0
\$\begingroup\$

I want to use an inverting amplifier to linearly shift and convert a signal with voltage range -10...+10V into 0 ... 3.3V for an ADC of a microcontroller. I wanted to use the circuit presented in this post. enter image description here

Instead of the 40 Hz source I would input the -10...+10V supply and instead of -15V I want to use a -12V supply.

1.) Which supply voltage range should the OpAmp have? I mean the output should only be in the range of 0 ... 3.3V, my resistor feedback network will take care of that. Or does it have to cover the sum of the two input voltage maximum and minimum (i.e. -22V and 2V)?

2.) Are there any constraints regarding the DC-bias supply (-12V)? Must it have a low output impedance (< 1 Ohm)? Does it matter whether I would use a Linear Voltage Regulator or a Voltage Reference (as long as the rated current of the device is not exceeded)?

3.) General question: Can I use an OpAmp that is specified for a supply range of +-5V...+-15V (i.e. dual supply) also as single supply? For example V_supply_- = 0V and V_supply_+ = 5V

Best regards and thanks in advance!

\$\endgroup\$
  • \$\begingroup\$ Embed the picture in your post. Not many of us will bother to follow a link just to understand the question. Even if you add the link one of us can fix it up for you. \$\endgroup\$ – Transistor Feb 20 '18 at 22:57
  • \$\begingroup\$ I fixed it. And also added a question. \$\endgroup\$ – Geralt Feb 20 '18 at 23:03
0
\$\begingroup\$

One thing you could do is use a summing amplifier to scale and offset the input signal simultaneously. This requires a little bit of math, you can reference the following image/equations from Sedra/Smith Microelectronics Circuits textbook.

enter image description here

enter image description here

enter image description here

enter image description here

This does invert the output, so you'd have to follow this amplifier with a -1 gain stage. If you plug in the values chosen (somewhat arbitrarily), you'll see how the math works out.

\$\endgroup\$
  • \$\begingroup\$ This circuit would be even better than mine because I would not need the -12V source. One question: Why did you place the 100k resistor (R7) at the output? \$\endgroup\$ – Geralt Feb 21 '18 at 23:21
  • \$\begingroup\$ The 100k resistor on the output is just so there is some load present. I would make sure you you select the right op-amp. One that is rail to rail since you don't have a split supply. You need to make sure that that output can drive to either supply, otherwise you'll see clipping on the output. \$\endgroup\$ – delanymichael Feb 22 '18 at 0:14
4
\$\begingroup\$

I want to use an inverting amplifier to linearly shift and convert a signal with voltage range -10...+10V into 0 ... 3.3V for an ADC of a microcontroller.

That's doable without much trouble.

Instead of the 40 Hz source I would input the -10...+10V supply and instead of -15V I want to use a -12V supply.

That should probably read:

  • Input signal: -10 V to +10 V.
  • Offset adjust current to be sourced from a -12 V supply.

1.) Which supply voltage range should the OpAmp have? I mean the output should only be in the range of 0 ... 3.3V, my resistor feedback network will take care of that. Or does it have to cover the sum of the two input voltage maximum and minimum (i.e. -22V and 2V)?

The important point to realise in this inverting amplifier op-amp configuration is that the feedback adjusts the output to make the two inputs equal. Since the non-inverting input is at 0 V the inverting input will be at zero as well. It seems strange but your op-amp will never see the input voltages. The two inputs will normally be extremely close to zero volts.

Since you only require 0 to +3.3 V out a rail-to-rail op-amp fed from those voltages would suffice but they generally can't sink or source much current when the output is close to the rail voltage. You have the luxury of having higher voltages available.

2.) Are there any constraints regarding the DC-bias supply (-12V)? Must it have a low output impedance (< 1 Ohm)? Does it matter whether I would use a Linear Voltage Regulator or a Voltage Reference (as long as the rated current of the device is not exceeded)?

With something like 10k in there you'll only be pulling a little over a milliamp. All that is required is that the source voltage is stable.

3.) General question: Can I use an OpAmp that is specified for a supply range of +-5V...+-15V (i.e. dual supply) also as single supply? For example V_supply_- = 0V and V_supply_+ = 5V.

Not always. Read the datasheets carefully. If the op-amp was specified with a minimum of -5 to +5 then it would work at 0 to +10 V, etc., but check how low the output voltage could go - i.e., how close to zero.

Best regards and thanks in advance!

Thank afterwards on this site by up-voting useful answers and accepting the best one.


You may not need to go to all that trouble. A potential divider with a little bias can do what you require without any op-amp in many cases.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple resistor bias arrangement.*

How it works:

  • R1 and R2 provide a 10:3.3 potential divider. Without R3 the ADC input would be -3.3 V to +3.3 V for a -10 to +10 signal range.
  • The equivalent source impedance of the potential divider is the value of the parallel resistors. AllAboutCircuits online calculator tells me that that 6.7k in parallel with 3.3k is 2.21k. If we set R3 equal to this value and tie it to +3.3 V it will pull the voltage at the R1-R2 junction half-way towards 3.3 V.

Let's consider that last line a bit more:

Vin      Vadc without R3    Vadc with R3
 0 V       0 V              1.65 V
+10 V     3.3 V             3.3 V
-10 V    -3.3 V             0 V

The built-in CircuitLab simulator lets us run a DC sweep on the input from -10 to +10 V and watch the output.

enter image description here

Figure 2. Bingo! The desired transfer function.

You haven't told us the source impedance so be careful that this circuit doesn't load it too much. If it's a problem scale all the resistors up, for example, by ten.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the detailed answers! I already considered a resistive divider. But the drawback on the one hand is that it only works properly if the output current (into the ADC of the microcontroller) is small enough to be neglected. On the other hand, a resistive divider network would have a faster response time than most OpAmps, right? So it could follow step changes in the input much faster. Do you know how much current an ADC usually sinks? In addition to that, I also read that the output impedance in front of an ADC should not be too high, otherwise the measurements become erroneous. \$\endgroup\$ – Geralt Feb 21 '18 at 23:11
  • \$\begingroup\$ All good points. You will find the ADC input impedance in the micro-controller's datasheet. \$\endgroup\$ – Transistor Feb 22 '18 at 7:12
  • \$\begingroup\$ I am so dumb! I can indeed use the resistive divider, I just have to put a unity gain buffer at the end! With this configuration, the OpAmp draws only some pA of current and the output impedance towards the ADC is very low. That might work as well. \$\endgroup\$ – Geralt Feb 22 '18 at 21:24

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.