Beginner: LM350 as constant current not working

[TBennettcc]

Okay...

I think I get it now...

You were telling him to put the resistor IN SERIES with (some part of the circuit... which part? The INPUT to the LM350?) and making the calculations using the current HE specified.... in order to only show 6 volts of potential to the LM350, AND to limit the current of the entire circuit to 700 mA. (But no, wait... the resistor is not limiting the current, we have the LM350 set up as a current regulator, and THAT is what is limiting the circuit to 700 mA, right?)

How does 6 volts give the regulator "plenty of headroom"? (i.e., what does that mean?) How would an absence of regulation be clearly visible?

(By the way, thank you so much for mentioning E12 values. Never heard of those before. I always wondered why there were 47 ohm resistors and 47 uF capacitors... Now I know! Any other obscure and hidden tidbits of information you'd like to share?! )

So you're using fault conditions to calculate the power dissipation of the resistor... this is assuming that the resistor is the only load on the circuit, that everything else has failed?

And if so, then under normal conditions, the resistor will only be dissipating ( 0.7 A * 5.74 V = ) 4 watts?

Posit:

Since our resistor only has 5.74 volts across it, that leaves 6.26 volts across the rest of the circuit. Since the current through the whole circuit is 700 mA, that means the total resistance of the rest of the circuit is 8.94 ohms? Minus the 2.7 ohm resistor leaves 6.24 ohms across the LM350? ...except that, according to the link from which Matt was working, R1 should be 1.8 ohms according to (I = V_REF / R1) in order to regulate to 700 mA... which makes the resistance of the LM350... 7.14 ohms? So then what happens when you add a load? The resistance across the LM350 decreases in order to compensate for the load?

Hmm...

Any thoughts?

In any case, thank you very much for your explanation, Steve. After reading through your post, I've been working through different questions I've had and come up with satisfactory answers. I have put some puzzle pieces together, and I understand a lot more than I used to.

 

[TBennettcc]

...(I just got a new electronics program, can you tell?)...

Care to tell us which one?

I've been using EAGLE CAD to do my schematics in, but honestly, it's quite a bit of a pain if you don't have EXACTLY the part you need in their library, as you have to go make it. I'd love to hear about the program you've found.

 

[(*steve*)]

You were telling him to put the resistor IN SERIES with (some part of the circuit... which part?

In place of the LED, as the load.

and making the calculations using the current HE specified....

Yep. If his circuit was operating correctly, the load needs to be in a sensible range.

in order to only show 6 volts of potential to the LM350, AND to limit the current of the entire circuit to 700 mA. (But no, wait... the resistor is not limiting the current, we have the LM350 set up as a current regulator, and THAT is what is limiting the circuit to 700 mA, right?)

I calculated a voltage drop at his designed current that would leave some margin for regulation within the regulator. Essentially, for this circuit you was 1.25V (across the current sense resistor) plus around 2V, so 3.25 is the minimum you'd want. I allowed 6 because if the regulator was not regulating the current the voltage across the load would appear significantly different. (possibly up to 10V or more)

How does 6 volts give the regulator "plenty of headroom"? (i.e., what does that mean?) How would an absence of regulation be clearly visible?

Kind of explained above, but if you imagine the regulator as having a transistor connected from the input to the output, there will be some voltage drop across it, and the base needs some current which means current from the input supply rail. If the output voltage is too high compared with the input voltage, insufficient base current can be delivered to the transistor and the regulator ceases to regulate. There are ways around this (low dropout positive regulators typically use a PNP pass transistor, but these designs are less stable and also require more current through the ground connection which can play havoc with using them as a current source.

Absence of regulation would show up as the output voltage (in this case the voltage across the current sense resistor) becoming unstable -- in this case probably dropping, and passing ripple from the input. The end result is that the current would no longer be stable.

(By the way, thank you so much for mentioning E12 values. Never heard of those before. I always wondered why there were 47 ohm resistors and 47 uF capacitors... Now I know! Any other obscure and hidden tidbits of information you'd like to share?! )

No problems. Tell me exactly what you don't know and I'll see what I can do

So you're using fault conditions to calculate the power dissipation of the resistor... this is assuming that the resistor is the only load on the circuit, that everything else has failed?

Yeah, it's the worst case condition.

And if so, then under normal conditions, the resistor will only be dissipating ( 0.7 A * 5.74 V = ) 4 watts?

Yep.

Posit:

Since our resistor only has 5.74 volts across it, that leaves 6.26 volts across the rest of the circuit. Since the current through the whole circuit is 700 mA, that means the total resistance of the rest of the circuit is 8.94 ohms? Minus the 2.7 ohm resistor leaves 6.24 ohms across the LM350? ...except that, according to the link from which Matt was working, R1 should be 1.8 ohms according to (I = V_REF / R1) in order to regulate to 700 mA... which makes the resistance of the LM350... 7.14 ohms? So then what happens when you add a load? The resistance across the LM350 decreases in order to compensate for the load?

I'm not going to check your figures, but they look pretty close.

Note that I told you that the pass element in the regulator is a transistor. A transistor can be modelled as a resistance (between collector and emitter) that is controlled by the current through the base.

So yeah, the regulator circuit is changing the resistance of the transistor to balance the circuit so that the appropriate voltage appears across the sense resistor (in this case).

In any case, thank you very much for your explanation, Steve. After reading through your post, I've been working through different questions I've had and come up with satisfactory answers. I have put some puzzle pieces together, and I understand a lot more than I used to.

Glad that my verbosity was of some use

 

[MagicMatt]

Care to tell us which one?

I've been using EAGLE CAD to do my schematics in, but honestly, it's quite a bit of a pain if you don't have EXACTLY the part you need in their library, as you have to go make it. I'd love to hear about the program you've found.

Well I am a complete beginner, so it's probably not as fancy as Eagle CAD... it does, however, have a fairly user friendly circuit simulator suitable for beginners like me.

It's "Livewire" from New Wave Concepts - retails in Maplin for £37 (I used some vouchers and got it for £20, so I'm happy!)
http://www.new-wave-concepts.com/ed/livewire.html

I expect Eagle CAD is far morecapable for layout, and Spice far more capable for simulation, but Spice is very expensive and it's the simulation I wanted.