J M Noeding wrote:
Hi

planning a power supply for NEC 36284 150W 2m linear amplifier with
2SC3286, see http://home.online.no/~la8ak/images/nec36284c.gif

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

Wonder what size capacitors I should use to keep ripple minimum above
+24V?

Work it out for sure, using Duncan Munro's PSU Designer program. Follow
the link from the 'In Practice' pages:
http://www.ifwtech.co.uk/g3sek/in-prac/best-of.htm#psud


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

On Sun, 30 Nov 2003 14:43:48 -0000, "Frank Dinger"
wrote:

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

Wonder what size capacitors I should use to keep ripple minimum above
+24V?
=====
As a rule of thumb not less than 2000 microFarad per ampere
Hence for a 20 Amperes supply a total of 40KmicroFarad or more.

Aren't we forgetting something, chaps? It's for Frank to tell us the
maximum level of ripple he's prepared to tolerate. You can't give a
meaningful answer without that, can you?
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

On Sun, 30 Nov 2003 14:43:48 -0000, "Frank Dinger"
wrote:

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

Wonder what size capacitors I should use to keep ripple minimum above
+24V?
=====
As a rule of thumb not less than 2000 microFarad per ampere
Hence for a 20 Amperes supply a total of 40KmicroFarad or more.

Aren't we forgetting something, chaps? It's for Frank to tell us the
maximum level of ripple he's prepared to tolerate. You can't give a
meaningful answer without that, can you?
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

On Sun, 30 Nov 2003 14:43:48 -0000, "Frank Dinger"
wrote:

planning a power supply for NEC 36284 150W 2m linear amplifier with
2SC3286, see http://home.online.no/~la8ak/images/nec36284c.gif

Is this really a 2m amplifier with a large number of striplines? I
guess that the amplifier would be quite bulky at such low frequencies.

Assuming 50 % efficiency, 150 W RF would require 300 W DC input power
or 12.5 A.

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

The secondary peak voltage Vpeak will be about 1.4*24 V or about 34 V.

Since the rectifier voltage drop is at least 1 V at large currents and
there are two rectifiers in series, the voltage drop is at least 2 V,
so the peak capacitor voltage is less than 32 V.


Wonder what size capacitors I should use to keep ripple minimum above
+24V?


As the lowest allowed voltage is 24 V, the capacitor voltage will drop
from 32 to 24 V (dV=8 V) between the half cycle peaks.

The hard part is to calculate when the rectifiers stop conducting
after the peak and when they start to conduct again in the next half
cycle.

The worst case capacitance could be calculated assuming that the
capacitor must deliver the current for the whole half cycle, thus 10ms
for 50 Hz full wave rectification. Since dV = I*t/C and rearranging C
= I*t/dV, thus the required capacitance would be C = 12.5 A*0.01 s / 8
V or 15.6 mF.

However, since the final capacitor voltage Vl is well below the
secondary peak voltage, the rectifiers start to conduct well before
the peak of the next half cycle. The conduction angle is given by
arcsin(Vl/Vpeak) so in this case arcsin(24 V/34 V) or arcsin(26V/34V)
= 50 degrees, if the rectifier voltage drop is included. 50 degrees is
more than 2 ms before the next peak.

Since the conducting after the previous peak occurred slightly after
the peak (due to the serial impedances in the transformer and
rectifiers), a good guess for the time when the rectifiers are not
conducting is 7,5 ms in this case.

Using this shorter storage time, the required capacitance can be
recalculated as 11.7 mF or 11700 uF.

It should be noted that in this case the 8 V is the peak-to-peak
ripple voltage and 24 V minimum output voltage during a cycle, which
is quite appropriate if series pass regulation is used. Some design
equations especially from the tube era talk about RMS ripple voltage
and average DC voltage, which was more useful when a choke was used in
the high voltage supply.

=====
As a rule of thumb not less than 2000 microFarad per ampere

That gives a voltage drop rate of 0.5 V/ms.

With discharge times from about 6 ms (60 Hz full wave) to 20 ms (50 Hz
half wave) the voltage drop would be 3 .. 10 V.

If this is much or not, depends on the voltage levels used. A 3-4 V
drop might be appropriate for a full wave 50/60 Hz system generating
12 Vdc, but for a 24 Vdc and higher systems a higher ripple can
usually be tolerated, thus the capacitor can be reduced.

Hence for a 20 Amperes supply a total of 40KmicroFarad or more.

Hopefully the amplifier will not consume 20 A (480 W), since the
efficiency would be quite low :-).

Paul OH3LWR

On Sun, 30 Nov 2003 14:43:48 -0000, "Frank Dinger"
wrote:

planning a power supply for NEC 36284 150W 2m linear amplifier with
2SC3286, see http://home.online.no/~la8ak/images/nec36284c.gif

Is this really a 2m amplifier with a large number of striplines? I
guess that the amplifier would be quite bulky at such low frequencies.

Assuming 50 % efficiency, 150 W RF would require 300 W DC input power
or 12.5 A.

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

The secondary peak voltage Vpeak will be about 1.4*24 V or about 34 V.

Since the rectifier voltage drop is at least 1 V at large currents and
there are two rectifiers in series, the voltage drop is at least 2 V,
so the peak capacitor voltage is less than 32 V.


Wonder what size capacitors I should use to keep ripple minimum above
+24V?


As the lowest allowed voltage is 24 V, the capacitor voltage will drop
from 32 to 24 V (dV=8 V) between the half cycle peaks.

The hard part is to calculate when the rectifiers stop conducting
after the peak and when they start to conduct again in the next half
cycle.

The worst case capacitance could be calculated assuming that the
capacitor must deliver the current for the whole half cycle, thus 10ms
for 50 Hz full wave rectification. Since dV = I*t/C and rearranging C
= I*t/dV, thus the required capacitance would be C = 12.5 A*0.01 s / 8
V or 15.6 mF.

However, since the final capacitor voltage Vl is well below the
secondary peak voltage, the rectifiers start to conduct well before
the peak of the next half cycle. The conduction angle is given by
arcsin(Vl/Vpeak) so in this case arcsin(24 V/34 V) or arcsin(26V/34V)
= 50 degrees, if the rectifier voltage drop is included. 50 degrees is
more than 2 ms before the next peak.

Since the conducting after the previous peak occurred slightly after
the peak (due to the serial impedances in the transformer and
rectifiers), a good guess for the time when the rectifiers are not
conducting is 7,5 ms in this case.

Using this shorter storage time, the required capacitance can be
recalculated as 11.7 mF or 11700 uF.

It should be noted that in this case the 8 V is the peak-to-peak
ripple voltage and 24 V minimum output voltage during a cycle, which
is quite appropriate if series pass regulation is used. Some design
equations especially from the tube era talk about RMS ripple voltage
and average DC voltage, which was more useful when a choke was used in
the high voltage supply.

=====
As a rule of thumb not less than 2000 microFarad per ampere

That gives a voltage drop rate of 0.5 V/ms.

With discharge times from about 6 ms (60 Hz full wave) to 20 ms (50 Hz
half wave) the voltage drop would be 3 .. 10 V.

If this is much or not, depends on the voltage levels used. A 3-4 V
drop might be appropriate for a full wave 50/60 Hz system generating
12 Vdc, but for a 24 Vdc and higher systems a higher ripple can
usually be tolerated, thus the capacitor can be reduced.

Hence for a 20 Amperes supply a total of 40KmicroFarad or more.

Hopefully the amplifier will not consume 20 A (480 W), since the
efficiency would be quite low :-).

Paul OH3LWR

If it helps....,
the amplifier is shown on http://home.online.no/~la8ak/d24.htm
but I've written text the text in Norwegian since material is
available for radio clubs in this area. But at least it is some
pictures and circuit diagram for the amplifier

73 LA8AK
-------
On Mon, 01 Dec 2003 12:07:57 +0200, Paul Keinanen
wrote:

On Sun, 30 Nov 2003 14:43:48 -0000, "Frank Dinger"
wrote:

planning a power supply for NEC 36284 150W 2m linear amplifier with
2SC3286, see http://home.online.no/~la8ak/images/nec36284c.gif

Is this really a 2m amplifier with a large number of striplines? I
guess that the amplifier would be quite bulky at such low frequencies.

Assuming 50 % efficiency, 150 W RF would require 300 W DC input power
or 12.5 A.

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

The secondary peak voltage Vpeak will be about 1.4*24 V or about 34 V.

Since the rectifier voltage drop is at least 1 V at large currents and
there are two rectifiers in series, the voltage drop is at least 2 V,
so the peak capacitor voltage is less than 32 V.


Wonder what size capacitors I should use to keep ripple minimum above
+24V?


As the lowest allowed voltage is 24 V, the capacitor voltage will drop
from 32 to 24 V (dV=8 V) between the half cycle peaks.

The hard part is to calculate when the rectifiers stop conducting
after the peak and when they start to conduct again in the next half
cycle.

The worst case capacitance could be calculated assuming that the
capacitor must deliver the current for the whole half cycle, thus 10ms
for 50 Hz full wave rectification. Since dV = I*t/C and rearranging C
= I*t/dV, thus the required capacitance would be C = 12.5 A*0.01 s / 8
V or 15.6 mF.

However, since the final capacitor voltage Vl is well below the
secondary peak voltage, the rectifiers start to conduct well before
the peak of the next half cycle. The conduction angle is given by
arcsin(Vl/Vpeak) so in this case arcsin(24 V/34 V) or arcsin(26V/34V)
= 50 degrees, if the rectifier voltage drop is included. 50 degrees is
more than 2 ms before the next peak.

Since the conducting after the previous peak occurred slightly after
the peak (due to the serial impedances in the transformer and
rectifiers), a good guess for the time when the rectifiers are not
conducting is 7,5 ms in this case.

Using this shorter storage time, the required capacitance can be
recalculated as 11.7 mF or 11700 uF.

It should be noted that in this case the 8 V is the peak-to-peak
ripple voltage and 24 V minimum output voltage during a cycle, which
is quite appropriate if series pass regulation is used. Some design
equations especially from the tube era talk about RMS ripple voltage
and average DC voltage, which was more useful when a choke was used in
the high voltage supply.

=====
As a rule of thumb not less than 2000 microFarad per ampere

That gives a voltage drop rate of 0.5 V/ms.

With discharge times from about 6 ms (60 Hz full wave) to 20 ms (50 Hz
half wave) the voltage drop would be 3 .. 10 V.

If this is much or not, depends on the voltage levels used. A 3-4 V
drop might be appropriate for a full wave 50/60 Hz system generating
12 Vdc, but for a 24 Vdc and higher systems a higher ripple can
usually be tolerated, thus the capacitor can be reduced.

Hence for a 20 Amperes supply a total of 40KmicroFarad or more.

Hopefully the amplifier will not consume 20 A (480 W), since the
efficiency would be quite low :-).

Paul OH3LWR


--
remove ,xnd to reply (Spam precaution!)

If it helps....,
the amplifier is shown on http://home.online.no/~la8ak/d24.htm
but I've written text the text in Norwegian since material is
available for radio clubs in this area. But at least it is some
pictures and circuit diagram for the amplifier

73 LA8AK
-------
On Mon, 01 Dec 2003 12:07:57 +0200, Paul Keinanen
wrote:

On Sun, 30 Nov 2003 14:43:48 -0000, "Frank Dinger"
wrote:

planning a power supply for NEC 36284 150W 2m linear amplifier with
2SC3286, see http://home.online.no/~la8ak/images/nec36284c.gif

Is this really a 2m amplifier with a large number of striplines? I
guess that the amplifier would be quite bulky at such low frequencies.

Assuming 50 % efficiency, 150 W RF would require 300 W DC input power
or 12.5 A.

It is only a simple psu, but using a 24VAC transformer
[http://home.online.no/~la8ak/images/4ua3.gif)

The secondary peak voltage Vpeak will be about 1.4*24 V or about 34 V.

Since the rectifier voltage drop is at least 1 V at large currents and
there are two rectifiers in series, the voltage drop is at least 2 V,
so the peak capacitor voltage is less than 32 V.


Wonder what size capacitors I should use to keep ripple minimum above
+24V?


As the lowest allowed voltage is 24 V, the capacitor voltage will drop
from 32 to 24 V (dV=8 V) between the half cycle peaks.

The hard part is to calculate when the rectifiers stop conducting
after the peak and when they start to conduct again in the next half
cycle.

The worst case capacitance could be calculated assuming that the
capacitor must deliver the current for the whole half cycle, thus 10ms
for 50 Hz full wave rectification. Since dV = I*t/C and rearranging C
= I*t/dV, thus the required capacitance would be C = 12.5 A*0.01 s / 8
V or 15.6 mF.

However, since the final capacitor voltage Vl is well below the
secondary peak voltage, the rectifiers start to conduct well before
the peak of the next half cycle. The conduction angle is given by
arcsin(Vl/Vpeak) so in this case arcsin(24 V/34 V) or arcsin(26V/34V)
= 50 degrees, if the rectifier voltage drop is included. 50 degrees is
more than 2 ms before the next peak.

Since the conducting after the previous peak occurred slightly after
the peak (due to the serial impedances in the transformer and
rectifiers), a good guess for the time when the rectifiers are not
conducting is 7,5 ms in this case.

Using this shorter storage time, the required capacitance can be
recalculated as 11.7 mF or 11700 uF.

It should be noted that in this case the 8 V is the peak-to-peak
ripple voltage and 24 V minimum output voltage during a cycle, which
is quite appropriate if series pass regulation is used. Some design
equations especially from the tube era talk about RMS ripple voltage
and average DC voltage, which was more useful when a choke was used in
the high voltage supply.

=====
As a rule of thumb not less than 2000 microFarad per ampere

That gives a voltage drop rate of 0.5 V/ms.

With discharge times from about 6 ms (60 Hz full wave) to 20 ms (50 Hz
half wave) the voltage drop would be 3 .. 10 V.

If this is much or not, depends on the voltage levels used. A 3-4 V
drop might be appropriate for a full wave 50/60 Hz system generating
12 Vdc, but for a 24 Vdc and higher systems a higher ripple can
usually be tolerated, thus the capacitor can be reduced.

Hence for a 20 Amperes supply a total of 40KmicroFarad or more.

Hopefully the amplifier will not consume 20 A (480 W), since the
efficiency would be quite low :-).

Paul OH3LWR


--
remove ,xnd to reply (Spam precaution!)