simple current divider - Page 1
Forum section: electronics / projects

author: ralphd
author_rank: Frequent Contributor
timestamp: June 17, 2015, 08:03:55 pm
content: I'm looking for a simple current divider to use for a power indicator for mcu projects. For 3.3V projects powered from 5V I use a 50ma-rated red LED and it is a convenient indicator of power use in the circuit. Id like to do something similar without the 1.7V drop.I tried reverse-biased tansistors like the 2n3904, but they still have a small positive gain when reverse biased.

author: moffy
author_rank: Super Contributor
timestamp: June 17, 2015, 11:50:46 pm
content: Sorry, I don't quite understand. How does a 2N3904 become an indicator? No matter how you divide the current you have to develop the voltage for the indicator e.g. LED.

author: jeroen79
author_rank: Frequent Contributor
timestamp: June 18, 2015, 12:31:47 am
content: I don't quite understand either.Are you now putting an LED between +5V and the MCu's vcc pin?The simplest power indicator would be an LED in series with a resistor between vcc and gnd.

author: ralphd
author_rank: Frequent Contributor
timestamp: June 18, 2015, 12:57:08 am
content: A normal transistor allows a multiple of the base current to flow from collector to emitter. I want a simple circuit that would allow a fraction of the collector-emitter current to flow through the base. So if the CE current is 10mA, the base current is 1 mA, and if the CE current is 100, the base current would be 10. I'd connect a LED to the base, so it would light up in proportion to the current consumed by the circuit.

author: moffy
author_rank: Super Contributor
timestamp: June 18, 2015, 01:17:51 am
content: Why does the current through the LED have to be proportional to load current?

author: uChip
author_rank: Contributor
timestamp: June 18, 2015, 01:54:25 am
content: QuoteUnfortunately transistors don't work in reverse. The physics is completely different than say gears. The Ibe current enables the Ice current. If there is no Ibe current there will be no Ice current.You can achieve what you want, but it will be a much more complex circuit. For instance, you could put a small resistor in the supply line then amplify the voltage across the resistor to drive the LED.The original configuration relies on the characteristics of the load (3.3V, xxmA) being exactly in the "linear" region of LED light output. It's kind of a neat application, but it only works in that one narrow space.- Chip

author: ralphd
author_rank: Frequent Contributor
timestamp: June 18, 2015, 02:35:12 am
content: I understand transistors don't work that way.I thought of a sense resistor and op amp, but want something simpler.

author: jeroen79
author_rank: Frequent Contributor
timestamp: June 18, 2015, 02:36:18 am
content: What about a small current sensing resistor between vcc and the MCU and a PNP voltage follower added:You choose REto be 10x Rsenseand put the LED between the collector and gnd.

author: ralphd
author_rank: Frequent Contributor
timestamp: June 18, 2015, 02:37:11 am
content: Quote from: moffy on June 18, 2015, 01:17:51 amI want a visual representation of power consumption; bright LED for high power, dim LED for low power.

author: void_error
author_rank: Frequent Contributor
timestamp: June 18, 2015, 07:44:45 am
content: How about a high side sense circuit likethis onefollowed by avoltage to current coverter?

author: ralphd
author_rank: Frequent Contributor
timestamp: June 18, 2015, 12:06:32 pm
content: Quote from: jeroen79 on June 18, 2015, 02:36:18 amI like that idea. I'll give it a try. I'm thinking a NPN voltage follower might work too.

author: Andreas
author_rank: Super Contributor
timestamp: June 18, 2015, 08:10:29 pm
content: Hello,You could also use a high side switch with current sense output like BTS443 (with 8200:1 ratio)http://www.infineon.com/dgdl/BTS443P_DS_v10.pdf?folderId=db3a304314dca389011537739e37155f&fileId=db3a3043183a9555011860442f5b3616&ack=tor a current monitor like LT6102 which needs 2 external resistors.With best regardsAndreas

author: ralphd
author_rank: Frequent Contributor
timestamp: June 20, 2015, 12:50:52 pm
content: Quote from: ralphd on June 18, 2015, 12:06:32 pmBefore I put it on the breadboard I realized it won't work. The PnP voltage follower keeps Vout at Vin-0.6. So no current would flow until the current across the sense resistor is >.6V.I thought of adding a diode and 100K resistor to ground to offset Vin by 0.6V, but Id need a way to match the diode with the transistor, as slight variances in Vf would throw off the circuit.

author: ralphd
author_rank: Frequent Contributor
timestamp: June 20, 2015, 12:58:06 pm
content: $2.5/qty 10 @mouser for the BTS443P makes it more expensive than the MCUs I'm using...

author: bson
author_rank: Unknown
timestamp: June 20, 2015, 06:51:31 pm
content: Quote from: ralphd on June 20, 2015, 12:50:52 pmIt keeps it at Vin+0.6V.However, the LED goes between thecollectorand ground so Voutdoesn't matter.REshould probably be trimmable since ICwill vary with hfe.(No it won't; need more coffee.)