What will happen if I will use an 100 ohm resistor instead of 120 ohm resistor for R1 (please have a look at the attached schematic) ?
It is clear that the minimum load current will be modified to about 12.5mA.
I will use an 1K ohm potentiometer and I will se the output voltage to 12V.
But this current (12.5mA) will not be too high ?


MOD: Created your own Thread

 

What will happen if I will use an 100 ohm resistor instead of 120 ohm resistor for R1 (please have a look at the attached schematic) ?
It is clear that the minimum load current will be modified to about 12.5mA.

You appear to be confusing "minimum load current" with actual load current.

When the data sheet says the minimum load current is 10 mA, it is telling you that the LM317's load current must be at least 10 mA or it may not regulate properly. The minimum load current is an intrinsic property of the part; it is not something you can change by changing resistor values.

By using 100Ω for R1, the only thing that will happen is that for any given value of R2, the output voltage will be 20% higher than it would have been with R1 = 120Ω. In either case, the minimum load current requirement is satisfied.

 

12.5mA satisfies the minimum load current requirement, even if the reference voltage is on the low side (1.2V) and the 100 ohm resistor is on the high side, assuming 5% tolerance. That wouldn't be the case if you used a 120 ohm resistor that was actually 5% higher.

In 4 decades of using LM317, I've only had one that wouldn't regulate with a minimum load current of 10mA. According to the OnSemi datasheet, typical minimum load current is 3.5mA.

 

Hello, I have a question regarding LM317AHVT from ON Semiconductor.
I made a power supply using this IC and I used 122 ohm resistor for R1 from the first post schematic.
This will result in a 10.2mA minimum load current.
This load current is sufficient for stabilization, regarding the datasheet (please see attached screenshot - 12mA) ?

 

I always use a 100 ohms for R1, and it always works.

 

This load current is sufficient for stabilization, regarding the datasheet (please see attached screenshot - 12mA) ?

Since the requirement is 12mA, then 10.2mA will not be sufficient worst-case.
You should use a 100Ω resistor.

 

I have a question regarding LM317AHVT from ON Semiconductor.
I made a power supply using this IC and I used 122 ohm resistor for R1 from the first post schematic.
This will result in a 10.2mA minimum load current.
This load current is sufficient for stabilization, regarding the datasheet (please see attached screenshot - 12mA) ?

You should have mentioned that you were using LM317AHVT in your initial post. LM317 have a minimum load current of 10mA. For the regulator you're using, it's 12mA.

 

I have some kind of a problem regarding the attached schematic. I tested with the scope a similar schematic, which was using 3.3k resistor instead of 3x8.2k in parralel and which was using a 122R resistor instead of 100R resistor. Should I test again the power supply with the oscilloscope or the first test is still valid ?

 

You are passing the values with a lot. The 3x8.2K are equivalent to about 2800R.

Check the current on R6 and if possible on all resistors and tell us, what do you see?

 

I have some kind of a problem regarding the attached schematic. I tested with the scope a similar schematic, which was using 3.3k resistor instead of 3x8.2k in parralel and which was using a 122R resistor instead of 100R resistor. Should I test again the power supply with the oscilloscope or the first test is still valid ?

Your circuit is far too overkill! !

Remove D1,D2,D5,R3,R5,R7,R8, not needed..

Increase R1 to 22 ohms.

What voltage range do you need, and what current output.?

 

I made some tests (I measured the output voltage, using different loads and different output voltages) using the attached schematic. I used a 7.3Amps load at 26Vdc output of the power supply. The input voltage was about 33Vdc.
I was some kind of inadvertently when I tested the power supply and probably (I do not know for sure) one of the 1R resistors was short circuited, so in reality we can imagine that one of the 1R resistors was replaced by a wire, while the others were still 1 ohm.
What can happen in this situation ? Are the tests still valid ?
I would like to know if the above described situation can affect the output voltage.
The utilized schematic is in attachement.

 

I made some tests (I measured the output voltage, using different loads and different output voltages) using the attached schematic. I used a 7.3Amps load at 26Vdc output of the power supply. The input voltage was about 33Vdc.
I was some kind of inadvertently when I tested the power supply and probably (I do not know for sure) one of the 1R resistors was short circuited, so in reality we can imagine that one of the 1R resistors was replaced by a wire, while the others were still 1 ohm.
What can happen in this situation ? Are the tests still valid ?
I would like to know if the above described situation can affect the output voltage.
The utilized schematic is in attachement.

By 1R, you mean 0.22?

Yes this will change the behaviour of the circuit. Normally by Ohm's law U = I x R. Not to mention that if its a wire and not a PCB, if you bend this wire or the soldering is bad that can change the resistance, which in turn affects Ohm's law.

 

I mean the resistors from the base of the transistors, which have the 1 ohm value.
What will happen if I remove one of those resistors and change it by a wire and the other 1 ohm resistors will stay there ?

MOD: added a clipped image of your circuit, easier for us to refer too.E

 

I mean the resistors from the base of the transistors, which have the 1 ohm value.
What will happen if I remove one of those resistors and change it by a wire and the other 1 ohm resistors will stay there ?

hi mike,
Why would you want to do that.?
E

 

I did not want to do that.
It was a mistake. I have all the power transistors and 0.22 ohm resistors and 1 ohm resistors mounted on the heatsink and it is very easy to touch each other.

 

hi,
The purpose of the 0.22R and 1R is to try to balance the output load current thru each TIP36.
If you replace a 1R with a wire link, it is possible that TIP would pass more more current into the load and it could over heat.
Ideally the TIP's share the output load current equally, but thats not the case in practice due to the spread in the TIP's characteristics, so those biassing resistors are required.

E

 

Thanks for your help. I managed to solve the problem.
I still have some questions:
I used 3 pieces of TIP36CW and 1 piece of BD250F as power transistors. Is that a problem because the transistors were not the same ? Can this influence the output voltage ?
I measured the voltage across the 0.22R resistors and it is the same voltage across all resistors.

 

hi,
If you use a different type power transistor, you could measure the voltage drop across its 0.22R and compare that to the TIP's 0.22R voltage drop.
So a add a load on the Vout of the PSU and do that voltage drop test, knowing the voltage drop and the resistor value you can calc the current.

If the BD drop is higher than the TIP's, you could increase the 0.22R, to say 0.33R.

Post the voltage drops and we can suggest a modification to the R values.
E

 

The voltage drops are the following:
BD250F 0.352V
TIP36CW 0.355V
TIP36CW 0.349V
TIP36CW 0.354V

The load was about 6A at about 19V output voltage.