Need help on Delta 1010 PSU mods

[mtl777]

The meter just shows what's there.

Parts have tolerances, and yours is well within that. A different reg will have a different voltage within that tolerance. The next one might be higher than 15 or even lower than your present one. Or maybe exactly 15... it's a non problem 🙂

Yeah, I tend to worry about little things, but I had actually replaced the +15V reg and the thing that concerned me was that the original one (an L7815CV) also had a lowish 14.9V output. The one that's there now is an LM340T-15. Well, if you say that it's just normal variation in the output of regulators then as long as it's within tolerance there's nothing to be concerned about. Thank you very much for this insight and for saving me from another needless worry! 🙂

 

[Mooly]

Look in the data sheet for the LM340-T15

http://www.datasheetcatalog.org/datasheet/motorola/LM340T-15.pdf

Available in two tolerances, 2% and 4%

 

[mtl777]

I had some fun with my new toy (scope) and tested the following capacitors for C13 (output smoothing cap for +15V reg U4) and C61 (output smoothing cap for +5V reg U23):

10uF Rubycon ZL
100uF Rubycon ZL
220uF Panasonic FM
330uF Nichicon HE
470uF Rubycon ZLH
560uF Rubycon ZL

I got the lowest AC ripple on the output of the regs with 330uF Nichicon HE. The 220uF Panasonic FM was a close second. So I'm going to use the 330uF Nichicon HE for the output smoothing cap of all the regs.

I think I have no more questions. I really appreciate everyone's help especially Mooly! Thank you so much for your suggestions, for being so patient with me and answering all my questions! I have learned a lot from you and couldn't have done this without your help! 🙂

 

[mtl777]

I spoke to soon. I still have a question. 😛

It was suggested to put a 100uF on the input of the regs in parallel with the small cap, for example, a 100uF in parallel with C15 on the input of the +15V reg U4. What if, instead of just a 100uF, I put a 100uF plus a 0.1uF ceramic in parallel with C15 (note: The 100uF and 0.1uF will be soldered directly on the pins of the reg)? So now there would be three parallel caps - 100uF electrolytic, 0.1uF ceramic, and the existing C15 on the input of U4. Would this be (a) better, (b) not as good, or (c) not much difference to be noticeable compared to just a 100uF in parallel with C15? I know I could test this with a scope, but I also would like to know the theoretical arguments for or against this.

Thanks again! 🙂

 

[Mooly]

It shouldn't make any difference as 100uf is small compared to the 2200uf main caps.
Only if the PCB layout were poor might it reduce the "noise" seen at the input to the reg.

It's noise at the output that matters though so IMO you won't notice or measure any difference.

Note... when measuring small noise voltages on the scope, correct grounding of the probe is essential and must be consistent from measurement to measurement. Moving the ground a cm or two along a "ground track" will give different results. So be consistent 🙂

 

[mtl777]

It shouldn't make any difference as 100uf is small compared to the 2200uf main caps.
Only if the PCB layout were poor might it reduce the "noise" seen at the input to the reg.

It's noise at the output that matters though so IMO you won't notice or measure any difference.

Thanks, that's good to know! 🙂

Note... when measuring small noise voltages on the scope, correct grounding of the probe is essential and must be consistent from measurement to measurement. Moving the ground a cm or two along a "ground track" will give different results. So be consistent 🙂

Due to the design of the 1010, it needs to be assembled in the chassis in order to power up (it has a relay switch that triggers on when the computer, which must be connected to the 1010 with the breakout cable, is turned on). This makes it difficult to access the power supply section because there are two boards, the AD board and the DA board, and the AD board is mounted on top of the DA board where the power supply section is, blocking access to it. Therefore, I had to solder wires to the regulator pins from underneath the DA board. The wires extend to outside of the chassis, and that's where I put my probes - on the wires. So each reg has three wires connected - one for input, one for ground, and one for output. I attach the black probe to the ground wire and the red probe to either the input or output wire. I hope this method is fine.

 

[Mooly]

That should be OK for what you are doing.

 

[mtl777]

That should be OK for what you are doing.

Thank you! 🙂

 

[mtl777]

I always thought that bigger is better for the filter caps C12 and C18 that are right after the diode bridge. They were originally 2200uF in the stock unit, and I changed them to 3400uF (actually paralleled 2200uF with 1200uF). But after reading this thread I'm not sure anymore if I did the right thing. Is 3400uF too large? What do you think is the optimum value for C12 and C18? What are the pros and cons of using a large value for these caps?

Thanks! 🙂

 

[Mooly]

Difference between 2200 and 3400 won't make any difference here. Larger isn't always better... but your circuit is different anyway by using 470 uf caps C14 and 17 which limit absolute peak currents anyway.

Larger caps mean the bridge rectifier conducts for less time (as the voltage on the cap falls less between each half cycle), but the same "energy" is used by the circuit it's powering, so that means the peak currents are correspondingly higher to replace the charge in the cap.
A smaller cap has larger ripple across it, but the angle of conduction of the bridge is longer, putting the energy back more slowly and at less peak current... which can be beneficial.

Large caps/short conduction angle can cause transformers to run hot due to copper and magnetic losses in the core... everything is a compromise... the regs do the job of reducing ripple... none should find it's way into the output rails or grounds.

 

[mtl777]

Difference between 2200 and 3400 won't make any difference here. Larger isn't always better... but your circuit is different anyway by using 470 uf caps C14 and 17 which limit absolute peak currents anyway.

Based on the advice in our past discussion I had already changed C14 and C17 to 2000uF (paralleled two 1000uF caps). I wanted 2200uF but couldn't find 1100uF caps to parallel, so I settled for 2000uF. Does this change your analysis then? Given that C14 and C17 are now 2000uF, what is the optimum value for C12 and C18?

Larger caps mean the bridge rectifier conducts for less time (as the voltage on the cap falls less between each half cycle), but the same "energy" is used by the circuit it's powering, so that means the peak currents are correspondingly higher to replace the charge in the cap.
A smaller cap has larger ripple across it, but the angle of conduction of the bridge is longer, putting the energy back more slowly and at less peak current... which can be beneficial.

Large caps/short conduction angle can cause transformers to run hot due to copper and magnetic losses in the core... everything is a compromise... the regs do the job of reducing ripple... none should find it's way into the output rails or grounds.

Thank you for explaining this! I'm still trying to get my head over it but I'm beginning to understand. So I guess the key is to find just the right value for C12 and C18 - not too small that the ripple will become too large, and not too big that the transformer will be stressed. My problem is to determine or calculate "just the right value", but I'm not sure how to do that. Could you help me please? Thanks again! 🙂

 

[Mooly]

The ripple voltage is given by Vr = I/CF where Vr is the ripple voltage, I is the current drawn, C is capacitance in Farads and F is the frequency.

The ripple voltage needs to such that the input to the regs never comes close to the "drop out" voltage of the regulators which you can find in the data sheets. Then you have to figure in worst case conditions... mains voltage at it's lowest permisable level etc.

If you have a "small" cap... and even something like 470uf would probably be OK for the currents drawn by the load... the problem then is that the ripple current in the cap will be "high" for the physical size of cap and result in a shorter life.

Your circuit draws so little current (in absolute terms) that it's all a non problem tbh

Electroylitics have a huge tolerance too +50% -20% is common.

So for 2200 uf caps if your circuit draws 200 ma from a rail then that gives Vr = 0.2/ 2200e-6*120 which is 0.75 volt ripple.

2200 uf caps are perfect 🙂

 

[mtl777]

The ripple voltage is given by Vr = I/CF where Vr is the ripple voltage, I is the current drawn, C is capacitance in Farads and F is the frequency.

The ripple voltage needs to such that the input to the regs never comes close to the "drop out" voltage of the regulators which you can find in the data sheets. Then you have to figure in worst case conditions... mains voltage at it's lowest permisable level etc.

What exactly does "dropout voltage" mean. The LM340 datasheet is saying that the +15V reg has a dropout voltage of 2V. Does it mean that if the input voltage falls below 17V (15 + 2) the reg will fail to function?

If you have a "small" cap... and even something like 470uf would probably be OK for the currents drawn by the load... the problem then is that the ripple current in the cap will be "high" for the physical size of cap and result in a shorter life.

Because you mention "470uF", I just want to be sure... You're talking about the filter caps C12 and C18 after the diode bridge, not the originally 470uF caps C14 and C17, right? You mean even as small as 470uF for C12 and C18 would probably work?

Your circuit draws so little current (in absolute terms) that it's all a non problem tbh

Yes, it's only 175mA as per my test with the 1-Ohm resistor, so the 200mA you mention below is pretty close! Amazing! I wonder how you knew that?

Electroylitics have a huge tolerance too +50% -20% is common.

So for 2200 uf caps if your circuit draws 200 ma from a rail then that gives Vr = 0.2/ 2200e-6*120 which is 0.75 volt ripple.

How did you get the 120Hz frequency?

2200 uf caps are perfect 🙂

So the original design of 2200uF for the filter caps C12 and C18 was already optimal after all. Thanks a lot! 🙂

 

[Mooly]

Drop out voltage is just as you describe, the minimum the regulator needs to work with. It's the voltage you measure across the input and output pins.

I was referring to C12 and C18, but have made a mistake keep glancing back at the original circuit... it's a half wave doubler, not full so the ripple frequency is 60 hz not 120 hz... line frequency 🙂

As you have a scope why not measure the ripple voltage across C12, C18, C22 and C60 and see what you actually get in practice with it working.

 

[mtl777]

I was referring to C12 and C18, but have made a mistake keep glancing back at the original circuit... it's a half wave doubler, not full so the ripple frequency is 60 hz not 120 hz... line frequency 🙂

I made a mistake, too. The current for the 15V is actually 151mA. The 175mA was for the 5V. Based on 151mA current and 60Hz frequency,

Vr for 2200uF = 0.151 / (2200E-6 x 60) = 1.14V
Vr for 3400uF = 0.151 / (3400E-6 x 60) = 0.74V

Based on calculation it seems 3400uF is better, but what is its effect on the transformer? How do you calculate the "stress" it will create on the transformer? I'd like to compare 2200uF vs 3300uF as far as the "stress" on the transformer is concerned.

As you have a scope why not measure the ripple voltage across C12, C18, C22 and C60 and see what you actually get in practice with it working.

Good idea, and will validate theory vs practice. Thanks, I'll do that! 🙂

 

[Mooly]

To calculate transformer characteristics, temp rise etc would need a mountain of data from the manufacturer.

All we can do is be sensible, and run the device within it's specs. You won't run into issues with your circuit and values... the problems come when you are running the thing at 100% capacity, then start to do something silly like add 100's of thousands of microfarads on a full wave bridge.

Remember your circuit also has those caps feeding the diodes, so it will be interesting to measure the ripple and see how it all looks.

 

[mtl777]

To calculate transformer characteristics, temp rise etc would need a mountain of data from the manufacturer.

All we can do is be sensible, and run the device within it's specs. You won't run into issues with your circuit and values... the problems come when you are running the thing at 100% capacity, then start to do something silly like add 100's of thousands of microfarads on a full wave bridge.

Remember your circuit also has those caps feeding the diodes, so it will be interesting to measure the ripple and see how it all looks.

With C12 and C18 = 3400uF, C22 and C60 = 2200uF, I measured the ripple and got the following:

C12 : 0.6V
C18 : 0.6V
C22 : 1.6V
C60 : 1.1V

I was going to measure the ripple with C12 and C18 = 2200uF. Unfortunately, one of the through hole pads for C18 broke as I was trying to desolder the cap. What a bummer! 😡 So I spent the rest of the day trying to fix it. The JB Weld glue I used takes long to dry so I'll have to wait until tomorrow to do the ripple measurement for the 2200uF.

I also measured the actual capacitance of the 3400uF for C12. It is 3536uF. Based on the formula for ripple that you gave, the current at the time of measurement was:

I = Vr x C x F
I = 0.6 x 3536E-6 x 60 = 0.1273A

Now the actual capacitance of the 2200uF cap for C12 that I'm going to test tomorrow is 2262uF. Assuming the same current, I would expect the actual ripple for this cap to be:

Vr = I / (C x F) = 0.1273 / (2262E-6 x 60) = 0.94V

Assuming this calculation is accurate, should I choose 3400uF then since its 0.6V ripple is better? Are there any other considerations that might make me want to prefer 2200uF despite its higher ripple? 🙂

 

[Mooly]

It's getting very involved is this isn't it 🙂

With the 'scope on DC coupling you need to be sure that the minimum voltage to the input of the regs is safely above the drop out voltage... so that means you are looking at the minimum point in the "ripple" superimposed on the DC.

Lower ripple is better, as long as you don't get silly and start slapping 1000's of uf across the rails. If the 3400uf fits the PCB use it.

 

[mtl777]

It's getting very involved is this isn't it 🙂

Yeah, that broken pad was difficult, but I came up with a good idea to fix it and it worked!

Lower ripple is better, as long as you don't get silly and start slapping 1000's of uf across the rails. If the 3400uf fits the PCB use it.

Good to know that 3400uF is not an insanely high value! 🙂

I have now measured the actual ripple with C12 and C18 = 2200uF:

C12 : 0.95V
C18 : 0.95V

So my calculation was quite accurate! Yeah, it looks like 3400uF is the better choice for C12 and C18. Thanks! 🙂

BTW, does the ripple current rating of a capacitor affect the ripple measurement on the capacitor? I'm wondering if a capacitor with a high ripple rating tends to have less ripple on it than one with a low ripple rating. Or does the ripple always obey the formula Vr = I / (C x F) regardless of what the ripple rating of the capacitor is?

 

[Mooly]

Ripple current rating doesn't effect the values, but if you are running close to the maximum for a given cap then choosing one that is higher will give significantly longer life... but you won't be anywhere near that with this design and values.