constant current sink understanding (10-50mA)

[Atom]

Hi to everyone, i need a little constant current sink circuit to put on the output of my bench psu so i can have always a little load for stability(that is not the main reason), but also for the turnoff, caps discharging ecc..

so i found this circuit , it a let's say crude but to me it's completly fine.

correct me if i'm wrong , if i understand correctly the 2 diodes in series generates a v drop of 1,4V .
by substracting the 0.7V of the vbe of the transistor we have a vdrop across R1 of 0,7V
we can calculate the current flowing with ohm's law so I=V/r.

i tried the circuit on a breadboard and it worked flawlessly but since the diodes v ref changes and the v drop of the transistor isn't constant the current taken varies let's say from 20ma to 30ma (also the transistor heats up)


then i remembered that the tl431 can be used as a control for a constant current sink but there isn't any kind of explanation on the datasheet of why it is used like that and since i don't like mindlessly copying schemes i wanted to understand it.
so could someone explain it to me or give me some hints ? any help is appreciated

[Benta]

An important piece of information is missing: what's the output voltage of your PSU?

[Atom]

An important piece of information is missing: what's the output voltage of your PSU?

sorry the output voltage is from 0 to 32 V .

[ahbushnell]

Hi to everyone, i need a little constant current sink circuit to put on the output of my bench psu so i can have always a little load for stability(that is not the main reason), but also for the turnoff, caps discharging ecc..

so i found this circuit , it a let's say crude but to me it's completly fine.

correct me if i'm wrong , if i understand correctly the 2 diodes in series generates a v drop of 1,4V .
by substracting the 0.7V of the vbe of the transistor we have a vdrop across R1 of 0,7V
we can calculate the current flowing with ohm's law so I=V/r.

i tried the circuit on a breadboard and it worked flawlessly but since the diodes v ref changes and the v drop of the transistor isn't constant the current taken varies let's say from 20ma to 30ma (also the transistor heats up)


then i remembered that the tl431 can be used as a control for a constant current sink but there isn't any kind of explanation on the datasheet of why it is used like that and since i don't like mindlessly copying schemes i wanted to understand it.
so could someone explain it to me or give me some hints ? any help is appreciated

The TL431 will hold the voltage on that resistor at 2.5 volts.  It does it by changing the amount of current shunted from the base of the transistor. 

[iMo]

The simplest current source/sink of around 10-20mA is a BF245C jfet.
Gate and Source wired together and to GND, Drain to Vcc rail (your PSU output).

[SiliconWizard]

An important piece of information is missing: what's the output voltage of your PSU?

sorry the output voltage is from 0 to 32 V .

Then it's not going to work down to 0V.

[Benta]

The simplest current source/sink of around 10-20mA is a BF245C jfet.
Gate and Source wired together and to GND, Drain to Vcc rail (your PSU output).

I agree completely that a JFET would be the simplest solution. The BF245 is only specified to 30 V, though.
Sadly, JFETs are difficult to find these days.
The 2SK170BL would be ideal, perhaps two in parallel to share power dissipation. The original Toshiba parts are obsolete, but Linear Integrated Systems offer them as LSK170B.

[Hiemal]

Arrow has these JFET's in stock which seem to be reasonably close to the specs of that JFET you guys mentioned.

https://www.arrow.com/en/products/2sk304e-spa/on-semiconductor

I bought these a few days ago and they're not surface mount, they're TO-92 esque package; just a little bit smaller. Not too terribly expensive either thankfully.

[not1xor1]

Hi to everyone, i need a little constant current sink circuit to put on the output of my bench psu so i can have always a little load for stability(that is not the main reason), but also for the turnoff, caps discharging ecc..

You might use the simple circuit below. With that a drift in the current value would be irrelevant as it would not affect the output current measure while still providing a minimal load and discharging the output capacitor (that is what HP used to call a down-programmer).
The lower compliance would be around 0.6-07V which IMHO is still reasonable and much better than the 2.5V (or 1.25V) of a TL43 (or LMV431).

BTW of course the relevant parts are just:
- R2 R4 you might reduce that to 47k or less if the current gain of Q3 is low (or the input voltage is low)
- C1 is for ripple rejection
- R1 sets the current to a value that is around 0.6/R1 (where 0.6 is Vbe of Q2)
- if the current is just a few mAs you might use a BC337 for both BJTs (Q2-Q3)
- the other components are there just to point where to insert the current sink, their values are irrelevant

[Zero999]

An important piece of information is missing: what's the output voltage of your PSU?

sorry the output voltage is from 0 to 32 V .

Then it's not going to work down to 0V.

No current sink will work down to 0V. Not without an external power supply. A J-FET will work closer to 0V, but the current will still fall, once the voltage drops below a certain point.

[Kalvin]

You can build a floating constant current load by using a [micropower] RRIO op amp, a MOSFET, a current sense resistor, a battery or a cheap isolated 5V dc-dc converter module. The external power supply will provide sufficient voltage so that the current sink will operate very close to the zero voltage. Typical topology would be something like this:

Just make sure that the MOSFET use select can be operated at the maximum load voltage (32V) and can dissipate the maximum power 50mA * 30V = 1.5W..

[Alex Nikitin]

For this application a JFET of better still a depletion type MOSFET (BSP129 for example) and a resistor between source and gate would be the simplest and best choice, as it will work as a current load down to a certain voltage (~1V for the BSP129) and as a resistive load below that point.

Cheers

Alex

[SiliconWizard]

An important piece of information is missing: what's the output voltage of your PSU?

sorry the output voltage is from 0 to 32 V .

Then it's not going to work down to 0V.

No current sink will work down to 0V. Not without an external power supply. A J-FET will work closer to 0V, but the current will still fall, once the voltage drops below a certain point.

Exactly, but that had to be pointed out to them as they might have not realized it. If that was obvious to them, that's all good. But that could definitely not fit their "0 to 32V" requirement.

With a single JFET, it'll still be at least 1V depending on the JFET, and there will be a wide dispersion, so a trimmer would be necessary.
There are also not many usual JFETs with a saturation current of at least 50mA, which was one of the requirements, and especially at low VDS voltages.

Actually, unless I've missed something, their approach with an NPN transistor should allow to get a lower operating voltage than with most JFETs as long as you use a low enough reference voltage IMO. You should get around 400mV or so minimum op. voltage with a reference voltage (base voltage) of 1V for instance (which is enough to ensure it will be above the max VBEsat), with an ADR510 for instance (not expensive). Current will also have to be adjusted of course, with a typical resistor value of 5ohm to 25ohm.

[Benta]

There are also not many usual JFETs with a saturation current of at least 50mA, which was one of the requirements, and especially at low VDS voltages.

Actually, the title of this thread is "10-50 mA", so where you got that from is anybody's guess.

Two 2SK170BL in parallel would do the job beautifully, and with much lower minimum voltage than the other solutions presented here.

[SiliconWizard]

There are also not many usual JFETs with a saturation current of at least 50mA, which was one of the requirements, and especially at low VDS voltages.

Actually, the title of this thread is "10-50 mA", so where you got that from is anybody's guess.

From the title. "10-50 mA" means from 10mA to 50mA to me, so potentially up to 50mA. Now if it meant anything in between as we see fit, nevermind.

(With this range, you'd need a JFET with at least 50mA saturation current to ensure a 50mA current sink. You can only adjust the current lower with a source-to-gate resistor, which is the common way of adjusting a JFET-based current source/sink. That was my point. You can also parallel several JFETs as you suggested, but that's more parts and more dispersion.)

Two 2SK170BL in parallel would do the job beautifully, and with much lower minimum voltage than the other solutions presented here.

Well, the 2SK170BL seems to have a ID vs VDS knee (@VGS = 0V) at a much lower VDS than with a lot of other common discrete JFETs, so that would be a decent fit regarding the operating voltage indeed.
And yes you can parallel several JFETs to sink more current. You'll also get more dispersion (and a lower output impedance if I'm not mistaken.) This transistor seems to be obsolete though and available only through old stocks. Maybe it has a more recent equivalent. Else the OP will have to resort to eBay.

The NPN version is more accurate as a current sink (this JFET's current will still vary up to at least 2-3mA over the whole voltage range, which may not be a concern here, I don't know), and as I said above, can be made to work at pretty low voltages given you use a low enough reference voltage. Yes, it will need more components. It will have some sensitivity to temperature, but the JFET as well. Not sure which of the two will be worse in that respect.

Yes the JFET solution is probably fine here if all the OP wants is to "bleed" some relatively constant current. That would not make a precision constant current sink, but I guess this is not what the OP is after.

In both cases, if they want to be able to set a defined current (even if not very accurately), they will need to make it trimmable. That's with an adjustable resistor between emitter and ground in the NPN case (and/or adjusting the reference voltage), or between the source and ground/gate for the JFET case. JFETs present a relatively large dispersion in their characteristics. If this is just a one-time project (or just a few), they can always hand-select them for a given current.

[Atom]

Well thanks to anyone , i think i'll for now stick to transistor based circuit because i don't have jfet in stock, for the range of 10-50mA it means that the current sink should be able to sink up to 50ma , again my fault here because i wasn't clear enough.
 
i'm not trying to make a super precise cc sink but if it is well...it's better, onestly i don't care as long as the drift is kept under a few mA. I don't want to sound rude but i knew that the cc source woudn't work with voltages that are too low, if i want it to work down to 0 i should have a negative rail right?

i think i'm going to use a to220 or a SOT32 transistor for the 1.5W or so of power and leave it flappin' in the breeze, or maybe a sot223 with enough copper area

By digging in the Art of electronics i found some cc sink/source, here they are not that i know how they are called i can google them and study them , thanks again