After further research, it seems that the R/2R ladder is better for me, especially when dealing with 12 bits.

Here is the issue.
I have a pcb setup so that the DAC circuitry itself (just resistors, and the cmos counter (4040)) takes up more space than my regen receiver :shock:

If I removed the resistors, I would save about 2cm * 2cm of space. In fact, that is all I need if I made the super regen without any trimmers or DAC's.

I was wondering, what is the best way to minimize the required PCB space for my 12-bit DAC?
and yes, I'm using resistors as jumpers as well.

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Below is the picture of my PCB

The grid is in the resolution of 1 mm.

The blue section is the DAC component.

If there is some way I can fit the entire circuit into a 4 cm by 3 1/2 cm space, (or 3 is even better), that would be great.
I bought small boxes, and buying bigger ones costs dollars more.

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simple answer = surface mount parts...

0805 package resistors are just millimeters long, and 0603 are even smaller... and if you can find the IC in a narrow SOIC, it will be about 1/2 to 1/3 the size of the through-hole version...

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EEgeek.net

 

The nice thing about R/2R ladders is that they only have two resistor values in them. Unfortunately, even a 12-bit R/2R DAC will require tolerances of better than 0.025% (12 bits=4096 steps, 1/4096~0.025%) if you want a monotonic output. This is the reason high-resolution DACs cost so much more than lower resolution ones. The resistors have to be laser-trimmed before the networks are encapsulated.
Have you seen Maxim's programmable capacitor? It only has 32 steps, but I thought you might like to be aware of it. Xicor and Intersil (and maybe others) also have them. Google "programmable capacitor".

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Ron

 

Other than surface mount?

Is there anything I can do so that resistors with 5% or 1% tolerance work?
like maybe changing or adding resistor values?

I'll look at programmable capacitors.

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No, you can only make a lower resolution D2A with lower quality resistors. You 'could' try matching resistors, but how are you going to match them to 0.025%?.

For a start I would check if they are suitable for VHF work?.

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Here is the "standard" equation many websites claim for a 4-bit R/2R ladder:

Vout = Vin * (1/16 + 1/8 + 1/4 + 1/2)

It seems that the only way that equation works is if the output of the DAC is either tied high or tied low. In my case, I tied it low, and the equation works.

I need to know this, because then I want to understand why high tolerances (> 1%) are a problem. I know Ron did a chart before, but I want to see (or make) a chart that illustrates the possible ranges of resistances, because I want to make the assumption that a 5% resistor is 2% off.

Would it be a benefit if many resistance values are low (< 10K) and one or two are high (> 100K)? or am I going the wrong way?

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from further research:

Take a look at:


http://www.evaluationengineering.com states that this is a String DAC.
Check this quote out:

"Figure 1" is the picture above.

Does it work for binary values?
and is that site telling the truth?

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Search for "digital pots". Lots of companies make them. I didn't mention them because they only seem to go up to 10 bits (1024 steps), and you said you wanted 12 bits. They have internal decoders, so they will work with binary. Some are up/down pulsed units, which might be better for you.

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Ron

 

No, and if you apply a couple of seconds thought to it you will soon realise why - if a couple of seconds of thought don't help, try ohms law!.

Yes, but I hope you are aware that it's ONLY a 3 bit D2A (8 possible outputs) - to make a 12 bit version would require 4096 resistors, and I though you were short on space? 8)

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Yeah, but you can probably get samples of the 10 bit digital pots, and increment-decrement will work fine if you don't need random access (and you won't need a 10 bit counter!). Here is a page that lets you look at Maxim's digital pots.

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Ron

 

it looks like I'll have to go with the R/2R network.

and I'll have to make use of my ohmmeter quite often.

I do plan to buy a large number of resistors (100+) for this and other resistor hogging projects.

I was wondering, could each 5% resistor (independent of manufacturer) vary as much as 2.5% above or below its value, even if I am buying all the resistors in the same strand? In other words, would resistor 1 equal 1% off the value, resistor 2 2% off, resistor 3, 3% off, or no?

I'm wondering because I want to avoid measuring 100 resistors with an ohmmeter, and finding 12 or more that are exactly the same value.

and looking at the bands won't work, because I'm interested in tolerance.

also, 1% resistors cost twice as much as 5% ones.

__________________
-=: The best low-priced components to troubleshoot with are the speaker and the LED :=-

 

No. They could be plus or minus 5%.
If they were all made at the same time (taped together) then most will be nearly the same value. You will need to measure and sort each one.

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Uncle $crooge

 

I would suggest buying 1% resistors, as you've not much chance of matching them that closely on your meter - which is probably no better than 1% anyway, and will be at least +/1 one digit as well.

Obviously this makes it pointless building a 12 bit DAC, but 10 bits would be plenty - and having 1024 different voltages is much more than enough for the full FM band.

But I would suggest you forget the DAC altogether for now! - use a simple potentiometer to provide the voltage for the varicap, and proof it works, and that you have sufficient tuning range. Once that's working fine, THEN start trying to replace it with a switched solution.

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I have a 4-bit binary-weighted DAC already working for me, but I realize that after it was built, the bits were backwards (the high bit as bit #0 instead of the low bit being bit #0).

I discovered that the 2N3904 transistor makes an EXCELLENT varicap which contributed to most of the success.

The reason why I wanted 12 bits is because I want to be able to fine tune a station and increase the range. I am using my receiver with my transmitter. You would obviously know that if the transmitter and receiver are on the EXACT same frequency, the transmitter can transmit to the receiver better.

I think that my 2N3904 works as an 8pF voltage controlled capacitor. I have the emitter and base tied together. Correct me if 8pF is wrong.

If I only used 4 bits, the steps would be in 1/2 pF increments, which is a little too large.
I was looking at 12 bits, because I have a counter that goes up to 12 bits.
Higher bits means finer resolution, and better chances of obtaining every single channel possible.
I think I'll have to do some math before continuing.

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