what is the analog ground plane?

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Hello, what is the analog ground plane? In the datasheet for the ATmega48 has this picture:

this figure showing how the AVcc pin should be connected to the digital Vcc supply voltage via an LC network, but not explained what is the analog ground plane and how do I accomplish it? Is it necesesairly?

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It is typically a separate region of the circuit board covering the area shown in the dotted line. It is connected to the ground pin next to AREF and the AVCC bypass cap is also connected. The grounds for the analog inputs are attached to this ground.

If done this way, it will be connected to the system ground inside the chip, so it is not "floating". The reason for doint this is to keep the currents associated with the processor clock and logic transitions away from the voltages being read by the ADC. The goal is to reduce the noise on analog input voltages.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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An analog ground plane is a separate plane of copper in the PCB under the sensitive analog pins. This plane attaches to the digital ground plane at only one point. The preferred point is right at the AVCC filter capacitor. The point of doing this is to prevent the digital ground currents from other signals from flowing under the chip in the area around the sensitive analog pins.
Remember your AC theory about induced currents? The legs and bond wires of the AVR act like antennas and pickup noise radiated from other current carrying conductors. For the analog stuff, these induced currents cause a lot of noise in the ADC readings. An analog ground plane shields the pins and gives quieter more stable ADC readings.
If you do not use the analog ground plane, the board will still work, but extra averaging of readings will be needed to try to eliminate the noise.

---- Steve

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Ok, I think I get it. If I understood clearly, there is two grounds (the digital ground and the analog ground) and they are connected together at only one point on the board, right?

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Right. As typically suggested in the AVR spec sheets, that point appears to be inside the chip via the analog ground pin (next to AREF).

Grounds and connecting grounds tends to be a bit of "black magic". Changes tend to be subtle and difficult to track down.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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exactly

Klaus
********************************
Look at: www.megausb.de (German)
********************************

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Ok, thanks.
I would like to see some circuit boards who has accomplished this.
Here is a part of my circuit board

if someone can tell me where to put the analog ground?

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I would put a back side flood plane around the ADC pins on the micro, around the adc input pins and under the traces from the input pins to the micro. There should be a bypass cap on AVCC pin with the ground end going to that flood plane. Flood plane also attaches to the ground pin next to AREF.

And, for best performance, you should have a general ground flood on the back, where you can, connected to the supply ground. It will save you much headache (and money for a new board)!

I tried to fill an example area but your image does not fill nicely in Paint. Understand that it would be behind the traces, with openings only for the pads that don't connect directly to it.

Jim

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Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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Ok Jim, now I get it, thank you very much for this :D, also thanks all folks for help in this topic :D

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Another point:

Your analog inputs all have to have a "ground" connection. You only show the "positive" connection but there has to be another one, somewhere. That connection should be to the analog ground.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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The ground of my analog inputs are directly connected to the digital ground in one point ( they are connected to the ground of my power supply). I have not knew that I must split the digital and the analog ground. In your above picture the blue area is the analog ground and I must connect it to my digital ground. After this procedure I again have only one ground, becouse they are connected together, this is little confusing me again.

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Here is the issue and the tradeoff.

On clock edges (of the micro), VERY high currents can flow in very short transients. A first order estimate of some of the older (and larger gate-count) AVRs suggests spikes on the order of 100mA lasting for 10-ish nanoseconds. That current flows from the supply, through the VCC pin, the digital ground pin, and back to the supply.

Now, imagine, for a moment, a ground system with an impedance (impedance because the frequency components are high enough that inductance is important) of 0.1 ohm between your supply ground point (where your analog signals are also connected) and the processor. That suggests a voltage drop (during the current spikes) of 0.1 ohm * 0.1A = 0.01V = 10mV.

This voltage is now inserted, in series, with the voltages you measure with the ADC. If you use a 5V reference and all 10 bits, you have a resolution of 5mV. You have potentially lost your whole least-significant bit! If your reference is lower, the penalty is even larger. With a 2.56V reference, you now loose the two least-bits!

Now, it is possible to argue that those are only spikes, but, because the ADC sample/hold is sync'd with the clock, one will almost always happen at (almost) the same time the ADC samples!

So, you can leave your signal ground connected where it is, and probably suffer higher noise. Or, you can split it and probably get lower measurment noise! Then again, your application may not need the lowest noise, and you can ignore all of this! Its your choice. And, its a very common choice to have to make in systems with analog-digital converters. It is NOT just an AVR "problem".

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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Only one more thing, the path (a part of coper in PCB) who connects those two grounds, is it the wide or is it the narrow path?

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From the AVR data sheet, it sounds like they suggest making the connection inside the processor. Those two grounds are connected, internally. So long as you don't have significant current though your signal ground leads (ie, sensor power), then I would let that be the connection. If there is a chance of current (more than a milliamp, perhaps, just a wild guess), I would connect them with an ordinary signal trace.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

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Now I get it all, thank you very much for help and explanation :D