R 2R ladder DACS

[large_ghostman]

I have a question, but i also want the following answer if possible " YES IT's FINE LG".

In most R 2R ladder DACS there is often a opamp at the end, if the implication isnt driving much (in this case a sub mA coil on a meter), is it ok to do away with the Opamp? The Resistors will be 10K and 5K or might end up 20K and 10K, Please see above correct answer in Bold .

Other answers welcome but likely to make me grumpy

Thanks
LG

 

[ronsimpson]

The answer is 42, Mr. Grumpy.

I think a meter coil will load the DAC down but will function just fine.
I have not tried it by my guess is that you should be able to drive the meter and all steps will show up.
If the DAC is loaded down a high level will be lower that it should be but high enough to drive the meter.

 

[MikeMl]

I always believed that there must be an opamp to provide a virtual ground (or call it a current-to-voltage converter) at the output end of R-2R ladder. Put it another way, the ladder produces a current output that is accurate only when the output voltage is zero... That type is discussed here.

PS, that is one type of R-2R ladder. The other type is here, which implies that its load should be very high...

 

[large_ghostman]

Thanks guys, I thought I would ask on the off chance! I rushed a board and didnt put an opamp in the circuit, lazyness dictated I ask if I could get away with it or have to make another board! Dosnt have to be super accurate but it will bug me if it isnt right. I will try with the board i have just on the off chance the gods of luck are smiling upon me, then when I discover there not I will go do another board and pay attention this time!! The meter at full scale is only 100uA but I spose being lazy is no excuse not to do it right! At least I can get away with one of those opamps we dont mention

 

[AnalogKid]

If you have an 8-bit DAC, and you want the output signal to accurate to 8 bits, then the impedance of the load on the DAC circuit output must be at least 512 times the output impedance of the DAC circuit. The odds of the meter coil having a resistance of 2.56 megohm are low. Whether as a transconductance amp or a voltage amp, the opamp is an impedance buffer.

ak

 

[large_ghostman]

Thanks.
Not that it makes a difference but I only really need 4 bits, its for a fuel gauge.

 

[MrAl]

I have a question, but i also want the following answer if possible " YES IT's FINE LG".

In most R 2R ladder DACS there is often a opamp at the end, if the implication isnt driving much (in this case a sub mA coil on a meter), is it ok to do away with the Opamp? The Resistors will be 10K and 5K or might end up 20K and 10K, Please see above correct answer in Bold .

Other answers welcome but likely to make me grumpy

Thanks
LG

Hello,

The answer is absolutely without a doubt definitely maybe there might be a possibility

To find the answer, simply experiment with a 1 bit model. That would be two resistors of equal value connected in series, one resistor going to ground and one resistor going to the 'bit' driver, and the center tap becomes the output. So this is almost like a two resistor voltage divider except we feed it with the bit driver not the Vcc power supply voltage.

When the bit is zero, we get 0v output, when the bit is +10v, we get +5v output. That's the entire range of our 1 bit DAC.
The question is, what happens when we load the lower resistor with another resistor.
If we start with 100k resistors and we load the lower resistor with 1k, suddenly we dont get +5v out anymore when the bit is high. We get a much reduced output.
If we start with 1k resistors and load the lower resistor with 100k (just the opposite) then we only loose a little voltage at the output when the bit goes high.
So the idea here is to keep the load a much higher impedance than the resistors are.
Dont forget though that here the bit drivers have to be two state voltage sources.

Since there are only 4 bits, another possibility is to use a weighted resistor ladder.
These resistors would each be set to deliver the required current for that bit.
For example, values 1k, 2k, 4k, and 8k, which would deliver current to a lower impedance load.
May have to go to 100, 200,400, 800 ohms, or somewhere in between to get the right currents, or maybe even 10k, 20k, 40k, 80k.
In this scheme the bit drivers can be one state voltage sources plus an open state (like an open collector output) which is much simpler than the two state drivers.

We could do a little circuit analysis to better understand all the options.

BTW, i guess you dont like op amps

 

[BobW]

For 4 bits I wouldn't worry at all. For 8 bits, maaaaaaybe, and for 16 bits, then an op amp is probably a good idea. Most design info that you see published is erring on the side of caution. But in a lot of cases you can simplify things a lot and still get the performance you need. This is called "engineering."
In other words, try the simplest practical circuit you can think of, and if it works, then you're done. If it doesn't work, then go back to the drawing board and add a few more complicated bits. This philosophy has worked well for me.

 

[large_ghostman]

Hello,

The answer is absolutely without a doubt definitely maybe there might be a possibility

To find the answer, simply experiment with a 1 bit model. That would be two resistors of equal value connected in series, one resistor going to ground and one resistor going to the 'bit' driver, and the center tap becomes the output. So this is almost like a two resistor voltage divider except we feed it with the bit driver not the Vcc power supply voltage.

When the bit is zero, we get 0v output, when the bit is +10v, we get +5v output. That's the entire range of our 1 bit DAC.
The question is, what happens when we load the lower resistor with another resistor.
If we start with 100k resistors and we load the lower resistor with 1k, suddenly we dont get +5v out anymore when the bit is high. We get a much reduced output.
If we start with 1k resistors and load the lower resistor with 100k (just the opposite) then we only loose a little voltage at the output when the bit goes high.
So the idea here is to keep the load a much higher impedance than the resistors are.
Dont forget though that here the bit drivers have to be two state voltage sources.

Since there are only 4 bits, another possibility is to use a weighted resistor ladder.
These resistors would each be set to deliver the required current for that bit.
For example, values 1k, 2k, 4k, and 8k, which would deliver current to a lower impedance load.
May have to go to 100, 200,400, 800 ohms, or somewhere in between to get the right currents, or maybe even 10k, 20k, 40k, 80k.
In this scheme the bit drivers can be one state voltage sources plus an open state (like an open collector output) which is much simpler than the two state drivers.

We could do a little circuit analysis to better understand all the options.

BTW, i guess you dont like op amps

Thanks for all the info. Its not I dont like Opamps, it was more a case of spending alot of time doing a board and totally forgetting to include the opamp in the board bit! I was using proteus and for some reason the opamp was ticked as not to be included in the board layout. No excuse I was rushing and not paying attention!!! I didnt want to try it and fail so i thought I would ask and cross my fingers that someone would say ' yeah thats fine I do it all the time'.

I knew it was a slim chance at best lol. The scale at the moment goes from 1 to 10 but has 100 divisions, if I am going to redo the entire board (it has other things on) then I might go 8 bits and have the option of two scales or kind of two resolutions for the gauge. I ummed and arghed about it at the start and still not got a preference as to one scale (4 bit) or two scales (8 bit).

I am struggling on the layout of the instruments, if I do them so it feels natural to me then it likely it wont feel natural to anyone else, then i start over thinking it . I will post a update in the week on my UAV thread as its part of that.
I have a couple of things to catch up on like take some pics of soldering stations etc for specs. I am also changing a bearing on a rotavator that goes on the little tractor, i got one side donr then discovered the harder side to do is the one making the noise!! Not a big job but its heavy and without an engine hoist lifting it on and off the bench is back breaker

 

[large_ghostman]

For 4 bits I wouldn't worry at all. For 8 bits, maaaaaaybe, and for 16 bits, then an op amp is probably a good idea. Most design info that you see published is erring on the side of caution. But in a lot of cases you can simplify things a lot and still get the performance you need. This is called "engineering."
In other words, try the simplest practical circuit you can think of, and if it works, then you're done. If it doesn't work, then go back to the drawing board and add a few more complicated bits. This philosophy has worked well for me.

I shaould have it all soldered up later, I had a smd part that refused point blank to solder correctly! Its a home made board and sometimes I get the odd track that seems to not solder well despite cleaning the board! Not convinced the copper is that great as the boards are ebay specials.

 

[large_ghostman]

Bah, whatever happens its a new board!! I did it so rushed I didnt reverse the parts for the smd stuff!! So all the SMD tacks are back to front At least it was home made and I didnt buy 20 boards!!!

 

[dr pepper]

No problem ghostie, you can even loose a resistor, design the dac with the meter as the lower resistor in the dac.

 

[large_ghostman]

Pulled the board a few days ago from the feric chloride, all chuffed I had designed and made it in under two hours blah blah blah...........that will teach me lol