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Re: [Amps] DC Blocking Capacitor

To: "Phil Clements" <philc@texascellnet.com>,"'Bob Easterbrooks'" <k6jqa@jasonsplace.info>
Subject: Re: [Amps] DC Blocking Capacitor
From: Karl-Arne Markström <sm0aom@telia.com>
Reply-to: Karl-Arne Markström <sm0aom@telia.com>
Date: Wed, 27 Apr 2005 07:50:33 +0200
List-post: <mailto:amps@contesting.com>
I wonder if you may have gotten the numbers "backwards"...

In my understanding of AC circuits, the requirement for the reactance of the 
blocking capacitor is
to decrease with decreasing plate load impedance, in order to keep the phase 
angle
of the series combination of load + capacitor within limits, say 5 degrees of 
negative phase angle

So, a 1000 ohms plate load at the same frequency would require an 8 times lower 
reactance, or 
8 times higher capacitance than the required capacitor for an 8000 ohms plate 
load.

Working out two examples:

(1)    Plate load 8000 ohms, frequency 1.8 MHz, allowed phase angle as seen 
from the tube output
        - 5 degrees. Maximum reactance of the blocking capacitor = 8000 * 
sin(-5) = -697 ohms
        which corresponds to C min = 1/(2*pi*f*697) = 127 pF

(2)    Plate load 1000 ohms, frequency 1.8 MHz, allowed phase angle as seen 
from the tube output
        - 5 degrees. Maximum reactance of the blocking capacitor = 1000 * 
sin(-5) = -87 ohms
        which corresponds to C min = 1/(2*pi*f*87) = 1014 pF

If the plate tuning could not be changed, the plate load would be somewhat 
higher than optimum, as 
Z = SQR (R**2 + X**2) and with capacitive phase angle, but when the network is 
adjusted for resonance, the contribution from the blocking capacitor will be 
absorbed into the C1 value, making the phase angle = 0 as seen from the tube.

This leaves only the reactive voltage drop over the blocking capacitor.

For 1500 W output, the fundamental component of the current in (1) = 
SQR(1500/8000) = 0,43A
and the voltage drop = 0,43*697 = 301 V. Compared to the fundamental voltage 
across the tube output this corresponds to 301/(SQR(1500*8000)) = 0,086 which 
should change the operating point of the tube only with a negligible amount.


73/

Karl-Arne
SM0AOM


        


----- Original Message ----- 
From: "Phil Clements" <philc@texascellnet.com>
To: <R@contesting.com>; "'Bob Easterbrooks'" <k6jqa@jasonsplace.info>
Cc: <amps@contesting.com>
Sent: Wednesday, April 27, 2005 1:17 AM
Subject: Re: [Amps] DC Blocking Capacitor


> 
> Amen, "R"! I think that so many guys over the years used whatever they had
> in their junk boxes that looked big enough that a proliferation took place.
> Sort of like parasitic suppressers...if they worked fine on tube type A they
> should work okay on tube type B.
> 
> A typical amp running max. smoke from a pair of 3-500Z's or a single 8877
> works out to a plate resistance of around 2200 ohms or so. Plugging in the
> formula, 5% of this means we need a capacitive reactance in our C/Block of
> around 110 ohms @ 1.8 mhz. Your understanding of 200 pf is enough is
> correct, as a 200 pf capacitor would have a reactance of around 440 ohms on
> 160 meters. At a lower plate load impedance, say 1000 ohms, you have even
> more headroom. You would have to exceed 8000 ohms plate load impedance to
> get even close to exceeding the 5% requirement with a 200 pf cap.
> 
> (((73)))
> Phil Clements, K5PC
> 
> 
> 
> 
> 
> 
> Subject: Re: [Amps] DC Blocking Capacitor
> 
> The value of the DC blocking cap is seeming less critical than is 
> commonly thought.  As I understand it, in most HF amplifiers, 200pF is 
> enough.  However, the I-handling ability become important on the 28MHz 
> band since most of the tank circulating-current passes through the DC 
> blocker on its way to and from the anode..
> 
> On Apr 26, 2005, at 12:26 PM, Bob Easterbrooks wrote:
> 
> > Gentleman, What determines the capacitive value of the DC blocking 
> > capacitor
> > used in a convential shunt fed PI, and can the same rules apply to a 
> > simple
> > link coupled tank circuit?   I thank you in advance.  Bob, K6JQA
> 
> 
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