Variable resistor

Thread Starter

jbord39

Joined Mar 31, 2010
41
Hey guys,

I am trying to create a variable current driver. Currently the voltage across two terminals is held constant at 5V, and I want to vary the resistance through it with a potentiometer. However, since the resistance needs to vary from like 3ohm-5ohm, most potentiometers cannot handle this power. I am hoping to use a power transistor instead (by varying the voltage at it's base or gate), but am having some problems. Mainly, the turn on is very exponential rather than a nice linear ramp.

Does anyone have any suggestions, other than an expensive rheostat?

Thanks,

John
 

tracecom

Joined Apr 16, 2010
3,940
What current levels are you working with? Is it the product of 3 ohms at 5 volts? Roughly 8 watts to dissipate?
 

wayneh

Joined Sep 9, 2010
16,102
You don't need an expensive component, except possibly a heat sink for your transistor. Can you diagram what you've tried so far? For a power transistor such as 2N3055, you can figure a current gain of 10-50X or so depending on conditions. It's no problem to get a smooth increase. However, the increase will be in base current, not base voltage. The base voltage will stay locked at about 0.65V over the emitter up to very high current.
 

Adjuster

Joined Dec 26, 2010
2,147
What kind of load are you driving? How is the voltage across the "driver" currently held at 5V, and will that always be the case? Do you really want a variable resistance, or is a variable current just as good? How quickly do you need to vary the current?

Please provide more information, so that people will be able to help you. Most likely a power transistor solution would work, perhaps enhanced with an op-amp feedback circuit if you want really precise control, but no-one can say if you don't comment further.
 

Thread Starter

jbord39

Joined Mar 31, 2010
41
Hey guys,

ALERT: The voltage source says 8V but it should be 18V. The 1 got clipped. Also, ground is at the bottom of the circuit.

Thanks for the quick replies. I am trying to drive a sensitive laser diode, which I want to take between 1A and 2A. I am hoping to not burn out any more :). The 4 diodes in series represent the voltage drop across the laser diode.
The basic operation is that the zener diode holds a constant voltage across the op amp terminal. Negative feedback causes the opamp to adjust its current output to make the voltage across the bottom darlington pair 5V. I am hoping to vary the resistance of the bottom darlington pair between 5ohms and about 3ohms. Any more help is appreciated.

John
 

kubeek

Joined Sep 20, 2005
5,598
Why would you want to drop 5V on the lower transistor? You should use very low resistance like 0.1ohm and use a voltage divider to get the desired drop voltage. Also something less antique that 741 would be nice, for example TL071 would do.
You should lower the supply voltage to cca 9 volts to lower the dissipation in the upper darlington.
POT represents the potentiometr, R9 is ther in case the wiper looses contact.
 

Attachments

wayneh

Joined Sep 9, 2010
16,102
I don't think your op-amp is doing much in that circuit. When the AC turns the laser on, it's full on at both darlingtons. When it's off, both darlingtons are fully off. (Is the AC a square wave?)

All you need to do is pass the current over a shunt resistor, say 2 ohms, and take the voltage at that shunt to the op-amp. 2V = 1 A. Pick a different resistor as needed. Adjust the reference voltage by dividing it down from your zener reference. The adjust voltage and ohms law as above will determine how much current is allowed to pass.

EDIT: I was responding to post #5. #6 looks good
 

kubeek

Joined Sep 20, 2005
5,598
Anyway this power scheme is quite inefficient and you should use switched regulator to get the desired current. I don´t remember the number of the chip that was made for current controlled white LEDs, but I hope someone will.
 

Thread Starter

jbord39

Joined Mar 31, 2010
41
Thanks a lot for that gif. The voltage division from the zener diode is a great idea (I knew there was a better way to do it). My power supply is set at 18.5V and up to 7A from a laptop battery charger. I realize that the increased Vce leads to more power on the top transistor, but I do not know how to regulate the voltage without ordering any parts. The reason is because the LM7812/789 are limited at 1.5A of output current, and I want to be able to safely go to 2A.

I am not really worried about efficiency at all because it won't be running enough to cost much anyway. (plus I don't pay for electricity living in dorms).

In the GIF, is V4 modeling the laser diode voltage drop?

Thanks again for the help,

John
 

Ron H

Joined Apr 14, 2005
7,012
Thanks a lot for that gif. The voltage division from the zener diode is a great idea (I knew there was a better way to do it). My power supply is set at 18.5V and up to 7A from a laptop battery charger. I realize that the increased Vce leads to more power on the top transistor, but I do not know how to regulate the voltage without ordering any parts. The reason is because the LM7812/789 are limited at 1.5A of output current, and I want to be able to safely go to 2A.

I am not really worried about efficiency at all because it won't be running enough to cost much anyway. (plus I don't pay for electricity living in dorms).

In the GIF, is V4 modeling the laser diode voltage drop?

Thanks again for the help,

John
What you have to worry about is how you are going to remove the heat from the Darlington. Do the math.
 

Thread Starter

jbord39

Joined Mar 31, 2010
41
What about a 3Ohm resistor between the darlington and the positive supply. At 2 amps this would bring a 6V drop; causing Vce to be much smaller, but still not saturated.

Thanks again,

John
 

kubeek

Joined Sep 20, 2005
5,598
Yes V4 stands for the diode, should actually have a diode in series to behave like one.
 

kubeek

Joined Sep 20, 2005
5,598
The current is boosted by the transistor, the chip is just the regulator. Using smaller sense resistor and probably a different transistor and schottky can get the current high enough.
 

Ron H

Joined Apr 14, 2005
7,012
What about a 3Ohm resistor between the darlington and the positive supply. At 2 amps this would bring a 6V drop; causing Vce to be much smaller, but still not saturated.

Thanks again,

John
If you are answering my post (you should quote the post you are responding to), then you can certainly use a BF resistor to reduce transistor dissipation, but the resistor will be big, and maybe expensive.
 

Ron H

Joined Apr 14, 2005
7,012
Found the circuit, not sure how easy it is to get that part, but you sure could make an equivalet circuit based on the internal diagram and description on page 5. http://www.diodes.com/datasheets/ZXSC300.pdf
Do lasers tolerate ripple? That datasheet does have one schematic with a cap across the LED (laser), but one would have to calculate the ripple and see if it is acceptable. A bigger cap will, of course, yield lower ripple.
 

kubeek

Joined Sep 20, 2005
5,598
Do lasers tolerate ripple? That datasheet does have one schematic with a cap across the LED (laser), but one would have to calculate the ripple and see if it is acceptable. A bigger cap will, of course, yield lower ripple.
Larger cap will also have larger ESR, also CLC filter could be used.
 
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