October 4, 2004

What does this circuit do?

...and how come "circuits" isn't pronounced like "biscuits"?

I've been fiddling with an ADS1251EVM and am a bit perplexed about what happens to the signal before the ADC. I would have expected something straight-foward, but the circuit is a bit unusual.

ads1251evm-input.PNG

We have here inverting amplifiers, U9 and U8, but the strange thing is that they take in (basically) the same signal at the negative terminal. The positive terminals have DC voltages coming from a voltage divider circuit. The first op amp gets 1.25 V and the second 2.5 V. Following the "golden rules" of op-amp behavior, U9- will also be at 1.25 V, which means Vin-1.25V across R13, And 1.25V-Vout across R12. The inputs "drawing no currents" means that

(Vin1-1.25V)/R13 = (1.25V-Vout1)/R12
Vout1 = 2.5V - Vin1
Hmm, I can see how this is going to work. In U8, we have the same situation, with 2.5V substituted for 1.25V, so that Vout2 = 5.0V - Vin2. But Vin2 is Vout1, so Vout2 = 5.0V - 2.5V + Vin1 = 2.5V + Vin1. So, when we do Vout2 - Vout1 = 2*Vin1. Why was this necessary? Buffering? I'm very confused.

Update. Sleep always help - and now it makes sense. The op-amps act to shift the inputs. Notice that Vout1 = 2.5 - Vin1 and Vout2 = 2.5 + Vin1. Since the ADC limits input voltage between 0 and 5 V, this lets Vin1 be between -2.5 and 2.5 V. Which is what you would want it to be anyway. This way you don't need to worry about changing the ground of your input signal. You can take input directly from an oscillator, for example. Good, good. I was just starting to worry about this.

Posted by torque at 11:27 AM | Comments (1) | TrackBack