How to Add a 5-Second Soft Start to LM334Z 2mA Current Regulator Circuit?

Hey all,
A bit of a newbie to creating circuits
I have a 12V battery regulated to 2mA with an LM334Z current regulator.
Here are the docs for the current regulator: http://www.ti.com/lit/ds/symli...
I would like to create a soft start, so that when I turn the device on there is not a rush of current.
I'd like it to ramp up over 5 seconds.
I've breadboarded a few schematics I found online but I can't get them working. Even if the battery power is slowly ramped up, the LM334 rams the voltage to get the current at exactly 2mA.
Here's a couple I've tried unsuccessfully.

Also I was using different transistors
I even tried this schematic to no avail:

But again was using different transistors, didn't think that mattered.
Can anyone help me figure this out? I've been stumped for days.
Thanks
Nick

Try this...

Will do

Should this take about 5 seconds to ramp up?

Not sure. You might have to tweak the values a bit. Specifically C2 and/or R3.

I couldn't get it to work. Can you help me understand how the circuit works so that I am not blindly following the schematic but I can debug myself with more understanding?

Appreciate your help.

Thanks

Sure... Before power is applied (i.e. the 20 volts), the capacitor, C2, will be discharged, and thus will have zero volts across it. As the output of the regulator rises it will pull the base of the transistor up through C2 and thus turn it on. This will create a low impeadence path to ground, around R1 (essentially shorting it out), This will hold the ref pin of RG1 to near ground thus holding the output of the regulator at around 1.26 volts (1.25 + the approx .1 Vce on Q1). As C2 charges through R3 and the base of Q1, the voltage across C2 will increase and thus allow the output of the regulator to rise, because it will take a higher voltage from the regulator to balance out the negative feedback that is applied to the adj pin via the transistor. The output voltage will continue to ramp up until Q1 is all the way off and R1 takes over the task of setting the output voltage of the regulator.

Thank you very much for replying on this. I helps a lot.

I'll debug the circuit and keep trying

BTW: This is basically an applied capacitance multiplier. You could do away with Q1 and R3 and place C2 across R1, but to get 5 seconds you would need something like a 4200uF capacitor. By using the transistor as a capacitance multiplier, you can do it with a much smaller capacitor.

Do you think there's any way to make this circuit almost 100% failsafe?

The current will be limited to 2mA max- but I have read that the regulators can fail.

I would like to design a failsafe for the circuit- however a 5mA fast blowing fuse is pretty expensive.

Curious to hear your thoughts.

Nick

See my last reply- but I have an update-

So I figured out why I was having so much trouble. The circuit still jumps to 2mA very quickly, but I started to monitor the voltage and saw that gradually rising. The problem is that the LM334 is a current regulator, not a voltage regulator- so it just raises or lowers the resistance to keep the current running at exactly 2mA regardless of voltage. Doesn't cause the amperage to raise slowly, just the voltage.

Am I using correct terminology?

Let me know your thoughts.

Thanks

What is your application of the current source? Given the problems you have encountered below, there may be a better way to solve your issues.

Nick, I'll need some specs before designing a soft-start circuit for the LM334.

Specs:

1. *Ramp shape*, exponential or linear? If exponential how close to 100% current do you need at 5 seconds? I.E. 99%?
2. Maximum *fail safe* current?
3. *Compliance voltage*? That is, what is the maximum voltage the circuit must deliver 2 mA into?

I'm realizing that I haven't been paying attention to the discrepancy between your original request, involving the LM334Z, and the schematic with the LM317, that you attached.

I assumed that the LM317 circuit was to precede the LM334Z current regulator and that the problem was focused on getting the LM317 output voltage to ramp up.

BUT, the question that I should have asked was: what is the "big picture" here. In other words, how, exactly, do you want the two circuits to interface and/or what is the overall intent? You want this to "soft start" in order to prevent an inrush of current, right? Then, why not focus on getting the current regulator to ramp up?

The attached circuit does this AND it limits the current to around 3ma, so if the LM334Z gives way, the current won't go any higher (it just won't be well regulated).

How the circuit works: Initially, C1 has zero volts across it, which holds Q2 in the off state. When the 15V is applied, C1 begins to charge through R2 and as the voltage across C1 rises, it gradually turns Q2 on. When the voltage across C1 reaches around 1.4V, Q2 will be fully on. During that time (when C1 was charging from zero volts to around 1.4V), the current to the LM334Z (represented by the constant current source), is being limited to something less than 2ma. Try as it might, the LM334Z can't produce more current than is available in the leg it is in series with, so Q2 will succeed in ramping the current up until it reaches 2ma, after which the LM334Z will take over. The timing on this is quite temperature dependent. But, you can adjust it by changing the value of C1. The reason it needs to be so high, is that it is being charged by a fairly high voltage (15V), yet has it's affect at a rather low voltage (.7V).


Q1 and R1 act as a current limiter. If the voltage across R1 reaches around .7V, Q1 will turn on and start to turn Q2 off. When Q2 turns off enough to limit the current to just as much as is needed to cause the right voltage across R1 to keep Q1 turned on just enough to keep Q2 limiting the current through R1...etc. Then the current will stabilize at around 3ma. This can be adjusted by changing the value of R1: R1 ~= .7/IL where IL is the limit current. Again, the actual limiting current will be quite sensitive to temperature and is rather imprecise, so not a good current regulator, but works great as a protection device.

Forgot to attach the diagram...

Also, I noticed that you refered to a 5ma fuse. To get this circuit to limit current to around 5ma, try replacing R1 with a 100 ohm to 120 ohm resistor.

But, that will limit the current to 5ma, not interrupt the current.

Not having specs from you I designed the following circuit that has an exponential current ramp. In simulation:

Tempco = 250 ppm/deg C
Voltage Compliance = 11 volts given a 12 volt power supply
Output Z = 95 k ohms