From cign–(at)–elios.phy.OhioU.Edu Sat Dec 21 21:33:19 CST 1996

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From: cign–(at)–elios.phy.OhioU.Edu (Dave Cigna)

Subject: Re: Tube amp design notes.

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Date: Sat, 21 Dec 1996 16:23:00 GMT

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Andy Moss

>Henry Pasternack wrote:

>

>Henry, you wrote a few messages back about two 10K bleeder resistors in

>your tube amplifier. I had mentioned that I found that rather suspect

>and you, if I recall, stated it was a generally accepted way of doing

>things.

>

>I still don’t believe. I would like you to provide some kind of *proof*

>to this end.

I had hoped that someone that understands the subtleties of power supplies

would answer this, but I’ll take a swing at it. The mathematics are too

involved to honor your demand for a proof, but I might be able to help

you understand what’s happening.

Consider full wave rectified AC as it comes from an ideal transformer and

rectifier. If you perform Fourier analysis on it, you’ll find that it

contains a large DC component and a series of even order AC components

(i.e. for 60 Hz input you get DC + 120Hz + 240Hz + 360Hz + …). It

happens that the DC component is close to .9 of the RMS transformer

voltage.

The wonderful thing about choke input filters is that they are great

low-pass filters. They almost completely block all of the AC components

so that all you’re left with is the DC part. Notice that this does

not depend on any capacitors charging and discharging, so the output

voltage is pretty much a constant .9 Vrms DC with a small AC ripple

independent of the load. Almost.

With very small loads (low current) the capacitor after the choke starts

charging and discharging so that the whole filter acts like a capacitor

input type. The output voltage is higher than .9 Vrms and approaches

1.44 Vrms at zero load. To make this plausible, let’s imagine removing

the choke so that we’re left with a simple capacitor input filter. We

know that the cap charges during some part of the AC cycle and discharges

during the rest of the cycle. No diodes are conducting when the cap is

discharging. Only when the transformer voltage is higher that the cap

voltage (plus any diode drop) does a diode conduct. This is called the

conduction angle. With a very light load the cap charges to almost the

peak transformer voltage, so the transformer voltage is higher than the

cap voltage for only a tiny fraction of the total AC cycle and the

conduction angle is only a few degrees.

OK, now let’s insert a *small* choke before the capacitor. The effect of

the choke is that the conduction angle will be stretched out slightly

(remember that inductors resist changes in current). At the same time,

the peak voltage at the output of the choke will be reduced slightly, so

the cap won’t charge to quite as high a voltage. Now imagine increasing the

size of the choke. At some value of inductance the conduction angle of

each of the two diodes will reach 180 degrees so that there will always be

least one diode conducting. As long we have at least this minimum value

of inductance the filter will behave like a choke input type and we will

have something close to .9 Vrms DC at the output.

Clearly, the critical value of of inductance depends on the load current.

What’s not obvious is that it also depends on the supply voltage. After

evaluating all the pi’s and square roots of two you get the relationship:

V

L = ——— (I is load current in amps)

.0008 * I

If we express the current in mA, then the denominator becomes .8 * I.

But .8 is close to 1, so most people use the easy to remember rule of

thumb for the critical value of inductance

V

L = — (I is measured in mA)

I

Note that we can rearrange this equation to give us the minimum current

for a given supply voltage and choke. For example, with a 440 volt supply

and a 10 henry choke we need to make sure that at least 440/10 = 44 mA of

current is being drawn from the supply at all times. As Henry discovered

though, filter chokes often exhibit higher inductance at low currents so,

depending on the specific components, you might not need to draw this

much in practice.

Hope this is the science you were looking for.

— Dave Cigna

From cign–(at)–elios.phy.OhioU.Edu Sat Dec 21 21:33:36 CST 1996

Article: 20555 of rec.audio.tubes

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From: cign–(at)–elios.phy.OhioU.Edu (Dave Cigna)

Subject: Re: Tube amp design notes.

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Date: Sat, 21 Dec 1996 16:53:02 GMT

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Dave Cigna

>Clearly, the critical value of of inductance depends on the load current.

>What’s not obvious is that it also depends on the supply voltage. After

>evaluating all the pi’s and square roots of two you get the relationship:

>

> V

> L = ——— (I is load current in amps)

>.0008 * I

>

>If we express the current in mA, then the denominator becomes .8 * I.

Oops. I was working from memory. That should’ve been

.0008 * V

L = ——— (I is load current in amps)

I

Now if we use mA the numerator becomes .8 * V.

My conclusion was correct though:

>But .8 is close to 1, so most people use the easy to remember rule of

>thumb for the critical value of inductance

>

> V

> L = — (I is measured in mA)

> I

— Dave Cigna