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Constant Current Source PROBLEM

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PRIYADHARSHINI PALANISAMY

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Hi all

I want to construct 1mA constant current source.Here I have attached my circuit.1mA current flows through 100E resistor.I change my resistor values upto 3.9K.If 1mA current flows through the resistor the voltage across the resistor

for 100E it will give 0.1V
for 3.9K it will give 3.9V

I will give this output to micro controller.But the problem with the circuit is the 1mA constant current decreases when I increased my resistor value above 2.5K.At that time the TL431 gives 3.3V instead constant 2.5V.What is the problem with this circuit?

Thanks in advance
 

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KlausST

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Hi,

obviously you need a higher supply voltage.
5V is not enough for this circuit with that high ohmic load.

You have some options:
* increase supply voltage
* decrease current
* optimize circuit for lower dropout
* limit load resistance
...

Klaus
 

SunnySkyguy

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PRIYADHARSHINI PALANISAMY

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Any LDO can be made into a 1mA CC source or sink. Some LDO's use 2.5V; this one uses 1.5V thus 1.5k gives 1.0mA

https://www.maximintegrated.com/en/app-notes/index.mvp/id/4404

to make your design work , the voltage drop on (1k+1.5k)*1mA=2.5V =Vref thus zener bias R must be changed to < 2.5k , pref <2.2k.

I have changed my resistor values to 2.2K.I get
1.04mA for 0.1k
0.81mA for 3.9k

actually in my circuit i should get 4V output for 3.9K
hence my current should be I=4/3.9k=1.025mA
My reference voltage 2.5V
To get 1mA current the resistor value should be 2.5K (i.e) (5V-2.5V/1mA)
A constant current source should give the constant current independent of varying its load.But it varies...
 

KlausST

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Hi
You have some options:
* increase supply voltage
* decrease current
* optimize circuit for lower dropout
* limit load resistance
As far as I can see you did none of those..
*****
Do a simple math:
You have 5V.
Across R28 and R23 there will be 2.5V. 5V - 2.5V = 2.5V remaining
Across Q you need at least 0.1V. 2.5V - 0.1V = 2.4V remaining
Across P1 expect at least 0.1V. 2.4V - 0.1V = 2.3V remaining
This 2.3V is the maximum value for R24.
Now you expect 1mA through R24. R24_max = 2.3V / 1mA = 2.3kOhms .
This says your circuit can't work with 3.9kOhms.

That's physics and mathematics you can't ignore.

Klaus
 
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PRIYADHARSHINI PALANISAMY

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Hi,

obviously you need a higher supply voltage.
5V is not enough for this circuit with that high ohmic load.

You have some options:
* increase supply voltage
* decrease current
* optimize circuit for lower dropout
* limit load resistance
...

Klaus

Are you asking me to increase the supply without changing its resistor values??
and optimize circuit for lower output---this means are u asking about to reduce the output voltage of TL431??
Thanks in advance
 

KlausST

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Hi,

Are you asking me to increase the supply without changing its resistor values??
and optimize circuit for lower output---this means are u asking about to reduce the output voltage of TL431??
Thanks in advance

You need 3.9V for R24 and you need 2.5V for R28 and R23. This means you need at least 3.9V + 2.5V = 6.4V.
Plus some 100mV for regulation.... Use at least 7V. (Currently you use about 5.2V instead of the shown 5.0V )

As long as the opamp can work with 7V and the Ref can work with up to 2mA you could apply 7V without any change.

****
It seems to me you don't understand how the circuit works....
The Opamp output is not 2.5V ... It is VCC - 2.5V - V_gs...
So with a 5V supply it may be around 1V. With a 7V supply it may be around 3V.
(I estimated about 1.5V for V_gs ... I didn't look into datasheet)

Klaus
 

Audioguru

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You are using a Mosfet that needs Vgs to be as high as 4V (its maximum threshold voltage) for it to conduct 0.25mA. Buy thousands of them and test them all then maybe you will find one that needs only 2V (its minimum threshold voltage) to conduct 0.25mA. Wrong device.
 

SunnySkyguy

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I forgot to mention... the reason for your R change is that in order for the Programmable Zener to regulate properly, is that you must ensure the bias current is above the worst case minimum requirement of 1mA.

Your cathode current must exceed the spec. limit of 1mA needed to stay in regulation.

I_min = 0.4mA typ, 1mA max


You must exceed the max.... This is called the maximum from the chip perspective, which is your minimum design requirement,
so do not to get confused. :thinker:

Good catch AG.. The Pch-FET needs to be a "Logic Level" device for Vgs in order to regulate as well. Vgs ratings depend on the Vcc-2.5V drop so a Vgs< 2.5V (max) @1mA is required.


Here are some candidates. CHoose Vgs near 1V.. or get a reliable LDO as I said initially and this time choose one with a dropout of < 50mV.
https://www.digikey.ca/product-sear...e=0&rohs=0&quantity=1&ptm=0&fid=0&pageSize=50
A good rule of thumb is to choose a current shunt that results in a 50 to 75mV drop and use a precision comparator to drive the FET with a low Vgs.


in your case you had a 2.5V drop for I sense and a Vgs of 4V
 

KlausST

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Hi,

as FET try a FDV304P.

It seems you made some changes on the circuit... parts, voltages, values....

****
If you want a LED that tells you when it is out of regulation (less than the desired 1mA), then use comparator and a 3.9V zener: cathode to VCC and a 10 kresistor form anode to GND.
Use a comparator and compare zener voltage (connection between 10 k and zener) against OPAMP output voltage. As long as OPAMP output voltage is higher everything is OK.
When OPAMP output voltage is lower then it is out of regulation...

Klaus
 

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