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Why can't you take the 12.6 volts supply for the 20a and at about 10
watts out the current draw is about 2 amps. Subtract about 300mas no
mod current and you get 12.6 x 1.7 = 21.42 watts. If you figure 50%
efficency you would have about 10.5 watts out.

If that doesn't work, let me know. It makes sense to me.
Leonard

--- In BITX20@... <mailto:BITX20%40yahoogroups.com>, arv
<arvid.evans@...> wrote:

Arnab - VU2BPW

This sometimes confuses me as well.

It seems that the key is in impedance relative to voltage and

current. If

we lower the impedance the current will increase and voltage will stay
the same.

In case of a push-pull output stage like the BITX20A, the output
transformer has a turns arrangement of 6 turns on the primary and 8

turns

on the secondary. Since the antenna presents a 50 ohm load and the
antenna LPF is designed for 50 ohms on each end, the impedance seen
by the MOSFETs would be lower than 50 ohms. This allows the current to
be more than would be the case if the MOSFETs were operating into a 50
ohm load.

In the original BITX and in the BITX Version-3, the final amplifier

is a

single
MOSFET operating into an autotransformer arrangement. This lowers the
impedance seen by the MOSFET, and allows higher current and higher
power than would be the case if the MOSFET were operating into a 50 ohm
load.

If we assume that a lower load impedance at the output device causes

more

current to flow, then increasing power is a matter of changing the

output

impedance transformation so that more current can flow and still

maintain

a 50 ohm match to the antenna. If the voltage remains the same, then the
impedance will be lower and power will be higher. Of course there

are some

limitations to this idea. Resistance in transformer windings, core
saturation
in transformers, on-resistance in MOSFETs, and non-linearity in MOSFET
devices all conspire to work against us in low voltage and high power
situations.

Arv - K7HKL
_._


On 06/20/2013 11:08 AM, arnab bhaumik wrote:

hi arv,

i am still a little bit confused. with 12volt and a single irf510
based design, how much rf power can we get??? 12volt rf across 50 ohm
load, so my calculation goes

12v * .707 = 8.484 rms

rms square = 8.484 * 8.484 = 71.978

rms square / 50 = 71.978/50 = 1.43watt.

it seems with the math formula we can get max 1.43 watt from 12volt
with single irf. thats why people increase the linear voltage to
24volt . so that they can get rf of around 5.75watt.

please clarify this.

arnab/vu2bpw

ps - this is the reason i tried irf push pull to get 5watt with

12volt

supply voltage. (but failed)

________________________________
From: arv <arvid.evans@... <mailto:arvid.evans%40gmail.com>>
To: BITX20@... <mailto:BITX20%40yahoogroups.com>

<mailto:BITX20%40yahoogroups.com>

Sent: Thursday, 20 June 2013 9:52 PM
Subject: Re: [BITX20] More Power Out From Bitx 20/3?



Rob

I think that it may take more than just increasing voltage to the

output

MOSFET. If you change the voltage the current will change and

impedances

involved will change. That would seem to require that the output
transformer
would then need to be re-designed for the new impedance in order

to match

the output LPF and antenna impedance.

Arv - K7HKL
_._

On 06/20/2013 07:43 AM, rob_kay14758 wrote:

Bitx is built and seems to be working o.k. on 20M,running on

external

13v supply. If i increased the voltage on the pa transistor only to
say 25v would we see much of an increase in power?
This circuit also looked interesting as an add on..........

Any thoughts....
Thanks
Rob