hi guys
i recently was trying to build a simple constant current source to use it in an arduino project and found this famous one transistor circuit :
i build it but there is a couple problems with it :
1- if i change the load from 1 led to 2 leds the change in the current is very small but when using a 33K or
some other high value resistor the current drops to a very small value ?! why is that,
the current should be independent from the load resistance !
2 - the measured value drifts a lot from the calculated value from the equation :
I = (Vb-0.6/Re )* β/1+β
i used a values of : R1 = 4.7K , R2 = 10K , Re = 100 ohm
You need to replace R2 with something that provides a fixed voltage on the base, for example an LED or 2 or 3 diodes in series. By fixing the base voltage you fix the voltage across Re. As the voltage across Re is fixed then so is the current. Most of the current through Re comes from the load via the collector, thus if Ie is constant, so is Ic and I load.
1 - The circuit is able to sink the specified current if the supply voltage is high enough. If the load is a large resistor you need high supply voltage.
2 - What is the measured current and values of beta and Vb you are using? Also the brackets in your equation are somehow wrong.
PerryBebbington:
You need to replace R2 with something that provides a fixed voltage on the base, for example an LED or 2 or 3 diodes in series. By fixing the base voltage you fix the voltage across Re. As the voltage across Re is fixed then so is the current. Most of the current through Re comes from the load via the collector, thus if Ie is constant, so is Ic and I load.
i replaced it with 2 of the 1N4148 diodes which supply 1.21 volt but the current yet drops as i replace the leds load with a high resistor !?
Smajdalf:
1 - The circuit is able to sink the specified current if the supply voltage is high enough. If the load is a large resistor you need high supply voltage.
2 - What is the measured current and values of beta and Vb you are using? Also the brackets in your equation are somehow wrong.
i am using a 6.2 volt power adapter and after the voltage divider it became 4.2 volt across the Re loop , i guess this is high enough ?
the equation right i think : I = (Vb-0.6/Re )* β/1+β = I = ((Vb-0.6)/Re )* β/1+β , or there is something wrong with it ?
6.2V through a 33k load is 0.18 mA. In your circuit a little less due to the CE drop of your transistor.
The values you give in the OP mean the circuit should put out some 28 mA (that's a lot for LEDs). If you want to get that 28 mA to flow through a 33k resistor you have to increase your supply voltage to over 900V.
The values you give in the OP mean the circuit should put out some 28 mA (that's a lot for LEDs). If you want to get that 28 mA to flow through a 33k resistor you have to increase your supply voltage to over 900V.
While probably true, it makes more sense to tell a newbie the resistor value they want to use is impractical
and refer to Ohm's Law to understand why), rather than tell them they would need a 900V supply to do it.
(which of course would result in the same conclusion).
The simple fact that the Re is series with the led current path and the fact that I = V/R
(I =0.028A*33000 = 924V)
The basic problem is that that's not a constant current circuit!
The simplest is to use an LED as a voltage reference:
With a red LED (Vf = 1.6V or so) you'll have about 1V across the emitter resistor to set the current at 10mA
for 100 ohms.
The problem is that LEDs don't have very well characterized forwards voltages compared to zeners and voltage reference chips, but its a simple approach and the supply voltage no longer affects the current directly.
wvmarle:
Then instead of the red LED use two regular diodes, should have a better defined voltage drop (about 1.4V in that case) than an LED.
Two pn junctions has twice the variation of one, both with current and with temperature. LEDs are used for this purpose specifically because they are better than stacked diodes for regulation. But yes, the need to calibrate is a little annoying, but for high accuracy you'd use a voltage reference and opamp anyway.
The temperature variation of the LED pn junction tends to cancel the transistor's BE junction tempco as well which wouldn't happen with two junctions.
MarkT:
The basic problem is that that's not a constant current circuit!
The simplest is to use an LED as a voltage reference:
With a red LED (Vf = 1.6V or so) you'll have about 1V across the emitter resistor to set the current at 10mA
for 100 ohms.
The problem is that LEDs don't have very well characterized forwards voltages compared to zeners and voltage reference chips, but its a simple approach and the supply voltage no longer affects the current directly.
While a LED doesn't have a super-rigid forward voltage, there's nothing fundamentally wrong with the approach. However, R2 has to be low enough to ensure that the LED is "ON". For a 6.2V power supply and about 1.8V for a red LED operating at 20mA, R2 should be equal to about 220 ohms.
thanks guys so much for helping
after reading all the comments i realized that my question was a little bit idiotic i guess , i mean i forget to apply ohms law and know that in order to derive this relatively high current through a high value resistor i will need a huge potential difference that my supply cannot provide .
i even came up with an equation that gives me the maximum load that i could apply in order to maintain a relatively constant current in case of using multiple diodes instead of a voltage divider :
I've been building constant current boards for LEDS for years (I have a large supply of NOS transistors) and here is the circuit I use and trust to run standard LEDS at voltages between 5 and 24 volts.
(not my schematic... but its the same)
In my case... I use 2 matched transistors in the same case because, well... they are in my parts bin. So unlike the schematic... it works fine with 2 of the same transistors, like 2N222A
I realize there is a temptation to assume everyone is familiar with the 2N2222 transistor but in
actuality I don't think that's the case and some newbies, unfamiliar with electronics conventions
would not realize that there is no 2n222 transistor and if they googled they wouldn't realize
Google is actually not smart enough to realize that either.
Obviously anyone who has been in electronics for any length of time knows that the transistor designation
system makes it almost impossible for there to be a 2n222 transistor. (AFAIK)
