I am trying make a current sink using a BJT transistor. In the diagram shown, the current across the V1 is changing by changing the variable resistance (VR7). LM385M is providing a fixed voltage of 1.24 V at the base.

I want to change the collector current (Ic) by changing VR but the current is fixed to around 100 uA. What is the mistake in the circuit.

How can Ic be changed by changing VR?

 

The current should be (1.24-Vbe)/(VR+20000) which should be approximately between 25uA and 5uA so 100uA is higher than the range that VR should be capable of setting. It could be that you are measuring the current incorrectly in some way or there's some additional component to the current by way of a leakage path.

 

Below is an LTspice simulation of you basic circuit.
The current goes from about 6μA to 34μA over the range of the pot resistance.

 

I am trying make a current sink using a BJT transistor. In the diagram shown, the current across the V1 is changing by changing the variable resistance (VR7). LM385M is providing a fixed voltage of 1.24 V at the base.

I want to change the collector current (Ic) by changing VR but the current is fixed to around 100 uA. What is the mistake in the circuit.

How can Ic be changed by changing VR?View attachment 89353

Most people use more than 1.25V for the base reference voltage, a common scheme is a pair (or more) silicon diodes in series to stabilise at multiples of 0.7V, some people favour a green LED, its close to 2.0V and allegedly pretty stable.

 

Thank you everyone for your answer. I have designed the circuit for 10uA (Ic). As shown in the attached file, I want to connect the Ic to another p-cmos current mirror circuit in the chip. Thus the node Ic will be connected to the drain of the pmos. I want to use 10 uA current inside my chip. Do you think I my connections are correct?

 

Thank you everyone for your answer. I have designed the circuit for 10uA (Ic). As shown in the attached file, I want to connect the Ic to another p-cmos current mirror circuit in the chip. Thus the node Ic will be connected to the drain of the pmos. I want to use 10 uA current inside my chip. Do you think I my connections are correct?

Without knowing what chip it is and more of a schematic I'm not sure anyone will really be able to comment. Originally you wanted 100uA was that the error in your circuit?

 

Your electronic circuit is not thermally stable.
Current with increasing temperature will increase
(0.4% per 1degree --> 10degree 4%).
Stable electronic circuit:

 

Without knowing what chip it is and more of a schematic I'm not sure anyone will really be able to comment. Originally you wanted 100uA was that the error in your circuit?

Hi. Chip is an integrated circuit I designed for my sensor (I cannot change anything there inside). I know its a very vague explanation but I wanted current to be 10uA but the current was stuck at 100uA.

I am measured current by breaking the path between Ic and collector node of the BJT which I believed it to be correct way but the current was zero. But when I am measuring the current from Ic to gnd with node Ic connected to collector current is 100uA. I can see the current at node of Variable resistance is changing.

 

Judging after all that you did not properly measure the current! You connect one probe of the ammeter to the ground! Obtain a current (5-0.7) / 47k = 100uA! You must use another source voltage of more than 1.2V or 5V. One probe of the ammeter (-) connected to a collector and a second probe (+) to power 5V (easiest way).

 

You designed your own chip! Doesn't sound likely for someone who can't get a current sink to work.

 

I explained why the author obtained 100uA. And the thing in the wrong connecting an ammeter. The current flows into the transistor, and it does not. And about the chip design. I did that for 20 years. Now I do current generators for lasers using transistors and microchips.
In my scheme, I added two transistor chip, to show how to work the current generator. Incidentally, if the transistors are too weak bipolar transistor is in saturation and the current will not be required.

 

And about the chip design.I did that for 20 years.

I was actually referring to the TS posting when I commented on this being an own design chip not one of your posts.

 

Judging after all that you did not properly measure the current! You connect one probe of the ammeter to the ground! Obtain a current (5-0.7) / 47k = 100uA! You must use another source voltage of more than 1.2V or 5V. One probe of the ammeter (-) connected to a collector and a second probe (+) to power 5V (easiest way).

It was very useful. Thank you.

 

You designed your own chip! Doesn't sound likely for someone who can't get a current sink to work.

I am a student and new to the circuit design. But yes I have designed this chip. This current sink is for the test board to test the working of the chip. FYI: This is my first chip design. Hope I can learn a lot from this design.

 

Hi. Chip is an integrated circuit I designed for my sensor (I cannot change anything there inside). I know its a very vague explanation but I wanted current to be 10uA but the current was stuck at 100uA.

I am measured current by breaking the path between Ic and collector node of the BJT which I believed it to be correct way but the current was zero. But when I am measuring the current from Ic to gnd with node Ic connected to collector current is 100uA. I can see the current at node of Variable resistance is changing.

If you have plenty of voltage headroom; you can make a current limit with 1 JFET and 1 resistor.

Just put a resistor in series with the source lead and return the gate lead to the bottom of that resistor.

When source current flows, it develops a voltage across the resistor - with the gate at the bottom of the resistor (N-channel JFET) it becomes negative compared to the source - this tends to pinch off the JFET channel and so stabilises the current.

JFETS have a fairly large parameter spread, so you can only calculate an estimate value of R within the spread of Vgs-cut off - you'll probably need a fixed resistor in series with a preset pot and trim to the exact current you want.