Capacitor types for 780x circuits

[edeca]

Up until now I've always used electrolytic capacitors for small power supply circuits. Is there any reason to only use them? I've got a box full of tantalum ones of varying values and not so many electrolytic ones. I'm assuming ceramic ones can't be used, but don't know why I assume that

I've found some good guides but most only seem to compare usage for high frequency or RF circuits, not what capacitors can be used for generally.

 

[ericgibbs]

Up until now I've always used electrolytic capacitors for small power supply circuits. Is there any reason to only use them? I've got a box full of tantalum ones of varying values and not so many electrolytic ones. I'm assuming ceramic ones can't be used, but don't know why I assume that

I've found some good guides but most only seem to compare usage for high frequency or RF circuits, not what capacitors can be used for generally.

Hi,
On the inp and oup of the 780X you should have a 100nF cap [ceramics are fine], prevents instabilty in the reg.

For reservoir and filtering you can use tantalums, if they are of suitable capacitance.

 

[edeca]

Thanks. Looking at the datasheet for an LM1117 it appears to have similar requirements to a 780x, 10uF on input and 22uF on output. I'm assuming these can be ceramic then, but is it worth having a reservoir capacitor if the input is a wall wart or USB?

 

[ericgibbs]

Thanks. Looking at the datasheet for an LM1117 it appears to have similar requirements to a 780x, 10uF on input and 22uF on output. I'm assuming these can be ceramic then, but is it worth having a reservoir capacitor if the input is a wall wart or USB?

hi,
When you say ceramic, do you mean the resin dipped type, they will be big.?

If the wall wart is regulated, the reservoir cap is optional, I would fit it, if the budget allows.
Check the LM1117 for the requirement for 100nF, maybe required.

 

[edeca]

Actually I was referring to the little ceramic ones, almost like a flat disc. I now understand you meant the dipped ones. The datasheet link below doesn't mention the 100nF capacitors, what makes you think of them?

**broken link removed**

As for a reservoir, how do I calculate a sensible size? Or a sensible minimum?

 

[ericgibbs]

Actually I was referring to the little ceramic ones, almost like a flat disc. I now understand you meant the dipped ones. The datasheet link below doesn't mention the 100nF capacitors, what makes you think of them?

**broken link removed**

As for a reservoir, how do I calculate a sensible size? Or a sensible minimum?

hi,
The 100nF is required for the 780X series, I just thought the same may apply to the LM1117.

The size of the res cap depends upon the load current.
About 470uF should be OK for the LM1117, 800mA limit.

 

[edeca]

Cheers Eric. Is there a simple rule of thumb for the reservoir capacitors? I'm not too interested in deep magical calculations, just something approximate

I'll get on and build tonight.

 

[ericgibbs]

Cheers Eric. Is there a simple rule of thumb for the reservoir capacitors? I'm not too interested in deep magical calculations, just something approximate

I'll get on and build tonight.

hi,
Not too much maths here, look at the online calc at the bottom of the link page.

**broken link removed**

EDIT: really depends a lot on the use of the psu, I use 470uF to 1000uF per 1 amp load current for smoothing and about 220uF to 470uF for reservoir cap.
I always fit 100nF either side of the Vreg to improve the high freq noise reduction.

 

[audioguru]

The LM1117 is a quasi-low-dropout regulator that is completely different from an ordinary 78xx and LM317 regulator. The LM1117 is recommended to have a 10uF tantalum input capacitor (the others need an input ceramic disc capacitor of from 0.1uF to 0.33uF only if the main filter capacitor is far away).

The LM1117 MUST have a 10uF (at least) tantalum output capacitor and it must have an ESR of from 0.3 ohms to 22 ohms to avoid oscillation. When the LM1117 has a Cadj capacitor then the output capacitor must be increased to at least a 22uF tantalum. Ordinary regulators don't need an output capacitor but one is recommended to improve transient response and to make the output have a low impedance at high frequencies.

 

[Speakerguy]

DO NOT USE A CERAMIC WITH THE LM1117!!!!

For standard regulators and most new LDO's, Eric is right - put a 100nF ceramic on it. On most all older and a few newer LDO regulators, you MUST follow the data sheet exactly to avoid oscillation. The capacitance and ESR value of the output capacitor must be within a certain range because it is used as a compensation cap to place a zero in the bode plot of the device, giving it more phase margin and ensuring stability.

Check the device data sheet and the data sheet for your capacitor and make sure everything matches up. Hopefully these older finicky LDO's will go away in time and the newer ones that are stable with ceramic caps on the output will take over.

 

[Speakerguy]

Dang, AG beat me to it.

 

[sniper007]

hi,
Not too much maths here, look at the online calc at the bottom of the link page.

**broken link removed**

EDIT: really depends a lot on the use of the psu, I use 470uF to 1000uF per 1 amp load current for smoothing and about 220uF to 470uF for reservoir cap.
I always fit 100nF either side of the Vreg to improve the high freq noise reduction.

Hi

I read this useful link and something don't understand

If we increase the capacity of smoothing capacitor, ripple will be lessen.
But we must not increase it in vain because we have to consider the charging current.
If so huge charging current runs, it will damage precision instruments.

Is capacity influence to current which charge capacitor ?

 

[ericgibbs]

Hi

I read this useful link and something don't understand
Is capacity influence to current which charge capacitor ?

hi,
Consider the voltage on the capacitor following the bridge rectifier.

If the capacitor is a high value, little of the voltage charge discharges thru the load.
So when the rectified waveform halfcycle voltage is less than the remaining voltage on the cap, the diode dosnt conduct as its reversed biassed, so no current flows thru the diode.

As the halfcycle voltage rises it reaches a point where it becomes forward biassed with respect to the voltage on the cap, so current flows thru the diode.

So effectively the diode only conducts when its topping up the voltage on the cap. If the cap is very large the diode conduction time can be very short, during which time the cap charging current is very high.

If you dont follow, please say, I'll post a sketch to explain..

 

[ericgibbs]

hi,
Did a sketch just in case.

Do you follow OK.?

 

[sniper007]

If the capacitor is a high value, little of the voltage charge discharges thru the load.

What does this mean? Sorry, but my English is not very well...

 

[ericgibbs]

What does this mean? Sorry, but my English is not very well...

hi,
It means the voltage on the capacitor will only fall by a small amount.

A simple numerical example would be:
Assume the peak of the positive half wave cycle input charges the capacitor to 20Vdc

As the cycle falls to zero, the capacitor has to supply the current to the load
and the voltage on the capacitor decays exponentially.

The capacitor has to continue supplying current to the load until the next positive half cycle input voltage
exceeds the voltage on the capacitor [allow for diode Vdrop].

When the positive half cycle exceeds the voltage on the capacitor the diode will conduct
and the capacitor will be recharged to say 20Vdc

The diode has to conduct a high current in a short time in order to recharge the capacitor. Look at my diagram

Another rough example, suppose the capacitor supplies a current of 1amp to the load for about 20mSec [50Hz mains]

If the voltage on the capacitor has only fallen by say 5V during the 20mSec discharge time, means that the diode has to top up [replace the capacitor charge] in about 2 mSec, so it has to conduct a 10amp current for 2mSec.

These are rough numbers for explanation only, the actual values will depend upon your circuit.

Do you follow OK.?

 

[sniper007]

OK, now understand

Thanks you very much

 

[sniper007]

Another question

Excuse me, if it nonsense but nevertheless...


Is there when signal cross 2 diodes (bridge rectifier) for first half period present any crossover distortion like this:

**broken link removed**

 

[ericgibbs]

hi,
As diodes have a nonlinear forward voltage/current characteristic distortion will occur.

You may know that a 1N4001 starts forward conduction at about 0.65V.

The rise in current after conduction starts as the applied voltage increases is not linear.

Is this what you mean.?

 

[sniper007]

Yes, tnx