# Constant current “driver”

I need constant current 100mA source (120 channels of them) with as much low output tolerances as possible (so I don't want to use for example LM317 +resistors).

I found Supertex CL7 chip (+- 5%)

But there is problem, because I need to circuit my load one one side to GND. Here in this Chip load is between Vdd and Vout. Do you think I can circuit my load between Vout and GND?

Do you maybe have any other solution?

• You reject the LM317 (tolerance 0.05/1.25 => slightly better than 5%), yet you want to use a chip that has a 5% tolerance?? – Wouter van Ooijen Jun 30 '12 at 12:20
• Yes because there are tolerance on resistors, also who can guarantee about reference 1.25V ? – Alex Knok Jun 30 '12 at 12:22
• The datasheet says that the reference is 1.25V +/- 0.5V. That is better than the 5% you seem to be happy with. Resistors can have 5%, 1%, or even 0.1% tolerance. No problem to meet your 5%, circuit is very easy, and can either source or sink. – Wouter van Ooijen Jun 30 '12 at 12:27
• No, you are wrong. Tolerance is +/- 0.05V (1.2V to 1.3V). To get 100mA, we need 12.5Ω. But there are no such resistor, so we must take 12.4 Ω (1206, 1%). So if you calculate worst scenario you can get output current (95.816mA-15.898mA), which is almost more than 5 % :)) – Alex Knok Jun 30 '12 at 12:46
• Be more creative: Mouser has plenty 25 Ohm 0.5% resistors, for instance 667-ERJ-P03D25R5V, which would give you a maximum error of 4.7%. – Wouter van Ooijen Jun 30 '12 at 13:15

Do you mean you want your Vout to be strictly 0v?

The chip you found have a constant current sink. If you have a current-driven device like LED, it's ok to let your Vout have some voltage like 0.2v or more. This does not affect your device. And for LED case the the current control the brightness and the internal resistance control the voltage drop across the LED. If u give 12v Vdd and voltage drop for LED needed is 3v the rest 9v must be eaten by the driver. This will make your IC very hot.

U can try TLC5940, it have 16 channel constant current sink and 120ma per channel.

If you want better than 5 % overall accuracy you're gonna have to pay for it. There are plenty 1 % regulators, but most of them are 2 % over temperature range, and you'll need a precision resistor as well.

The SPX1117 has a 1 % initial output voltage precision, 2% over current and voltage range, and a low adjust current of 120 µA maximum. If you take the 1.5 V version with R1 = 15 Ω/0.5 %, then your current will be between 97.51 mA and 102.63 mA. 0.1 % resistors will bring you closer to 98 mA to 102 mA, but again, those are expensive.

Yes, you certainly can put load between Vdd and GND. To eliminate a user mistake completely mark the load as negative polarity powered: say "GND" and "-100mA source".

BTW. how to get rid of roundy points in this editor ?

• Hmm, are you sure? Did you check "typical application circuit" of CL7? I didn't find anywhere that I can put load between Vout and GND. – Alex Knok Jun 30 '12 at 12:17
• Follow the schematics in your datasheet, extend the led alone and mark anode as new GND and cathode as -100mA, then box the rest of parts as a single source in red on my picture. It will be same schematics, but with different names for nets. – user924 Jun 30 '12 at 12:22
• But what about logic inside chip (control and resistor look schematic of chip). So you said that I will have negative current on my load ? – Alex Knok Jun 30 '12 at 12:47
• The chip and surrounding logic, power supply is what you control better than the load and external environment, so just make the source + board with schematics completely insulated from AC mains and anything other than load. Then you can have "normal" real GND inside your schematics, which is not the same as external GND. Having 2 different "grounds" is the whole idea. – user924 Jun 30 '12 at 12:51
• You add the IC operating current in when you use the IC he mentions in this manner. – Russell McMahon Jun 30 '12 at 14:50