Why isn't this circuit working? - Page 1
Forum section: electronics / beginners

author: Mint.
author_rank: Frequent Contributor
timestamp: April 28, 2012, 02:32:00 am
content: I attempted to build a constant current source out of transistors, similar to this oneI build this circuit it doesn't work, I troubleshoot it, it still doesn't work, I simulate the circuit on pc, it works. Grrr...Is there anything wrong with my wiring?(EDIT: The image is inverted, sorry about that, seems to be some error when I uploaded the image here)

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 28, 2012, 03:19:48 am
content: I don't see any obvious problems.Have you checked voltages? The voltage at the base w/ respect to ground should be ~1.4V. The voltage across the emitter resistor should be ~700mV.What's the part number on the transistor? Did you swap emitter and collector, perhaps? Is it an NPN type? Have you checked the transistor with your meter's diode check feature?What voltage is your source?What is the to value of the resistance going from Vcc to the base? I see you are using 3 resistors there.What is the resistance of your pot? Can your constant current source stay in compliance for all possible values?What's the value of the emitter resistor? Are all the resistance values what you expect them to be?

author: Unknown
author_rank: Unknown
timestamp: April 28, 2012, 03:26:35 am
content: Additionally check connectivity of the prototype board. They play up sometimes.

author: Mint.
author_rank: Frequent Contributor
timestamp: April 28, 2012, 03:38:41 am
content: Fixed! The problem ended up to being a faulty diode. I measured the voltage across the RE and it was at 11.30v which was definatley looking suspicious, afterwards I changed the resistors, still same voltage so then I measure the first diode and it shows the correct ,but when I measure the second one and it shows no drop. Thanks for helping me out!

author: MikeK
author_rank: Super Contributor
timestamp: April 28, 2012, 04:49:30 pm
content: Why would R=0.7/I? Won't there be more than 0.7V across the resistor because of the diodes?

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 28, 2012, 04:54:51 pm
content: Quote from: MikeK on April 28, 2012, 04:49:30 pmThe two diodes will raise the base up to 1.4V. The transistor's Vbe will then drop that voltage down by 0.7V. 0.7V remains to drop across the emitter resistor.

author: MikeK
author_rank: Super Contributor
timestamp: April 28, 2012, 05:00:10 pm
content: Thanks, that makes sense. And the two diodes are needed to forward bias the base?

author: IanB
author_rank: Super Contributor
timestamp: April 28, 2012, 05:00:23 pm
content: Quote from: MikeK on April 28, 2012, 04:49:30 pmThe transistor will try to conduct when Vbe > 0.7 V.Since the two diodes nominally have a 0.7 V drop each, that puts Vb at about 1.4 V.Since we are looking for (Vb - Ve) > 0.7 V we want (1.4 - Ve) > 0.7 => Ve < 0.7 V.From the diagram Ve = Rset x ITherefore when I < 0.7 / Rset the transistor will turn on and start conducting and the current will rise. When the current reaches 0.7 / Rset the transistor will start to turn off and will stabilize the current at that value.

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 28, 2012, 05:04:45 pm
content: Quote from: MikeK on April 28, 2012, 05:00:10 pmYes, to forward bias it and set it at a specific voltage.

author: MikeK
author_rank: Super Contributor
timestamp: April 28, 2012, 05:36:34 pm
content: And using a resistor in place of the diodes wouldn't be ideal, because its voltage drop wouldn't be stable? Now I'm curious why simple radio amplifier circuits use a resistor divider to bias the transistor instead of diodes...Is it because of the single direction of the diodes?

author: IanB
author_rank: Super Contributor
timestamp: April 28, 2012, 05:43:39 pm
content: Quote from: MikeK on April 28, 2012, 05:36:34 pmA partial answer might be that radio amplifier circuits are working with small AC signals, whereas this circuit is working with a large DC signal.

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 28, 2012, 05:50:53 pm
content: In the case of this particular circuit a voltage divider wouldn't work unless a very specific input voltage is used. Using diodes allows for a wide range of input voltage.In the case of biasing a transistor for amplification, a voltage divider is actually more ideal since you really want the transistor's quiescent point to be relative to Vcc (usually in the neighborhood of 0.5 * Vcc.) This is so the output voltage 'swing' is maximized.

author: Tube_Dude
author_rank: Regular Contributor
timestamp: April 28, 2012, 08:15:52 pm
content: Quote from: MikeK on April 28, 2012, 05:36:34 pmYou don't see two diodes used to bias a transistor amplifying stage, because in this case we need a voltage swing at the input of the transistor stage. With the use of two diodes, all the compliance given by the resistor voltage divider is lost...

author: slateraptor
author_rank: Frequent Contributor
timestamp: April 30, 2012, 12:30:00 am
content: @Mint.: Now that you've gotten this circuit to work, I recommend looking into a transistor current mirror and Widlar current source.