# Slow start circuit for PSU.

Discussion in 'Electronic Design' started by Cyber Vagrant, Dec 3, 2003.

1. ### Cyber VagrantGuest

I'd appreciate it if you could check my work. This is a slow start
circuit for my Gain-Clone amplifier. It uses the coil of a 24V spst
relay in series with a voltage dropping resister and a zener diode
across the output rails to short out a current limiting resistor on
the primary of the transformer.

The PSU has a working voltage of +-34V, that 68V across the rails

The 24V relay has the following specs:

1440R, 16.7mA, 19.2V Pickup, 1.2V Dropout, 36V max.

19.2V / 1440R = 0.0133A (pickup current).

A 28V zener seems to be about right.

So, 68 - ( 24 + 28 ) = 16.
16V for the resistor to drop.

16 / 0.0167 = 958. Call it 960 Ohms.

So, pick up rail voltage will be:

( 960 * 0.0133 ) + 28 + ( 1440 * 0.0133) = 59.9

Call it 60V(+-30V). 88% of working voltage.

Now, for max dissipation:

73 - 28 = 45 and 45 / ( 1440 + 960 ) = 0.019A no-load

0.019 * 1440 = 27.34V max drop across the coil, good!

0.019^2 * 960 = 0.35W 1 Watt resistor should do, aye?

And 5W for the zener just to be safe.

Whatcha think, will it work?

I was wondering if 88% of working voltage was too high. However, I
tested the amp with maximium sine wave input and the rail stayed at
68V. So, I'm thinking 60V pick up on the relay will be OK.

Just for the sake of completeness, the dropout rail voltage should be:

2.4 / 1440 = 0.00167 (dropout current)

( 960 * 0.00167 ) + 28 + ( 1440 * 0.00167 ) = 32.008 Call it +- 16V.