Re: [vox-tech] [no linux] High voltage circuits?
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Re: [vox-tech] [no linux] High voltage circuits?
On Tue, 20 Mar 2001, Peter Jay Salzman wrote:
> jeff,
>
> that makes total sense -- an increase in kinetic motion causes an increase in
> resistivity of the filament.
>
> i think adding a small inductance would also work, and has the benefit of not
> drawing power from your source. the "back EMF" would kick in only when the
> current initially flows (and stops).
Two different time scales are at work here... the thermal filament warm-up
period, and the 60Hz ac time period. Even if the thermal time constant is
much shorter than one 60Hz cycle (which I doubt), an inductor probably
won't be a practical solution. Choosing a larger relay sounds more
appropriate.
>
> pete
>
> On Tue 20 Mar 01, 8:53 AM, Jeff DeFay said:
> > Mark,
> > If you are using incandescent light bulbs, your current will be much higher
> > during the
> > brief time that the filaments are cold. The 60 watt power consumption is
> > calculated
> > with the filaments at normal operating temperature; introducing resistance
> > into the circuit
> > would also keep the effective resistance of the filament low by preventing it
> > from reaching
> > normal temperatures.
> >
> > Regards,
> > Jeff DeFay
> >
> > "Mark K. Kim" wrote:
> >
> > > Anybody here ever work with high voltage circuits? I'm creating a
> > > low-volage / high-voltage hybrid circuit for the first time; a digital
> > > circuit (low voltage, 5DCV) controls the logic portion, and uses two
> > > relays to control two light-bulbs (high voltage, 110ACV). And I'm
> > > wondering:
> > >
> > > 1. How much current usually goes through those 60W light bulbs?
> > > How much current is required to make it look bright enough?
> > >
> > > My guess would be:
> > >
> > > 110ACV * I = 60W
> > > => I = 0.55A (max)
Standard voltage is 120+/-10%, so call it 0.5A.
Also, R = 120/0.5 = 240 ohms.
> > >
> > > And I'm using a 1A relay rated at 120ACV, so my guess would be
> > > that I can hook up this relay to the light bulbs directly.
> > > But when I did that, once the light bulbs were turned on,
> > > they stayed turned on (I couldn't turn off the relay connection
> > > -- apparently the connections got fused together.)
What a great practical experiment.
> > >
> > > 2. So my idea was to force-limit the current. I got new relays
> > > (same kind) and bought 1W resistors (200 Ohms... actually, two 100
> > > Ohm resistors soldered together.) Once I got them in circuit,
> > > it had two problems:
> > >
> > > a. The light bulbs were too dark, and too slow to lighten up.
> > > Apparently too much resistance?
The bulb expects 0.5 Amps, and you are only giving it 120/440=0.27A
> > >
> > > b. I started seeing smokes come out around the resistors.
> > > I don't know if this is because the resistors need to be
> > > higher wattage, or if the current is melting the solder.
> > > I'm pretty sure the electrical wires are good enough for
> > > this project (it's rated at 700V), although if the solder
> > > is melting (meaning it's at least 400 degrees Ferenheit)
> > > then it could be also melting the electrical tape.
> > > Ideas?
Voltage rating is for insulation breakdown. Current rating is the
critical issue here, since small wires have higher resistance.and
dissipate more power.
> > > I need to know if I need to switch to higher rated relays, and if so, to
> > > what.
I haven't done it myself, so I can't say what the actual rating should
be. You are a student... see if you can get a storage oscilloscope and
look at the current through the bulb as you light it. :)
Might also consider a 40W bulb? or a flourescent bulb?
> > > Also I need to know if I need to use a different resistor and/or do
> > > not use solder. I'd really appreciate any suggestions from anyone
> > > knowledgeable in this field. Thanks!
I don't think the resistor is the right solution, since the bulb is
designed for 120V, and the resistor solution creates a voltage divider.
Compression connections are best for power transmission.
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