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constant current source...!!

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hussain_kiet

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howland constant current source

how can we construct a constant current source(CCS) using OP AMP
And should a CCS never be left open..:?:
 

howland current source

i think a ccs is used in the construction of an opamp.(and i dont think the vice versa is true).

a constant current source needs a path for it's current to flow. so the circuit loop must be closed(this is obvious).
 
constant current source with grounded load

You can easily build a current source with an opamp.
Simply use the opamp to maintain the voltage constant across a resistor in the current path through feedback.
If the load can be floating, then the opamp can simply be used in the inverting amplifier configuration. The current through the feedback resistor is going to be constant and equal to:

I=Vin/Rin

Where Vin is the input to the amplifier and Vin is the resistance in series with the opamp's input.

If the load must be connected to Vcc (or grounded) then you need a transistor. I can post schematics tomorrow.
 

howland constant current source + .pdf

i too agree with VVV a constant current source can be designed as he has said......

also many more constant current source circuits like howland current source can be made only using op-amp,...... circuits of this sort has been given in art of electronics chapter about op-amp circuit.....
 

constant current howland

using opamp i don't know, but LM317 is a perfect constant current source.
 

lm317 constant current source using op amp

how do you say LM317 is a good constant source.......... actually it is a constant voltage source and hence the output current is constant only if the load is constant and any variation in the load will change the current.........
 

howland constant current

The LM317 is indeed an excellent current source. You can search the NS website for LM317 and look for 1A current regulator at page 17. What the internal circuit of the LM317 does is trying to keep the voltage at the adjust pin 1.2V below the voltage on the output pin. So, the voltage across the resistor in series with the load is also 1.2. Since I= U/R a current of 1A will flow in the load regardless of the voltage across the load. For an LM317 to maintain a constant current the input voltage has to Vout + 2.5V.
 

ns constant current source

inventor u say that

The LM317 is indeed an excellent current source. You can search the NS website for LM317 and look for 1A current regulator at page 17. What the internal circuit of the LM317 does is trying to keep the voltage at the adjust pin 1.2V below the voltage on the output pin. So, the voltage across the resistor in series with the load is also 1.2. Since I= U/R a current of 1A will flow in the load regardless of the voltage across the load. For an LM317 to maintain a constant current the input voltage has to Vout + 2.5V.
Click to expand...

but how do u say that the R in I=U/R is a constant...... the resistor and the load are in series and hence when the load varies it means that the value of the R is going to change........ hence the I is going to change........

Inventor can u explain ur circuit design with a circuit diagram
 

1a constant current source lm317

Yes I can. Like I wrote, you can search for LM317 on the national semi website and take a look at page 17. The circuit inside the LM317 holdes the voltage over the series resistor constant regardless of the load resistance. I can post a drawing. But not today.
 

grounded constant current source

inventor can you atleast paste the link of page no.17.... I have problems finding downloading the data sheet and other things about LM317 in NS site........
 

constant current source grounded load

Here are the circuits I was talking about.
The first one requires that the load be floating.
The second has the load to Vcc, while the third works with a grounded load, but the input voltage is referred to Vcc.

In all three the principle is the same: the opamp attempts to keep the voltage between its inputs close to zero, which means it will force the voltage across a resistor to be equal to the input voltage. Since the current through a resistor is proportional to the voltage across it, if you keep the voltage constant the current will remain constant, too.
 

constant current source circuits 1a

A.Anand Srinivasan
Here is a link to de document in question
h**p://www.national.com/ds.cgi/LM/LM117.pdf

Here is how it works:
on page 17 you can see a 1A constant current source.
The load current will create a voltage drop across R1.
What happens is that the LM317 tries to regulate the voltage across R1 to 1.25V. If the voltage across R1 is constant then we know that the current trough R1 is constant. In this circuit the load is in series with R1 so the load current is constant.
 

here is a problem i face with consant current source (pulsed output) :-

whatever the current i set for, there is a spike in the voltage waveform across the load. It needs about 200uS to settle up.

This happens for every switching of the current pulse.

Regards
 

to be precise:

there is a ringing in the waveform for about 150us until it settles.

any one have some thoughts on it?
 

@Cameo_2007

what you see is the settling time of your circuit. You can try to add a small capacitor across the load.

Regards
 

inventor, Exactly, it is the Settling time of the current source.

Capacitor across the load?

Any ways to reduce this settling time..? will the capacitor help it?
 

Can you post a schematic? And what kind of load are you using?
 

I could give a brief about the schematic.

It's basically an opamp based circuit. The input ofthe opamp comes from a DAC which sets the required current.

The load is a simple resistor. The current passed is just used for this load resistance measurement.
I would look to post the scematic in a while.


Thanks.
 

The speed of your circuit depends on the settling time of the DAC and the speed of your op-amp. The capacitor in parallel with the load I suggested would dampen the 150nS ringing/overshoot you where seeing but the capacitor will also increase the settling time of your circuit.
 

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