Hello

First, I must let you know that I am fairly new to electronics circuits so please have patience with seemingly simple or stupid questions/comments. I have been reading the data sheet on a 3A 5V regulator, MC78T05, and found that I could use it as a current regulator. This setup might help me lower the power requirements for a circuit I have.

I am currently using primitive voltage/current divider circuits to control my power requirements. I have need of 5V @ 1.5A and so my questions about this regulator configuration (see picture) are:

What exactly is meant by 'output voltage compliance'? That is, what is this 'input voltage less 7.0V'? Does this mean that if I supply 5V that my output voltage will be -2V? How is this value calculated?

I have looked through the data sheet and did not find an equivalent circuit to try to figure out and so I don't know how this is calculated or how it will change for different regulating resistors. Thanks.

 

If you give the chip less voltage than it needs it can not do its job, in other words it ceases to regulate. You will get less than the correct voltage out of the device. Typically these chips drop around 2-3V to power the internal circuitry.

A better 3 terminal regulator for a current regulator is a LM317. This is a programable voltage regulator, and because it outputs 1.25 volts (as opposed to the 5V your regulator outputs) it drops less voltage. The 5V regulator drops 2V, as well as the 5V it regulates at, for a total of 7V internal drop for a current regulator, not good at all. The LM317 does the same thing, but only drops around 3V or so, which is much better.

Here is how the current regulator is done...

All the voltage regulators should have a 10µF and 0.1µF cap directly on the output of the device, and if there are long wires going to the input a 10µF or bigger on the input, to prevent oscillation.

 

Thanks very much for the explanation.

 

Bill has it mostly right.

A 78x05 regulator is a fixed voltage regulator.
It's minimum "dropout" voltage is 2v. That is measured from the source on the input to the output. So to get 5v out of a 7805 regulator, you need to have 7v at it's input.

Voltage regulators are by definition low impedance; capacitors are required on their inputs and outputs in order to keep them stable and to suppress transients.

Current regulators are by definition high impedance; you should not have capacitors on the output of a current regulator, as that would tend to lower the impedance.

The LM317 has a minimum "dropout" of 1.7v from the input to the output terminals.
Vref, measured from the output to the ADJ terminal, may vary anywhere between 1.2v and 1.3v, and still be within manufacturer's specifications. Nominally, it's 1.25v. However, when used in current regulation mode, you should plan on a minimum dropout from the input to the ADJ terminals of 3v.

The LM317 attempts to maintain a nominal 1.25v Vref between the output and ADJ terminals by sourcing current from the OUT terminal. There is a slight additional current (somewhere between 50nA-120nA) that is sourced from the ADJ terminal.

78xx and 79xx fixed voltage regulators CAN be used as current regulators, but they are a very poor choice for such, as much power will be wasted.

 

So, that is where the 7V is involved. Thanks very much again! I will do some more reading on this and see if I can find a suitable combination of components.

 

I have need of 5V @ 1.5A

Are you saying you need a constant current of 1.5 amps at 5 volts?

Constant voltage regulators hold their output voltage to (5volts in this case) whatever the load so supply only the current required (drawn) by the load.

Constant current regulators supply (force) a set current, whatever the load, and adjust their voltage output to cause this to happen. If that requirement needs an ouput voltage of greater than 5 volts (in this case) regulation will fail.

Not sure what you want a constant currrent of 1.5 amps for, this is unusual. So whilst constant voltage regulator ICs abound, there are few constant current ones.
One such IC is the L200 which has built in, but accessible, current control circuitry so can be used in CV or CC mode.

Battery charging circuits usually have additional control circuitry as pushing 1.5 amps into a near full battery can cause an explosion, so the rate is reduced to a trickle at the end of the charge cycle.

So run you final circuit past some experts here, before committing to hardware.

 

Sorry for the misunderstanding. I have a motor that is rated at 5V and 1.5A. The stator windings have about 3.2 ohms resistance and so it will of course simply use about 1.5 ampere if connected to a 5 volt potential. Since Ampere's Law relates the magnet field directly to current (not voltage), I was entertaining the notion of creating a limited current supply (maybe use 1.2 ampere) for the H-field of the stator windings with this supply being more safely drawn from the larger 3A regulator. If my understanding of this regulator setup (as a current regulator) is correct, I should be able to use it in this way, albeit very inefficiently as you guys have pointed out.

Another question: Since this device uses 2V for internal power and 5V through a 2.5 Ohm resistor to get 2A (from the picture in my first post) assuming 7V at the input. If I supply 5V at the input, will I still get 2V for internal power and then 3V through 2.5 Ohm for 1.2A? Or will the device not work because the internal power is competing with the output? Or what?

 

...
Another question: Since this device uses 2V for internal power and 5V through a 2.5 Ohm resistor to get 2A (from the picture in my first post) assuming 7V at the input. If I supply 5V at the input, will I still get 2V for internal power and then 3V through 2.5 Ohm for 1.2A? Or will the device not work because the internal power is competing with the output? Or what?

Here is a 7805 used as a 1A current-source, where the load resistance it is driving is swept from 0Ω to 10Ω (the X axis in the plot). The input voltage is fixed at 15V.

First note the LtBlue trace. That is the current through the load resistance R2. Everything is going great until the load resistance gets to ~8Ω; then things go to bad. For loads less than 8Ω, the current through R2 is 1.005A (the extra 5mA comes out the 7805 GND pin, and adds to the current programmed by the 5Ω resistor, R1.

You can say the "compliance" of this constant-current generator is slightly less than 8V, because the current is constant up to the point where the CCgen is asked to put out more than 8V! btw- 1A through 8Ω requires 8V.

At the 8V point is where the CCgen runs out of compliance. Why? Look at the Red trace, V(out). Note that things go bad just as V(out) reached 13V. What happens at 13V? The difference between V(in) 15V and 13V is 2V, which is the DropOut Voltage Vdo of a 7805! Further, note the Blue trace, Vadj, which is the voltage at the load and at the GND pin of the 7805. Also note that the delta between the Red and Blue Traces is a constant 5V until the 8Ω point...

Now, let's look at power dissipation in the regulator and the programming resistor. Note the LtGreen trace which is the power dissipation in the 7805. Note that it is highest when the load resistance is minimum, and it decreases as the load resistance increases. At 15V input, the peak dissipation is ~10W. Note that the Purple trace shows that the dissipation in the programming resistor R1 is a constant 5W as long as the current is 1A, but then decreases after that.


Now scale this up to your desired 2A load current. You will need a big heat sink!

 

That is some awesome software (is it expensive?). Thanks, btw, for showing this to me. Once I see all the voltages and currents in this manner it becomes a little clearer. It seems that I am wasting power no matter which way I go. And also, I have stacked two heat sinks together to take the heat from my voltage regulator which is simply running through a series of resistors to lower the final current through the windings. I apologize for not posting a schematic of my project (which is more or less finished and being used) as it would make your efforts easier. A cursory glance at my circuit would allow you guys to point out many mistakes and better methods so I will suffer the embarassment to aquire the lessons but I am waiting for some software to do this. In the mean time, thanks again.

 

That is some awesome software (is it expensive?)...

Free!!! LTSpice, see Linear.com