Comments on: Design a Bipolar to Unipolar Converter to Drive an ADC

By: Adrian S. Nastase

Adrian S. Nastase — Thu, 14 Jan 2010 04:14:07 +0000

ChiTownRick, there should be no loading unless your R1 and R2 resistors are really small. Please send me your schematic. I will send you my email address at the address you left me.

By: ChiTownRick

ChiTownRick — Thu, 14 Jan 2010 00:23:16 +0000

After testing the circuit from the discussion above, I am noticing some loading on the circuit behind my V1. I have an AC-coupling buffer amplifier that supplies V1. When I place an o-scope on the V1 line, prior to R1, I see a voltage offset proportional to the reference voltage, V2. Is there any way to stop this circuit from loading the circuit that is supplying V1?

By: ChiTownRick

ChiTownRick — Tue, 29 Dec 2009 20:13:14 +0000

Thank you very much for the quick feedback! Much appreciated.

By: Adrian S. Nastase

Adrian S. Nastase — Tue, 29 Dec 2009 00:27:01 +0000

Indeed, if you calculate the resistors with V2 = 3.3V and the input range of -5V to +5V, one of the resistors is negative. A negative resistor means that the circuit needs an extra voltage source to be able to output 0 to +3V.

There are two solutions:

1. Reduce V2. A non-inverting summing amplifier cannot have a sub-unity gain, so it cannot attenuate signals by changing the gain resistors R3 and R4. From an input range of 10V, the output range has to be 3V. This is attenuation. One can always attenuate the input range by making R2 smaller than R1. As a consequence, V2 has to be reduced as well. Therefore, as an example, use two 10k resistors to get 1.65V from your 3.3V source. Use another op amp in a voltage follower configuration to feed 1.65V to R2. With 1.65V as V2, the resistors are: R1=10k, R2=3.32k, R3=10k, R4=2.1k.

2. Use a 3 input non-inverting summing amplifier. Two inputs are V1 and V2 and the third is connected to ground. The third input resistor will create an extra attenuator for both V1 and V2. The advantage is that you need just one op amp. The disadvantage is that the calculations are more complex, but not unbearable. There is more to this than I can write in this comment, so I will write an article about this method. Please subscribe to this website using the RSS feed. The article should be up soon.
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Update: I published this article. You can find it here:
Design a Bipolar to Unipolar Converter with a 3-input Summing Amplifier

By: ChiTownRick

ChiTownRick — Mon, 28 Dec 2009 21:56:01 +0000

Having trouble converting an input signal with the range of -5V to +5V to an output range of 0V to 3V. V2 has a value of 3.3V.

By: jorgecosta

jorgecosta — Thu, 05 Nov 2009 21:27:26 +0000

Thank you very much!

By: Adrian S. Nastase

Adrian S. Nastase — Sun, 01 Nov 2009 02:56:05 +0000

jorgecosta, start with the summing amplifier transfer function (equation 3 in this article). Write 2 equations, one for 0V output and one for 5V output as follows:

0V = (-2V*R2/(R1+R2) + 5V*R1/(R1+R2)) * (1+R4/R3)
5V = (+2V*R2/(R1+R2) + 5V*R1/(R1+R2)) * (1+R4/R3)

This system has 2 unknowns: R1/R2 and R4/R3. Solve, choose R3 = 10k and R2 = 10k and calculate R1 = 4.02k and R4 = 7.5k.

If you need the output voltage range to be 0 to 4.9V just replace 5 with 4.9 in the previous equations. The new equations are as follows:

0V = (-2V*R2/(R1+R2) + 5V*R1/(R1+R2)) * (1+R4/R3)
4.9V = (+2V*R2/(R1+R2) + 5V*R1/(R1+R2)) * (1+R4/R3)

The result is R3 = 10k, R2 = 10k, R1 = 4.02k and R4 = 7.15k.

By: jorgecosta

jorgecosta — Fri, 30 Oct 2009 00:42:54 +0000

Can you also show the math method to design the converter? I am interested in the converter with input -2V to +2V and output 0 to +5V. I also would like to know how can I design the converter so the output reaches just below +5V, maybe 4.9V.

Thank you for your time.