Wednesday, 14 November 2018

A better time delay startup circuit

The amplifiers I've built so far have all incorporated a delayed turn-on circuit for the high voltage supply. The intention is to allow the 6 volt supply to turn on first and allow the valves to reach operating temperature before turning on the high voltage supply.

This is accomplished with a simple circuit based around a 555 timer IC in monostable mode, set up to a delay of around 25 seconds.

The circuit I've been using, while functioning, had a few problems. Driving a relay directly from the output of a 555 IC resulted in a lot of voltage drop through the IC and the relay coil voltage being low, for a 5V relay it was getting around 3.5 volts, fortunately this is still enough to trigger it, but less than ideal.

My re-design of the circuit was prompted by my addition of a 2-colour LED to the design, to glow red at initial turn-on but change to green when the timer activates and the HT voltage is turned on.

These LEDs are 2-pin, they work by reversing the polarity into them. So they're 2 LEDs in one envelope, and depending on the polarity of the applied voltage, one will be forward biased and glowing, the other reverse biased and dark.

After breadboarding it and measuring carefully, this is the circuit I designed:

Click to enlarge if necessary.

Note in this diagram my symbol library for the MOSFET is wrong... if you're gonna use this same MOSFET be very aware its pinout (viewed from top) is S-G-D instead of G-D-S. So my pin numbers are wrong. Sorry about that.

The 7805 voltage regulator is not strictly necessary but it does result in a nice 4.9V across the relay coil.

The 330K and 68µF cap provide the time constant for the timer IC. The formula in this mode is:

T = 1.1 x R x C

The MOSFET Q1 buffers the output of the IC switching the negative on or off to the relay based on the voltage at the gate, which comes from the output of the IC at pin 3. This starts low until 25sec elapses then goes high and stays high until power down.

The two 330R resistors form a voltage divider, at the mid-point the voltage is 2.5V. When the relay is off, the + voltage will flow through the coil (which is around 62 Ohms) and then into the LED, then to ground through the lower 330R resistor. This results in a voltage drop of around 0.2 volts across the relay coil, not enough to turn it on.

When the IC turns on, the voltage appears at the gate of the MOSFET, switching the transistor on. This effectively shorts the Drain and Source, causing the negative to connect to the relay and the LED. At which point the return path for the LED is through the top 330R resistor, so this reverses the polarity across the LED causing it to change colour.

The reverse-biased diode across the relay is for flyback suppression. 

After breadboarding, I've designed a single-layer PCB layout for this circuit which is 35mm x 35mm utilising a W02 rectifier.

On my board design I've also added a header for a regulated 5V power supply, in case it's needed elsewhere (such as a tone control bypass relay for example)

The current and dissipation is such that no heatsinks are necessary on either the voltage regulator or MOSFET.

Be sure to put the relay on the AC side of the rectifier diodes, relays have a much easier time switching AC than DC and this is reflected in the voltage rating on the datasheet.