# How does this opamp output -4, +4 output?

I'm new to electronics. This circuit is supposed to take 0-2 V MCU signal and convert that to a range of -4 V and +4 V, so when input value changes, the output changes as well till maximum.

I think I understand the left side opamp U3.2, that it's outputting an inverted value of 0 to -4 V voltage with 0-2 V input, so it's inverting and amplifying the voltage. My question is, there is no power supply connected to it, then how is it generating that voltage?

On the second one, it's completely confusing. I don't know how it's taking the input of the first and then mapping the input to a range of -4 to 4 V.

I apologise if this is too basic, I've watched many opamp tutorials, but I can't figure it out.

Also, what would I have to do to make it work with 0-3.3 V signal from MCU and +4 V to -4 V?

My aim is basically to drive a galvanomirror that moves a mirror up and down, so that's why the negative voltage is required, and it accepts a voltage range of ±5 V, but I'd like to stay below the maximum.

• "there is no power supply connected to it..." that's because the TL084 package has 4 opamps, and they share the same power pins.
– Gos
Commented Aug 3 at 22:12
• "the second one it's completely confusing..." the amplification and mapping is done in the 1st opamp. The 2nd opamp is just inverting the output of the 1st one. So, in the J1 output, in one pin you have the -4V,+4V signal, and in the other pin you have the same signal inverted.
– Gos
Commented Aug 3 at 22:18
• @Gos Thanks so much for taking the time to answer. helped! Commented Aug 13 at 18:20

The reason that no supply is shown connected to U3.2, is that it shares the power supply of U3.1, which is part of the same integrated circuit. The TL084 contains 4 op-amps, all sharing the same supply on pins 4 and 11, as shown here (from elprocus.com):

Therefore, it is powered, even though the schematic seems to show that it isn't.

Op-amp U3.2 has its non-inverting input connected to some non-zero potential $$\V_{S\_REF}\$$, which causes its output to be offset (centered about) some non-zero value also.

As I explain in this answer, for the first (U3.2) stage, the relationship between input and output is:

$$V_{OUT} = V_{S\_REF} \left( 1 + \frac{R_{10}}{R_8} \right) - V_{DAC\_X}\frac{R_{10}}{R_8}$$

Plugging in known values gets you:

\begin{aligned} V_{OUT} &= V_{S\_REF} \left( 1 + \frac{12k\Omega}{3k\Omega} \right) - V_{DAC\_X}\frac{12k\Omega}{3k\Omega} \\ \\ &= 5 V_{S\_REF} - 4 V_{DAC\_X} \\ \\ \end{aligned}

As you can see, the inverting input $$\V_{DAC\_X}\$$ is subject to gain −4. The non-inverting input $$\V_{S\_REF}\$$ is a constant potential, subject to gain +5, and the goal now is to determine what fixed potential there would produce the required offset. First rearrange the above equation to have $$\V_{S\_REF}\$$ the subject:

\begin{aligned} V_{OUT} &= 5 V_{S\_REF} - 4 V_{DAC\_X} \\ \\ V_{S\_REF} &= \frac{V_{OUT} + 4 V_{DAC\_X}}{5} \end{aligned}

Then plug in some known values to reveal $$\V_{S\_REF}\$$. We know, for instance, that when the input is $$\V_{DAC\_X}=+2V\$$, the output should be $$\V_{OUT}=-4V\$$:

\begin{aligned} V_{S\_REF} &= \frac{(-4V) + 4 (+2V)}{5} \\ \\ &= +0.8V \end{aligned}

That's why you require R5 and R6 to produce $$\V_{S\_REF}=+0.8V\$$.

The second stage (U3.1) is a simple inverting amplifier, with gain $$\-\frac{R_{14}}{R_{11}}=-1\$$ so that as the prior stage output goes from −4V to +4V, the second output does the exact opposite, going from +4V to −4V.

To have the system produce the same output voltage range for inputs between 0V and +3.3V, you can use the same algebra I used above. One trick you can use to get started is to recognise that input $$\V_{DAC\_X}\$$ is subject to gain $$\-\frac{R_{10}}{R_8}\$$. If you can find that gain, then you can choose R10 and R8 to suit. Gain $$\A\$$ is the ratio output swing to corresponding input swing:

\begin{aligned} A &= \frac{(-4V) - (+4V)}{(0V) - (+3.3V)} \\ \\ &= \frac{-8}{3.3} \\ \\ \end{aligned}

Use that value to obtain R10 and R8. If I leave R10 unchanged:

\begin{aligned} -\frac{R_{10}}{R_8} &= -\frac{8}{3.3} \\ \\ R_8 &= 3.3\frac{R_{10}}{8} \\ \\ &= 3.3\frac{12k\Omega}{8} \\ \\ &= 4.95k\Omega \\ \\ \end{aligned}

From there you can use the same procedure as above to determine the required $$\V_{S\_REF}\$$.

• Thanks for taking the time to write such a detailed reply. i'd like to read up more on this theory. i'm not sure how the equations came up or how things are working. could you recommend a book where i can find more about this? I tried to look in "Art of electronics" but not alot of details was present there. certainly not this application. Commented Aug 7 at 12:11
• @ShoaibKhan It's not so much a theory as just a derivation, using KVL, KCL, Ohm's law and op-amp behaviour with negative feedback. Commented Aug 7 at 13:56
• @ShoaibKhan The formula I gave is a generalisation of both inverting and non-inverting op-amp configurations, which I explained in another answer. Commented Aug 7 at 13:57
• @ShoaibKhan All those concepts are covered in nearly every modern beginner/intermediate book about electronics ever written (including AoE), but those links should get you started at least. AoE is so comprehensive, and probably a bit advanced, that finding any individual snippet to support the pricinples I employed here might be difficult. Commented Aug 7 at 14:03
• @ShoaibKhan At the risk of blowing my own trumpet too hard, here's another answer that attempts to explain the whole "op-amp with negative feedback" thing, and will go some way to explaining the fundamentals prior to reading the other answer I linked, which is more directly related to this question of yours. Commented Aug 7 at 14:22

To analyze U3.2, you can look at the gain, which will be -12K/3K = -4 so the input range of 3.3V will result in an output range of 13.2V, an 8V range would require the input to change 2V

Next, look at the zero. For 0V out, the inverting input must be at 0.8V so there will be ix= 0.8V/12k current flowing through R10 and the input therefore will be at 0.8V + ix*3k or 1V.

So the equation for the output voltage in the area of linear operation is Vout = -Vin*4+ 4V

Note that for 3.3V you'd get 9.2V out if the op-amp output did not saturate, but with an 8V supply it won't get that high. The input range for +4 to -4 out is 0 to 2V.

To reverse the action, you could simply add an inverting stage afterward.

simulate this circuit – Schematic created using CircuitLab

It's also possible to get the desired output with a single op-amp, with a bit of calculation.