# Single-ended to differential conversion and ADC application

I am not quite familiar with the things I am going to describe. So, I need some help over here.

I want to integrate an AD4001 ADC at a board I currently design. The thing is that my analog signal is single-ended and I need it to be differential to enter the ADC. The datasheet of the AD4001 proposes an application circuit, the one at the figure I attach. Based on the AD4001 datasheet, I selected the ADA4940 differential amplifier for converting my analog signal to its differential one. I downloaded the SPICE model from its site and ran it using LTSpice. Attached is its circuit.

My analog input signal is 0-5 Volts, so I used a voltage supply with 2.5V DC offset and 2.5V magnitude. Then, 2.5V as VOCM and 5V as VS. The ouput signals I get are the following: Is the blue-red differential output the right one of the green input? How is it connected to the VOCM? I cannot understand how the initial signal can be formed based on its differential one.

Is it the correct input for the ADC for my analog input? I have read that the ADC converts the difference between positive and negative signals, but their difference never reaches 5V.

Also, I would like a single supply system (5 V DC). But my analog signal is 0-5V. Does this mean that it is not feasible to implement such a circuit since Vref reaches ~4 V or I could connect 5V to Vref and V+ of the amplifier?

• Why are you using a differential A/D when the signal you are trying to measure is already ground-referenced? This makes no sense. – Olin Lathrop Jul 27 '17 at 13:18
• I want an SPI interface integrated in the ADC and a throughput as close to 5 MSPS as possible. I found the AD4001 with SPI and 2 MSPS and thought it matches... – Konstantinos Tsoumanis Jul 27 '17 at 13:21
• That's a silly reason to pick a otherwise inconvenient part. There should be many SPI A/Ds out there to chose from. Also, can you simply ground the negative differential input of the A/D? I haven't looked at the datasheet to see if this would be possible. Often differential input is a option, not a requirement. – Olin Lathrop Jul 27 '17 at 13:45

## 1 Answer

I have read that the ADC converts the difference between positive and negative signals, but their difference never reaches 5V.

Your single-ended to differential amplifier has a voltage gain of 1, so with 5Vpp input the differential output is 5Vpp. To get 5Vpp on each ADC input you need a voltage gain of 2, producing a differential output of 10Vpp.

I would like a single supply system (5 V DC). But my analog signal is 0-5V. Does this mean that it is not feasible to implement such a circuit since Vref reaches ~4 V or I could connect 5V to Vref and V+ of the amplifier?

A single 5V supply will work as the reference voltage if the amplifier has rail-to-rail output. However there will be some nonlinearity close to the supply rails and it may be difficult to get sufficient reference voltage stability. Therefore it might be better use a lower reference voltage and reduce the ADC input voltage range to match.

Generally it is not necessary (or even desirable) to 'max out' the ADC input range. Using a 4.096V reference with a differential signal range of 5Vpp (rather than the theoretical maximum of 8.192Vpp) loses less than 1 bit of resolution. However if you must have the full range of codes then simply adjust the amplifier gain to produce it.

• Thank you Bruce. So, if I could have a stable 5V voltage reference (e.g. output from a voltage regulator), then my reference voltage should be ok, right? And if I want the input range of the ADC to be 0 to 5 V as single-ended, how should I configure the whole circuit? Should I input the single-ended 0 to 5V signal at the +IN of the differential amplifier and increase its gain to 2 in order to get 10Vpp for the differential input of the ADC (5Vpp for each input)? Is this how it works? Should each one of the differential inputs of the ADC have 5Vpp in order to get the right result? – Konstantinos Tsoumanis Jul 29 '17 at 18:21
• Yes, 5V supply is OK if well regulated. Increase R1 and R3 to 2k to get x2 gain. A single-ended input of 0-5V is +-2.5V relative to Vcc/2, therefore I think R4 should be connected to +2.5V, not ground (unfortunately I can't verify this because I don't have an LTspice model for the ADA4940). Each ADC input should get 5Vpp, which is what occurs with x1 gain and 10Vpp (+-5V) single-ended op amp input. – Bruce Abbott Jul 29 '17 at 23:24