Most Analog to Digital Converters have a unipolar input that can be a problem when designing bipolar circuits. Some common ADC input voltage ranges are 0 to 2.5 V, or 0 to 5 V. However, the analog circuit that drives the ADC can have voltage swings of, –1 V to +1 V, –2 V to +2 V , –5 V to +5 V, and so on. Bringing the ADC input below ground is a big No-No, because the current from input will flow through the chip substrate creating irreversible changes in the ADC and damage it. So, how do we connect a bipolar front end circuit with a unipolar ADC? Enters the bipolar to unipolar converter. Let’s design one.
The converter can be designed with a summing amplifier, as in Figure 1. How do we calculate the resistors?
Figure 1
For this, let’s take a numeric example. Let’s say we have a –5 V to +5 V input, and we need the converter to output a voltage range of 0 to +5 V for an ADC with the same input range. If the converter output has to swing in the same direction as the input, which means no phase change, the design requirements are as follows:
If Vin = –5 V, then Vout = 0 V,
If Vin = +5 V, then Vout = +5 V,
where with Vin I noted the converter input voltage and with Vout, the output voltage.
One can use the mathematical approach I described in this article, Solving the Summing Amplifier. However, in this case things are simple. We can easily determine the gain and offset of the bipolar to unipolar converter, from the design requirements. This method is more intuitive than the math approach, adding to the idea that Electronics design is an art form.
Any linear circuit has a gain and an offset. The general equation is as follows:
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(1) |
There are two unknowns here, Gain and Voffset. But, since the output swing is half the input range, the gain has to be 0.5. Let’s note that down: Gain = 0.5.
Now, to determine the circuit offset, we can replace in equation (1) one of the input/output conditions, say Vin = -5 V and Vout = 0 V and determine Voffset. Even so, let’s make this more fun and reason to find out what the offset should be. If we use a gain of 0.5, the output will swing from –2.5V to +2.5V. To make the output go from 0 V to +5 V, we need to add an offset of +2.5 V. Hence, Voffset = +2.5 V.
With the gain and offset known, let’s write equation (1) again:
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(2) |
It should be clear now, why I chose a summing amplifier. There is a plus sign between Vin and offset. Now, let’s look at the summing amplifier transfer function:
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(3) |
We can rewrite equation (2) as follows:
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(4) |
If V1 = Vin and V2 = 5 V, it is clear that we do not need 1 + R4/R3. This means that R4 is zero (just a short circuit) and R3 is infinity (disconnected). Also, if R1 = R2, the ratios R1/(R1+R2) and R2/(R1+R2) are equal and equal with 1/2.
We can choose any practical value for R1 and R2, say 10 kOhm. With this, the bipolar to unipolar converter design is done and it is shown in Figure 2.
Figure 2
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Categories: Analog Design, Electronic Circuits Examples, Mixed-Signal Design, Summing amplifier
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.
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.
Thank you very much!
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.
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
Thank you very much for the quick feedback! Much appreciated.
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?
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.