Hi All ,
I have an requirement to convert the 8 channel analog input (Single ended - bipolar) signal with the range of 10Vpp, The waveform will be sin, ramp, square,and Sweep.
Hence I am planing to use AD7768. In this I need a clarification of following points:
1. I have 8 channel single ended bipolar signal inputs (10Vpp, +/- 5 ), So I will connect this signal to the correspoding port of (AIN0+ to AIN7). My question is, May I Ground it the remaining ports of AIN-0 to Ain7- ?
2. My input signal should support (+5V to -5V , 10Vpp), so in REF1- and REF2- may I connect to Ground or -2.5V should give ?
3. If I ground the REF1- and REF2- , The AD7786 support the input signal swing 10Vpp?
4. Even though My Signal is single ended bipolar , Is there I want to make this signal to converted into differential bipolar (True Biploar signal)?
Please clarify the points as soon we are going to finalise the design stage.
Thank you very much for your inputs !!!
I am going with AVDDx, REF+x = +2.5V using ADP7118 ; and AVSSx , REF-x = -2.5V using ADP7183ACPZN2.5-R7
Is this LDO Current is enough for AD7768 Supply ?
I have an rearranged architecture of my 8 Channel Receiver card.
Before the architecture is :
Re-arranged architecture :
Can you give me your clarification of following parts.
Is this right way of attenuate my signal range ? or do you have any recommended part for this design?
This single ended to differential end conversion is there right part to do the conversion instead of you recommended LTC6363.
Please confirm may I go ahead with this part ?
Thanks and Regards,
I just stumbled across this post, and wanted make some comments on the signal chain you have posted on August 21.
You show the ADA4922-1 driving the AD7768 - The ADA4922 will cause performance degradation of the AD7768 at the maximum sample rate. As the AD7768 samples the output of the ADA4922, it injects a small amount of charge onto that node at the beginning of each acquisition. The amplifier that drives that node needs to be able to recover from that disruption quickly, and drive that node back to the proper voltage.
Niall is recommending the LTC6363 or the ADA4940 b/c they are capable of driving the AD7768 at it's full sample rate.
The ADA4922 *might* be able to drive the AD7768 when using a much lower fmod than the maximum value. Here's a snippet regarding fmod from the datasheet:
One of the specs that's critical in this application is GBWP:
Faster is not necessarily better, but for this ADC, 40MHz GBWP is probably too low for the driver to fully settle between samples.
On the other hand, if you're willing to use a lower fmod, the ADA4922 might work fine.
Yes - Looking at differential drivers, it looks like there's a limited selection for higher supply voltages.
If you are willing to slow down the sample rate of the AD7768, then the ADA4922 will settle fine. You may run into performance issues at the fastest sample rate.
I am looking for 24 bit ADC. Then I saw this post.
I can see many things happing in signal conditioning before ADC. I am curious and want to know the Signal conditioning and ADC that is coming for this Post.
1. ADC7768- Since in the post, we are Differential input in the range of ± 1.25/2.5V.
a. It means we are attenuating the signal. So is this the right way to do condition monitoring?
b. Can we use this ADC for ±5V range?
c. Can we use AD7176-2 in place of AD7768 for ±5V range (http://www.analog.com/en/design-center/reference-designs/hardware-reference-design/circuits-from-the-lab/cn0310.html)?
In Signal conditioning-
2. We are Passing Differential signal to AD8253 (PGA) that is converting differential signal into Signal ended. Then We are using LPF- LT1697IS (Dual/Quad Low Noise, Rail-to-Rail, Precision Op Amps-) But it an Op-amp.
So how can we use this as an LPF? Can we not use differential LPF?
then We are attenuating the signal using Attenuator (LT1678)- But As per Mohhamad suggested that why not we can go for LT6363 (It can take differential Input (±5V range), work as an LPF and generate Differential Output). So if we use this IC (LT6363) we don't need to add LPF, attenuator, and Single-ended to a differential ended amplifier like- LT1944/ADA4922-1.
So As we are making the DAQ system we should use minimum component for better performance. Kindly suggest and if I missing any point for the DAQ system. Do we have any other Differential ADC to full fill above criteria?
I changed my Old architecture which I posted earlier,
The updated design architecture I mentioned in below:
Please let me know your comments if anything to be update in the schematic.RECEIVER_CARD_CH0_WITH_ADC7768_1.pdf
Mohamed: Now, I am not attenuate the whole range as per the (10 Vpp Differential Input from external world which I mentioned in the Block diagram) . The range I reducing to 5vpp with differential bipolar to matching the ADC differential input.
2.Can we use this ADC for ±5V range?
Mohamed: NO, ADC maximum range of supply AVDD1 to AVSS = 5.5V as max. if you use ±5V as supply, AD7768 will damage due to exceeding the maximum range. Instead of using ±5V you may be us ±2.5V Supply. Which will be the range of AVDD1 = +2.5V - AVSS = -2.5V = (+2.5 – (- 2.5V )) = 5V.
3.Can we use AD7176-2 in place of AD7768 for ±5V range
Mohamed: I am looking for 8 Channel ADC which have sampling rate more than 125Khz , Simultaneous sampling.
Hence AD7176-2 wont match my requirement.
Hello Mr. Mohammed,
Thanks for the reply and sharing the details for Analog chain.
As you replied that we can not use the AD7768 with Input voltage range from +5V to - 5V. But in the datasheet, it has given that We can use the ADC with +5V to - 5V range. Please see figure 40 and figure 41.
In the Figure, it is clearly mentioned that We can use the ADC with +5V to - 5V range. But the Integral Non-linearity (INL) will increase. Since as per the requirement of Sampling frequency we can go for the median mode.(Fs>108kHz). So please suggest.
Please see in the electrical characteristic page no.7 for VRef range and page no.14 for AVDD – AVSS range,
In that External Vref mentioned , 1V to (AVDD1-AVSS).
For AVDD- AVSS range
Hence you cannot use -5V as VREF range.
In Figure 40 and 41 tells that
If you use VREF = 5V (+VREF with respect to - VREF = 5V) the INL error will come this,
If you use VREF = 4.096 (+VREF with respect to - VREF = 4.096) the INL error will come this,
If you use VREF = 2.5V (+VREF with respect to - VREF = 2.5V) The INL error will come this,
You cannot use VREF is less than 1V.
Your VREF range should be with respect to - VREF.
For My application I used (+ VREF = +2.5V and – VREF = -2.5V ). = 5V
In datasheet not clearly mentioned that we can use the ADC with +5V to - 5V range.
Hope you understand my answer.
Thank you so much for the clearing the points.