LT Dual Tracking 3A Regulated Power Supply 1.25V to 20V
In the data sheet for the LT1033 negative adjustable regulator there is a typical application schematic for a tracking regulator that i am interested in but don't understand. I like the simplicity of the design as all the tracking regulators i am familiar with use opamps.
here is the schematic from the data sheet
here is the more common implementation as well as the formula for calculating resistor values to set the output voltage.
my two questions are:
1. how do you adapt the formula to calculate resitor values for the tracking regulator?
2. how does the circuit match a voltage change at the output of one regulator in the other regulator?
Re: LT Dual Tracking 3A Regulated Power Supply 1.25V to 20V
2) As you adjust R3, then the resistance between the adj pin and gnd is symmetrically changed for both regulators (look at the resistor network formed by R2 R3 R4).
i believe the resistor values determine the output voltage - it is an adjustable regulator - so i need to at least know what the output voltage is for the resistor values in the schematic.
i think the second resistor in the voltage divider is made up of r4 and r3 in parallel. the effective value of R3 is acutally 2.5K not 5K as i originally thought.
using the above logic i calculate the output to be 22V. Still 2 volts too high so i must still not be accounting for something.
I just cannot get my head around what you are trying to say. You do realize that R3 is a potentiometer? When R3 is adjusted to 5K, the voltage is +/- 20V and when it is adjusted to 0 the voltage is +/- 1.25V.
ok - i think.
i do recognize that r3 is a trim pot.
can you run through a calculation for me with r3 set to 5k so i can better understand?
by my (i assume incorrect) calculations i get a voltage output of about 42 V when r3 = 5K
Essentially, you don't need to change the formulas in the single supply schematic to adjust it for dual tracking. Actually, I think its a really nice application note.... The key is to identify the current flowing through the resistor divider network (R1 - R5).
If R3 = 0, R1 and R5 are tied together and the reference pins are pulled together, putting them to 0V offset through the now parallel connection of R4 and R5.
If you want to fix R3, calculate half its value in parallel R2 and R4.
prototype up and running
leadbelly and bakmeel, thank you for your replies.
i now have a working prototype
i have not tested it's tracking but i will do so soon.
i have to admit that i still don't quite understand the circuit but, i do at least know how to calculate the output voltage based on the value of the trim pot.
i am planning on increasing the stability of the supply by incorporating the following typical implementation circuit from the data sheet into the tracking regulator circuit.
i don't have the voltage reference in hand yet so it might be a while before i get the chance to try it.
after i build a point to point version i plan on designing and ordering circuit boards. This supply will form the basis for the buffers i have in the works (B1 and a LME49710 based buffer).
if i can find a negative regulator that is pin compatible and can handle higher output amperage it might work for my chip amps as well.
Good work okapi,
Try to approach the regulator as follows: It will work to maintain a voltage difference between OUT and ADJ of 1.25V, based on its internal reference. It will adjust the current passing to the output as a function of the current flowing into the ADJ pin.
Work your way through the resistor network based on the constant 1.25V, and you'll find how the remaining currents flow and why. It's not difficult at all.. ;)
The additional voltage reference is a nice trick to improve stability.
For your benefit: put some ceramic capacitors (typically 100nF) parallel to the output capacitors. Put them physically close to the regulator, and minimise the track distance to the ceramic from the OUT pin and GND. Keep GND close to the input and output (or use a double sided PCB with a ground plane).
The ceramic cap will short high frequency noise generated by the regulator
in regards to your capacitor comment the data sheet for the lt1033 says the following:
"An output capacitor, C3, is required to provide proper frequency compensation of the regulator feedback loop. A 2uF or larger solid tantalum capacitor is generally sufficient for this purpose if the 1MHz impedance of the capacitor is 1W or less. High Q capacitors, such as Mylar, are not recommended because their extremely low ESR (effective series resistance) can drastically reduce phase margin. When these types of capacitors must be used because of other considerations, add a 0.5W carbon resistor in series with 1uF. Aluminum electrolytic capacitors may be used, but the minimum value should be 25uF to ensure a low impedance at 1MHz. The output capacitor should be located within a few inches of the regulator to keep lead impedance to a minimum"
i don't know the what the important differences in specifications are for tantalum and ceramic caps with the same capacitance and voltage ratings.
Should i substitute the ceramic for the tantalums or use the ceramics after the tantalums? I also have hundreds of 0.1 uF polypropylene caps, could i use those?
i also noticed that some power supplies also have much larger caps after the regulators, 1000 uF or more. when is this sort of configuration necessary?
Solid Tantalum generally has a poor impedance (high ESR) and works well as buffer capacitance. It is true that in some cases a high Q capacitor can cause loop instability, but a 100nF ceramic is not large enough to make a difference in the loop stability.
You may keep the electrolytics you're using now, or replace them with tantalums. It won't make a difference. Film caps instead of ceramics is no problem either, as long as you keep them small enough.
Power supplies with large capacitances are best when you're expecting large variations in load currents. They are also commonly used in hi-fi audio as an eye-catcher or just to exgagerate :D
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