Capacitor Value Calculations

Is that full or half-wave recification? 50 or 60Hz? What current are you pulling from the 6V line? These all affect capacitor selection.

... as a rough guide, with 50Hz FW rectified, you'll get about 1V droop per amp per 1,000μF.

... think that should be per 10,000μF (d.p. slipped in by brain).

Whatever's wrong with suck-it-and-see? Bung in a few μF, check the circuit with a 'scope, and make a judgement as to whether to increase it, that sort of thing.

This is what the breadboard approach is for.

It depend how much current you are drawing...!

If you want a quick rule of thumb I'd say about 330uF for each 100mA you are drawing... add a 0.1uF either side of the regulator too to stop any oscillation...Of course it there is any logic or microcontrollers you will need a good sprinkling of 0.1UF s around the board.

This is suposed to be a helpful post...If someone wishes to get into loads of calculation and longwinded discussion...fine, but this is just a quick dirty guide.

Oh...yes and over rate the capacitor votage by at least 50% e.g go to say a 35v cap'.

Del

Blimey..I just re-read it.... a 7806 !!!?? I hope you aren't drawing too much current else you will have a lot of heat to get rid of! 18v seems rather a lot to be starting with....

As others here have warned, you may want to downsize the input voltage to your 78L06 a bit.

-e, that's a great answer. I, myself, have never tried calculating capacitor values for power supplies.

Back when I was studying, it was a "bigger is better", sort of thing. There was also the "if you get hum and it's not because of incorrect grounding, put in a bigger cap". This is one reason why I have a box full of extra capacitors that my wife is always asking "don't you want to throw this away? You haven't used them since we were married"?

Back when I was studying, it was a "bigger is better", sort of thing. There was also the "if you get hum and it's not because of incorrect grounding, put in a bigger cap". This is one reason why I have a box full of extra capacitors that my wife is always asking "don't you want to throw this away? You haven't used them since we were married"?

What??? I coulda sworn that was my wife. At least it sure sounded like her.

Excellent answer!

My one comment: as I read 'Δt is the period of one cycle. Here in the US it is 1/60Hz', I read ...one sixtieth of a Hertz, whereas you (obviously, with a bit of thought) meant ...one divided by sixty Hertz. Since ∆t must be a time, I would prefer 1/60sec.

I don't mean to nit-pick - only to attain maximum clarity.

Dick

Good point! Yes, it would have been clearer to express Δt as (1/60) sec, etc.

Thank you for answering my question in such detail. Your answer has been extremely usful. I will come back you regarding this more time since i am in the middle of changing my design and a few other things.

God loves you because you help others.

An output of 6VDC requires a minimum of 8V input at pin1, 2V being the maximum voltage drop on the regulator. Maximum value of the supply voltage on the input filter capacitor being 18V, one can have a ripple voltage of up to 10V at the input.

CV=Q, Q =It where C = Capacitance in farad, V = the ripple voltage in volt, Q = charge on the capacitor, I = load current, and t = frequency of the ripple voltage.

Assume that the rectified output is full wave. Then t = 1/100 second for 50 Hz supply or 1/120 for 60Hz supply.

Assume the load current is 0.1A. V = ripple voltage = 10V. Then for a 50Hz mains supply,

C = (0.1 x 1/100 )/10 farad

This gives C = 10^-4 farad or 100 micro farad.

Any fluctuation in mains voltage or variation of load current is not considered in the above treatment. If the space allows, use a capacitor with a value higher than 100 micro farad, let us say 470 microfarad@50VDC.

I agree with you completely, its always good policy to use higher voltage capacitors than are theoretically needed (as you point out), they last longer if not being used in a critical part of there characteristics.

Price hardly makes a big enough difference to be worth talking about generally! As you said, space can be a problem, but there again, one should design for the bigger cap in the first place whenever one can.

My rectified output voltage is 18V.

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The absolute maximum power dissipation for this device determines the maximum load current you can draw before toasting your part. The power dissipation varies by package:

SOP-8: 300 mW

TO-92: 500 mW

SOT-89: 350 mW

Although the 78L06 can supply a max current of 100 mA, you still must take into consideration the device's power dissipation. This, in turn, will determine the maximum current you can draw. At 18 volts input, the max current you can draw (vs package style) will be less.

I = P/ΔV,

where,

I = device current, in amps

P = device power dissipation, in watts,

and

ΔV = the difference between the input voltage and the output (regulated) voltage. With 18 V input, ΔV = 18V (input) - 6V (output) = 12V.

Computing the max current we can draw while staying within the device's max power dissipation rating we have, for each package style:

I_SOT-8 = 0.300 W/12 V = 25 mA,

I_TO-92 = 0.500 W/12 V = 41 mA,

and

I_SOT-89 = 0.350 W/12 V = 29 mA.

These are absolute maximum values and are not recommended for continuous service! Your average load current should be less.

You can attach a heat sink to the regulator, and heat sinks are commonly available for TO-92 packages, but you may have trouble finding sinks for the others.

If you expect your load to draw more current (but less than or equal to the max of 100 mA), you need to reduce the input voltage to your regulator.

Let's say your load draws the maximum of 100 mA spec'd for this type of regulator. To ensure the power dissipation doesn't exceed the maximum, your input voltage (vs package style) should be no more than:

ΔV = P/I, where

ΔV_SOP-8 = 0.300 W / 0.100 A = 9 V,

ΔV_TO-92 = 0.500 W / 0.100 A = 11 V,

and

ΔV_SOT-89 = 0.350 W / 0.100 A = 9.5 V.

At 18 V input and 100 mA load current your device will dissipate 1.2 watts - but only for a few seconds!!!

Hope this has been helpful.

-e

Correction! That last section should read:

V_in(pkg) = V_out + P/I,

where,

V_in(SOP-8) = 6 V + (0.300 W / 0.100 A) = 9 V,

V_in(TO-92) = 6 V + (0.500 W / 0.100 A) = 11 V,

and

V_in(SOT-89) = 6 V + (0.350 W / 0.100 A) = 9.5 V