OK here's an update. First, thanks for the generous feedback.
[...]
The 170deg heatsink says you're taking about 70ma from the +5V.
How did you get that number (temperature -> current)?
24V AC would rectify to approx. 35VDC. 70ma on 56ohm would drop only
4V. You still have 31VDC to the 7805.
In the arrangement shown assuming DC current of 70 mA, the RMS current
upstream of the filter capacitor is a lot more.
Although I would decrease the value of the filter capacitor to 470 uF
(to nearly half the RMS input current to make things easier on the
transformer).
Let me go a bit further as best as I know...
70 mA over 1/120 of a second will discharge a 470 uFcapacitor by only
about 1.24 volt, and a 100 uF one by only about 5.83 volts.
Assuming no resistance nor other reactance in the way to stretch out the
input current pulses to the filter capacitor, *and* taking a first order
approximation assuming that the filter capacitor has output load with no
input for 1/120 of a second every half-cycle that is!
Continuing with same assumptions, a 470 uF capacitor sucks in current
from 34-peak VAC (24V RMS) about 15.5 degrees out of 180, if I worked
this out right! (ratio to 1 of cosine of 15.5 degrees is about
[34-1.244]/34!) This means that assuming within-conduction-cycle-portion
RMS-average ratio close to the 1.11 of a sinewave and RMS-average ratio
throughout the cycle being the within-conduction-cycle-fragment times the
square root of duty cycle of conduction at all...
34V peak 470 uF 120 Hz 70mADC means 262 mA RMS through the transformer
secondary winding.
Repeat the calculation for 100 uF and 5.8 volt "ripple" - .07 amp over
1/120 second discharges a 100 uF cap by 5.8 volts over 34 degrees of a 120
Hz cycle. Correcting as I would do pulling straws from a hat for now)
for time needed to recharge the filter cap from 180 degrees to 150 degrees
70 mA over 1/144 second (5/6 of 1/120) discharges a 100 uF cap from 34V by
4.86 volts - so I would expect a 34V peak rectifier output 120 Hz sinewave
to a 100 uF cap with a 70 mA load to have peak-to-peak ripple of close to
5 volts!
And with conduction 34 degrees out of 180, I would assume that the
conducted current waveform during the conduction time fragments is close
enough to half a sinewave with the half chopped at the peak from a
sinusoidal half-cycle. During that conduction fragment, my assumptions
predict RMS/average ratio of the usual 1.11. To account for conduction
duty cycle within a 1/120 second half-cycle (assuming 60 Hz AC), divide
this by the square root of the duty factor of the rectifier passing
current from the transformer secondary. With 30 degrees, RMS current
through a fullwave rectifier would then be about 2.7 timesthe DC load
current. (Just be thankful that rectifier heating is determined more by
the average current!) Transformer winding resistance will stretch out the
curent conduction portion of each AC half-cycle and reduce the RMS current
somewhat...
This RMS current value through transformer secondary winding will:
* Roughly, by first order approximation, vary directly with square
root of load current from load current of above-assumed 70 mA. Doubling
the load current from 70 to 140 mA will only increase the 189 mA RMS
transformer secondary winding RMS current (assuming 100 uF main filter
capacitor value) to 267 mA. I would say real-world guessing from a hat
more like increase from 150-155 to about 230 mA with doubling of load
curerent from 70 to 140 mA from 24V transformer. And that use of a 470 uF
capacitor rather than a 100 uF one will close to or at least almost double
(and assuming no resistance will worsen more rather than the
real-world-usual-less compared to multiply by square root of 4.7) these
RMS values, which I figure for now being indicating 70 mA from 470 uF to
mean transformer secondary current even real-world likely exceeding .3 amp
and theoretically maybe .38 amp or so.
* Roughly, by first order of approximation, vary nearly enough inversely
with the square root of value of the filter capacitor.
* Have peak value close to 1.4-1.45 times the actual RMS value, while I
have been predicting RMS current values easily enough to be seen as a bit
on the alarmist side.
This arrangement should run much cooler
john
1N4004
___
26Vac o---|___|---->|---o-----o 18V to 7805 in
|
56ohm 1 Watt |
|
[Transformer] |
---
--- 2200u
|
26Vac o-----------------o-----o 0V
(created by AACircuit v1.28 beta 10/06/04
www.tech-chat.de)
- Don Klipstein (
[email protected])
