the lower end of R2 through an additional negative voltage source V-. We can calculate its voltage or set it experimentally in the simulator; either way it gets 909 mV. It is higher because of the divider's attenuation.

schematic

simulate this circuit

As we can see in the graph below, Vout begins changing linearly from 0 V to 9.5 V.

STEP 2.1

Split-supplied potentiometer

This is how we solved the specific OPs problem, but in order for the circuit to be universal, usually the two sources are the same, and the schematic is drawn in this symmetrical and beautiful way (like a bridge circuit).

K = 0.5: The "bridge" is balanced.

schematic

simulate this circuit

K = 1: The "bridge" is positively unbalanced.

schematic

simulate this circuit

K = 0: The "bridge" is negatively unbalanced.

schematic

simulate this circuit

The graph is symmetrical; Vout linearly changes from -8 V to 8 V.

STEP 3.3

More observations

If we are observant enough, we will notice that:

  • Regarding the ground, the two voltage sources have opposite polarity and yet their voltages are summed. This is because they are connected, regarding the voltage divider, in series but unidirectional, and the divider is supplied by V- + V+ = 20 V.
  • Regarding the load (Vout voltmeter), the two voltage sources are oppositely connected in parallel (through the top and bottom resistances), and their voltages are subtracted.
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