# Independent Current Source

1. Apr 16, 2010

### SpartanG345

I still can't visualise them.

They deliver the same current all the time, however there is no potential drop across them. Is it just a theoretical device that doesn't exit?

I am sure current sources do not add new charge into the circuit, but rather move charges around but this requires a potential difference.

2. Apr 16, 2010

### Staff: Mentor

Yes, absolutely. A current source in the real world has to have some minimum voltage drop across it. The term used is "compliance". At a minimum, the current source is a transistor, controlled to deliver a (relatively) constant current. The transistor needs to operate in the linear region, and you probably know what kind of minimum voltage drops that entails.

3. Apr 20, 2010

### cabraham

A generator can be operated in current source mode by forcing a constant torque. In commercial power generation, the turbine is run at constant speed, in order to produce constant voltage. Load variations require that fuel burning varies to keep constant speed. Thus a CVS requires operation at constant speed & variable torque. A CCS requires constant torque & variable speed.

The voltage would be simply the current value I, times the load resistance R. For a dead short, power is near zero and speed is minimal. As load resistance R increases, the same torque results in higher speed and higher voltage.

The reason it is never done this way is that of losses. I^2*R is much greater than V^2*G. With CV, the full voltage is always generated, but current is that of the demanded load, 35%, 51%, etc. With CC, the current is always generated at 100%, with voltage varying with loading. Losses are greater in this case.

In addition, the CV is stepped up w/ xfmrs to increase V & decrease I. Again, losses are cut this way. If high temperature superconductors ever become real & affordable, constant current might be realized.

However, CV has another advantage - constant frequency. CF makes synchronous motors run at a fixed speed, and that is useful. CC results in variable frequency, not bad, but synchronous operation of clocks would be lost. Make sense?

Claude

Last edited: Apr 20, 2010
4. Apr 20, 2010

### sophiecentaur

If there were no PD across the source then it would have zero impedance. A constant current source will have whatever PD is needed to make the required current to flow through the load - the load resistance must have no effect on the current through it. So, for a 'passive realisation of this, the source resistance and source voltage need to be as high as necessary to achieve this over the desired range of load and accuracy of Current value.

With an active device involved (an amplifier) you can get away with a much lower source voltage and just vary the PD across the source resistance to get the current you want (over the desired range of load resistances). You need the source voltage to be such that the current through the Highest required Load resistance is what you want. At this point, the series resistance would be zero (if you had a suitable device in series). But a practical device will have a certain minimum resistance. So the source voltage needs to be enough to obtain the desired current through (maximim) load resistance and (minimum) source resistance in series.
At the other end (zero load resistance), the source resistance needs to be
RS = source voltage / desired current.
The source resistance must be controlled to be some value between the two extremes by suitable feedback control.

5. Apr 20, 2010

### Bob S

An ideal current source has an infinite dynamic output impedance, so the output current is independent of the termination. For example, a reasonably good 1-milliamp current source is 1000 volts in series with a 1 megohm resistor. But this is an awkward solution, so using active transistor feedback is easier. See simple 1-milliamp current source with feedback in thumbnail. In this circuit, Q2 controls the base input to Q1 such that the base-emitter voltage of Q2 is 0.75 volt. R2 pullup should be as high as possible to minimize the Q2 emitter current.

Bob S

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Last edited: Apr 20, 2010