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Lm388 1.3 - 30v power supply

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Sceadwian

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I have no idea why they suggest using such a massive cap, the data sheet only recommends a .1u for regulation at 5amps on the input unless you really need the hold up time the cap is wasted money and space. You could use the 100u but there's no reason you can't completely omit the 14000u one.
 

waterbug

New Member
Thanks for reply

Baring waste of time (money is not an issue), does the 140000 or 10000uf spec have any bearing on the control of the circuit? What is the purpose of the caps in the circuit. I'm just beginging to learn theory. Any help in learning is greatly appreciated.
Tom
 

audioguru

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The LM338 has an output of up to 5A. When the current is as high as 5A then the rectifier's filter capacitor must be huge.
With "only" 10,000uF the ripple voltage is 4V peak-to-peak and a larger capacitor is better.

The circuit from Aaron's Homepage has many errors.
 

Sceadwian

Banned
Ahh, I think I might understand now, sorry for my ignorance =) The PDF show's a .1u cap, but that looks like it's for a straight DC incoming voltage. Can you get linear regulators that have better ripple removal at higher currents without using such big caps, or is that just the way it is with most linear regulators?
 

MikeMl

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MikeMl

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I have no idea why they suggest using such a massive cap, the data sheet only recommends a .1u for regulation at 5amps on the input unless you really need the hold up time the cap is wasted money and space. You could use the 100u but there's no reason you can't completely omit the 14000u one.
BzzzzTTTT! The big cap has to be there to store energy between full wave rectified cycles from the bridge rectifier. At 60 Hz, the pulses out of the rectifier recharge said capacitor every 8.33ms, so all of the current delivered to the load COMEs out of capacitor until the next pulse.

Suppose that the power supply is trying to deliver 24V at 5A. The big capacitor will charge to about 1.2 times the rms voltage on the transformer secondary (1.2 because of transformer impedance and voltage drops across two silicon rectifiers in series). 1.2*24 = 28.8V. Since the supply is trying to deliver 24V to the load, that leaves only 4.8V of headroom.

Some of the headroom is used up by the dropout voltage of the regulator, which per the datasheet is over 2.5v @ 5A. That leaves an allowable sag of 1.7V of the capacitor voltage during the 8msec before the next current pulse gets there.

So how big does the capacitor have to be to not let the voltage sag more than 1.7V with 5A of current discharging the capacitor? q=C*ΔV=i*Δt. Rearranging, C=i*Δt/ΔV= 5*0.008/1.7=0.024F (yes, that is Farads), or 24,000uF.

You can parallel as many caps as you want, but the total capacitance must be > than 24,000uF if you want your supply to deliver 5A at 24V. This supply will not deliver more than a few mA at 35V. btw- the voltage rating of your capacitor(s) should be ≥ 40Vw
 

Hero999

Banned
[FONT=Arial, Helvetica, sans-serif]I've never seen a 14000µFcapacitor before, it's not standard value.

The regulator will not be able to supply 5A at low output voltages.
[/FONT][FONT=Arial, Helvetica, sans-serif][/FONT]
 

waterbug

New Member
Great feedback

you guys have helped alot. I have another question. I would like to install an amp and volt meter in the circuit. would the ampmeter be in series on the top leg? And the volt meter across the two?
 

mananshah93

New Member
BzzzzTTTT! The big cap has to be there to store energy between full wave rectified cycles from the bridge rectifier. At 60 Hz, the pulses out of the rectifier recharge said capacitor every 8.33ms, so all of the current delivered to the load COMEs out of capacitor until the next pulse.

Suppose that the power supply is trying to deliver 24V at 5A. The big capacitor will charge to about 1.2 times the rms voltage on the transformer secondary (1.2 because of transformer impedance and voltage drops across two silicon rectifiers in series). 1.2*24 = 28.8V. Since the supply is trying to deliver 24V to the load, that leaves only 4.8V of headroom.

Some of the headroom is used up by the dropout voltage of the regulator, which per the datasheet is over 2.5v @ 5A. That leaves an allowable sag of 1.7V of the capacitor voltage during the 8msec before the next current pulse gets there.

So how big does the capacitor have to be to not let the voltage sag more than 1.7V with 5A of current discharging the capacitor? q=C*¦¤V=i*¦¤t. Rearranging, C=i*¦¤t/¦¤V= 5*0.008/1.7=0.024F (yes, that is Farads), or 24,000uF.

You can parallel as many caps as you want, but the total capacitance must be > than 24,000uF if you want your supply to deliver 5A at 24V. This supply will not deliver more than a few mA at 35V. btw- the voltage rating of your capacitor(s) should be ¡Ý 40Vw
Understanded the explaination for value of C1,but why C2 is there for 100uF 50 V??
 

audioguru

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Understanded the explaination for value of C1,but why C2 is there for 100uF 50 V??
The project originally came from Aaron Cake's site Aaron's Homepage where there are many errors with projects.
This one has the 100uF capacitor wrong (maybe it should be 100nF) and it has the 240 ohm resistor instead of a 120 ohm resistor. It might have more things wrong.
 

Hero999

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MikeMl

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Understanded the explaination for value of C1,but why C2 is there for 100uF 50 V??
I just went and looked at the schematic in the link by the OP. I'm guessing that the author was concerned that the regulator might oscillate because it is mounted several cm from the "big" ripple filter capacitor, which is huge, and is mounted away from the regulator.

The guidance from the regulator maker suggests placing an "input" capacitor close to the reg if the wires are longer than X (I cant remember what X is, go look at the data sheet). I have built a similar supply (for 15V), and I didn't bother with C2. No instability detected.
 

mananshah93

New Member
I am trying to make, 7.5V, 2Amp. power supply using LM338, after calculating, my C1 comes out 3300uF, for 9V transformer, (50Hz, 230V) supply, drop out voltage for LM338 at 2Amp. is 2.1V...C3 can be 0.1uF, no C4..
Is it correct?
 

MikeMl

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Peak voltage(loaded) = 9*1.2=10.8V
Peak voltage(unloaded) ≈ 9 * 1.5 = 13.5V
Allowed ripple ΔV=10.8-7.5-2.1=1.2V
Δt @50Hz full-wave is 0.01s
i = 2A
C=i*Δt/ΔV = 2 * 0.01/1.2 = 16.667e-3 = 16667uF @ 15Vdc
 

MikeMl

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The 10% ripple factor used in the reference is somewhat arbitrary. My method estimates how low the filter cap can sag without the voltage regulator dropping out of regulation.

Both methods could be improved if you knew more about the transformer. Transformers have an effective internal series resistance. The 1.2 factor in my method allows for some internal voltage drop inside the transformer at rated current.
 

Mr RB

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That schematic is also missing the critically important cap from the LM388 adjust pin to ground. 10uF should do it, but 10uF with a 0.1uF in parallel is better.
 

audioguru

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The cap from the ADJ pin to ground reduces noise that is not noticeable to 20db less which is barely measureable.
 
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Mr RB

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If they were 2 fixed resistors, and it was a little power supply I might agree with you. ;)

Imagine what happens if (when?) the pot gets a momentary open circuit (typical of dust in any crackly pot) while the 30 volt 5 amp PSU is powering a delicate low-voltage device... Remember its got a zillion uF input cap at 30+ volts. :eek:
 
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