Thanks for the help.

 

If the output falls to 0, the transistor comes out of conduction and the loop current goes to 0. In a 4 - 20 ma loop, that is a failure indication.

 

4-20mA is a common method of measurement in industrial applications.

For example, pressure inside a pipe. Lowest reading would be 4mA, highest would be 20mA. If it falls out of that range, there is a failure somewhere. Using the current source instead of voltage source allows for multiple instruments in series without seriously degrading the signal.

 

If the output falls to 0, the transistor comes out of conduction and the loop current goes to 0. In a 4 - 20 ma loop, that is a failure indication.

How would I cause the transistor to go out of conduction in this circuit? I'm interested in what i can do to cause this. Thanks.

I hope I'm not asking you guys too many questions.

 

The source of the current is the supply on the right with the battery symbol, labeled "External Loop Supply 7V-36V"

 

The source of the current is the supply on the right with the battery symbol, labeled "External Loop Supply 7V-36V"

Hmmm .... so the OpAmp plays no role in supplying any current? So what is the "real" controlling element?

The DAC is setting some output voltage and causing a voltage drop across RS. If the source of the current is coming from the Loop Supply then how is this thing driving current to the load?

Is it just taking whatever current that comes in from the supply and sending out the 4-20 mA on the return? The reason I am asking is because I thought the base current was suppose to be almost equal to the emitter current so I though that the OpAmp was driving current out at first.

Any explanation would be greatly appreciated.

 

The drive to the base of the transistor controls the current in the loop from the external battery, through RS.

The feedback voltage across the sense resistor, along with the output of the ADC are inputs to the amplifier which varies the drive to the transistor, using Rs as feedback for the correct amount as the ADC needs.

 

This circuit can be a bit tricky since it floats on top of RL.

The triangle (SG) is a floating ground, just connect all triangles if this makes it easier.

The opamp's non-inverting input has the same potential as the SG.

The DAC can swing 0-5V with respect to SG which means it can inject current through R3. This will lower the potential of the lower side of R3. This means a voltage drop across RS.

If you assume that the opamp is active then the voltage is zero between it's inputs.

 

The drive to the base of the transistor controls the current in the loop from the external battery, through RS.

The feedback voltage across the sense resistor, along with the output of the ADC are inputs to the amplifier which varies the drive to the transistor, using Rs as feedback for the correct amount as the ADC needs.

I see what your saying. But what ADC? Do you mean the DAC?

 

I see what your saying. But what ADC? Do you mean the DAC?

My Bad, Typo. The keys are like, right next to each other.

 

The subject as you can read is:
4 mA to 20 mA Process Control Loop Using the AD5662 DAC

It's a DAC. You have quite a lot info at the link you have mentioned also. If you are in doubt you could also read in the datasheet for the part in mind.

 

Look up "precision current source" or "high side current source" or "voltage to current converter". National has an example in their Op Amp Circuit Collection ap-note. There's many examples of designing a circuit exactly as you describe. The op-amp essentially acts as a high side voltage to current converter. The transistor/s on the output of the op-amp enable higher current capability. The calcuation of the output usually simply comes down to Vin / Rsense which makes it easy to calculate values.

 

I knew it was an ADC and I've read the application note quite a few times but was not understanding how this circuit was doing this regulation of current.

I'm actually interested in changing this circuit from a 4-20 mA circuit to a 0-20 mA circuit. Any ideas? I'm thinking of just adjusting the resistor values but would that be enough to get me 0 mA, i.e. no transmission whatsoever?

How would I go about turning this thing off?

The voltage to operate this circuit is generated by the voltage drop across Q1 and D1. Since the voltage regulator is 5V then the voltage drop across Q1 an D1 needs to be greater than 6V. Going to 0 mA is not possible because the circuit need the current below 4mA to create the drop across Q1 and D1. This circuit here is 2 wire transmitter you would need a 3 wire in order to modify it for 0mA output.

 

The lowest current you can get is the current consumption of the whole circuit which I don't know how much that can be.

What exactly are you up to? It's possible you can settle for a much simpler solution.

 

you could take it to zero with some resistor adjustment and a dual polarity supply on your amp.

 

I'm coming into this discussion really late.

Low-pass filters generally need at least three connections; input, output, and ground. The undesired frequencies need to be dumped somewhere. Guess where? Ground.

If you need a low-pass filter, we can probably whip up something. If it's really low frequency, you might look at using an active filter rather than an LC filter.

Texas Instruments has a program called FilterPro that's available for free, here:
http://focus.ti.com/docs/toolsw/folders/print/filterpro.html

Filters aren't cure-alls, but they can certainly make a big dent in big problems.

 

The LPF notwithstanding can you help me in determining the proper feedback connection to get this circuit working like the original.

I thought that's what I was doing already?

The reason the LPF is there is because I can have some AC riding on the DC that I want to feedback. I can design a LPF and I probably should have not put that in there but lets just assume I dealing only in DC at the moment.

Thank you for replying though.

Try experimenting with the TI filter designer. It can do lots of good things.

 

OK, I'm looking at your circuit. The AD8627 that's driving the base of Q8 - I'm wondering if you meant to just have the cap in the feedback path? Except for that, basically, the opamp is running open loop. Did you mean for it to be like that?