Bill wrote...
> I'm in the process of building the power supply for my amp project. I am
> having problems getting the correct delay in the step start circuit. I
> decided to use the same circuit as the amp project by G. Daughters, K6GT,
> that appears in the most recent ARRL handbook.
Oops... that puts me on stage, I fear.
> I am using a 24 V DC relay that has a coil resistance of approx. 280 ohms
> (Rcoil) and I want to get about a 3-4 second delay in the RC circuit. When
> I got all the parts together I knew the circuit wouldn't work just looking
> at the values, so I mocked it up and it indeed didn't work.
I used a physically small relay, and I think the resistance was much
higher... but I'm not at home now, so I'll have to wait to confirm
this.
> The circuit consists of a diode (D5) and 3.9k resistor (R13) in series
> connected to the coil of the 24 VDC relay with a 600 uf capacitor (C11)
> connected across the coil of the relay. 120 volts AC is connected to this
> circuit. This circuit is basically a half-wave rectifier and R13 and the
> Rcoil act as both a voltage divider and resistance for the RC time
> constant. I am only getting about 4 volts across the coil where this relay
> needs a minimum of about 14-15 volts to kick in. This makes sense looking
> at the circuit equations:
>
> Vcoil = Vin * Rcoil / ( R13 + Rcoil ) ~= 55 * 280 / (3900+280) ~= 4.0 V
>
> where Vin was measured and is about 0.45 * 120 volts which makes sense for
> a half-wave rectifier circuit.
This isn't quite right. A lightly-loaded 1/2 wave rectifier will
tend to charge all the way up to the peak voltage of the
half-cycle... more like 1.4 * 120 = 160 volts. Heavily loaded, it'll
never get to that level, though.
> Am I doing something really stupid and missing something?
Probably not. You certainly understand the circuit. Note that your
relay is a big current user:
I = 24v/280ohms = about 86ma.
Mine can't be like that... 86ma X 3900 ohms would require 330 volts!
I'll check my P.S. and report.
73,
George T. Daughters, K6GT
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