On Dec 6, 2:39*pm, tricky > wrote:
> On 06/12/11 20:42, davygrvy wrote:
>
>
>
>
>
>
>
>
>
> > On Dec 6, 1:27 am, > *wrote:
> >> On 05/12/11 21:32, davygrvy wrote:
>
> >>> Why any heater wire at all? *What if I could just bolt a 15 amp
> >>> (TO-3P) transistor right to the manifold itself and cook the metal
> >>> directly? *Hopefully there's enough material to grind a flat. *Using a
> >>> PNP transistor, collector is at ground potential so no need for any
> >>> mica insulation.
>
> >>>https://www.facebook.com/photo.php?f...et=a.111195867...
>
> >> Remember, with the transistor fully on, there will be little voltage
> >> drop across it. *Therefore there will be little power dissipation. ....
> >> Little heat .
>
> > Not true when emitter to collector is right across the battery. *10A @
> > 12VDC = 120W
>
> >> you need to bias it half way between the rails to get the most power /
> >> heat, out of it.
>
> > not true at all without a load. *I had nearly the same convo with a
> > guy I work with. *Like you, he normally thinks a transistor drives a
> > load. *What if it is the load and is operated as just a variable
> > resistor?
>
> >> It's early, so I might have missed something there ! ...
>
> > paradygm shift
>
> >> Like I said on FB, a plain old power resistor would be much simpler
>
> > I'll post the heat chart from the simulator now so you can see how
> > it'll work.
>
> >> Rich
>
> I must admit, my interest is audio. Different classes of output have
> different heat characteristics.
> Usually for a fully on transistor, the power/heat is dissipated in the
> load. *If you have no load ... then I guess you have to over current the
> transistor to get the same max power, for the rating of the transistor ?
>
> However you figure it , P = VI

I think google dropped my other reply. lemme do it again.

When emitter of the heat making PNP is connected to battery plus and
collector to battery minus, Vce is constant (at least to the
capabilities of the battery which is good assuming start current is
about 80 amps). To get a current of 1A through a transistor (Ic) you
need 1/Hfe current through the base. For a 2SA1943, Hfe is about 50.
Therefore, I need to pull 100mA to get 5A. 5A*12VDC = 60 watts of
heat.

> If the transistor is shorted (fully on) then you only have the junction
> voltage drop to put in the equation .

Ouch! That full capabilities of the battery. That's a crazy amount
of current. You mean saturated? Doesn't get there. There's a clamp,
it can't happen. Vary R11 to set the clamped base current.

> Therefore you cant get the full
> power if you stick to the max current rating .

Replace Q5 with a higher current type, R11, and maybe use three diodes
in the clamp to get you the 300mA so you can pull 15A through Q4.

15A is 180W @ 12VDC. Do you or I need that much? Can my ignition
switch handle it if I use the coil feed? I dunno yet until I make one
and see what I like. I'd like to get in the car, place the key to the
on position and pause waiting for maybe a minute or so while the temp
comes up. Colder it is outside, longer I wait until I kick it over.
Would 60 watts get the temp of the manifold from freezing to 20C in a
minute? no clue until I try. 60 watts is just an initial guesstimate
for me until I know better.

> A simple resistor will give you a set amount of heat without any
> controll needed - bar a thermistor circuit to switch it off. -

Naww, not tunable. well it is, but I'd have to replace a $10 (or so)
power resistors compared to 2 cent resistors in a circuit board.