# Cheap, low-quality constant-current sink over wide voltage range

I'm looking for a cheap (as in: low component count, low PCB area, low component cost) way to put >= 1mA of load on a voltage that may be anything from 1.2 to 17 volts (minimum required load on a LDO output).

The easiest solution is a simple 1.2kOhm load resistor, but the power consumption becomes enormous at higher voltages (240mW at 17V, which leads to a huge package size).

A JFET-based current-limiting diode seems ideal - but I can't find a vendor that actually sells one.

So my idea is to use a BC817-16 NPN transistor, which has current gain between 100 and 250, send 10-15µA flowing into the base from a nearby 3.3V rail via a 220k resistor, and just watch 100 to 250 times that (1mA - 4mA) flow from the collector to the emitter, no matter the collector voltage:

simulate this circuit – Schematic created using CircuitLab

The circuitlab simulation indicates a current of about 4mA for 17 volts, which gives a more than acceptable power dissipation.

However, one of my (much more experienced) colleagues strongly advises against this technique, claiming that there will be runaway effects and the transistor will blow, so... will this circuit work reliably? Any other suggestions?

• current sink is pretty much the same as current source... – Eugene Sh. Apr 26 '16 at 16:55
• Maybe an LM337 as current sink. But I think it will need more than 1.2V to run. – George Herold Apr 26 '16 at 19:23

No, don't try to use the constancy of HFE to run your transistor, it varies with temperature, voltage, device, too many things.

As you have 3.3v available, use emitter degeneration like this. This is a standard technique for biassing transistors and creating current sources.

simulate this circuit – Schematic created using CircuitLab

The two diodes maintain around 1.4v on the base, or 0.7v across R1. With RE=130ohms, that's around 5mA. It will vary slightly with VBE and the diode drops. Change the value of R1 for other sink currents.

R2 needs to be low enough to conduct at least Iout/HFE with 1.9v across it, ideally 5x that, so 5mA Iout, 100HFE, IR2 should be at least 250uA, R2=6.2k.

Depending on the VCEsat of the transistor, this will sink current down to around 1v. The transistor needs to be heatsunk to tolerate 16.3v * 5mA of power dissipation within it.

You can replace the two diodes with another resistor if you want, which gives more flexibility for Vre, at the cost of a little thermal and HFE stability. If you are happy to sink current only over the range 3v to 17v, then you can connect the base directly to the 3.3v, with 2.7v across R1, resized accordingly.

• That's quite a few components, unfortunately... could the diodes be replaced by a simple voltage divider, by any chance? Remember that I don't need an exact current draw, anything from 1mA is fine... – mic_e Apr 26 '16 at 17:09
• You could use an LED instead of the two diodes. – The Photon Apr 26 '16 at 17:17
• Yeah, I second the comment about the LED. If your device has an LED indicator already then you are getting a crude voltage reference for free! Just use Neil_UK's circuit but the LED and its bias resistor replaces R2, D1, D2. So really all you are adding is Q1 and R1. Neat, eh? – Vince Patron Apr 26 '16 at 17:44
• @Vince Patron: Wow, great point! Unfortunately, the LED is on the opposite end of the device, so a lot of PCB real estate would be wasted by the trace... – mic_e Apr 26 '16 at 17:48
• C'mon man. PCB traces are free and its routing is so non-critical. Go min width .006" even :) – Vince Patron Apr 26 '16 at 18:08

Yes, you can do that, you can even buy transistors with more closely defined hFE. For example, the 2SC1815Y is guaranteed to have an hFE within a 2:1 range (120~240). It might vary another 2:1 from -25°C to 100°C, so that's a 4:1 range total, plus perhaps 50% for Vce variation. In SMT, the BCX70x is even more tightly specified (< 2:1 at 2mA).

It's generally frowned upon by folks who have never done the math, but it will work if you're not that picky (5-6:1). Expect a lot of clucking. This crappy method of biasing transistors has been used in consumer electronics in the past many times.

The JFET regulator diode probably won't work because the 1.2V is too low to get it to regulate well.

Using a dual transistor (eg. MBT3904) with the below circuit gives you a lot better control. It's best to use a dual so they will track thermally. The current will probably change 10-15% over temperature (simulate it if that matters to you). I did the below for 2mA nominal. The self-heating will cause the current to increase somewhat over time, depending on the voltage, but it still should be pretty good.

simulate this circuit – Schematic created using CircuitLab

The J505 is still available I believe here; Link to RS. It's a JFET device with gate connected to source like this: -

And has a VI response like this: -

According to the data sheet it typically draws 1 mA and works across a range of voltages from about 1.2 volts: -

All the way up to 100V

• Thanks for the tip; unfortunately I'm looking for tiny SMD parts... the only suitable component seems to be the SST505, which is listed as obsolete. – mic_e Apr 26 '16 at 17:42
• Then do some research on JFETs, choose a resistor (gate-source) and make your own. – Andy aka Apr 26 '16 at 17:45
• BTW Newark say they are getting stock in July 2016 – Andy aka Apr 26 '16 at 17:50