On Sun, 9 Dec 2007 15:31:13 -0800 (PST),
wrote:

thanks. Got that bit about the grounds connected to the source.

Was reading art of electronics last night (rocket science to me!) and
then I saw something that you said as well:
"Always use a supression diode when switching an inductive load"

They had the following diagram:
+Vcc --+---COIL----+----C E----- GND
| | B
+---|------+ |

Ok. So I unleash the 72V through my favourite coil. When I stop the
voltage.... The magnetic field subsides. But according to you and the
book it generates somesort of spike.

---
Yes.

the magnetic field doesn't just subside, it collapses very quickly
and when it does it cuts the turns of the coil and induces a voltage
across the coil opposite in polarity to the voltage impressed upon
it by the battery. That is, with the transistor turned ON the
voltage on the collector side of the coil will be close to ground,
but when the transistor turns OFF The voltage on the collector will
rise as described by:

dI
E = L ----
dt

where E is the voltage appearing across the coil, in volts

L is the inductance of the coil, in henrys and

dI
---- is the time rate of change of current in the coil.
dT

For example, I don't remeber all the details, but let's say your
solenoid has an inductance of 0.1 henry, there's 3 amperes of
current in it when you open the switch, and it takes 1 millisecond
for the current to decay to zero amps from 3 amps.

Then:

dI 0.1H * 3A
E = L ---- = ----------- = 300 volts
dt 0.001S

Which will more than likely punch a hole through the transistor if
the diode isn't there.

BTW, just for grins, take a look at what happens if dT is 100µS.
---

So the way I figure it, is that the current must be generated in the
opposite way. So previosly the C of coil in my diagram was the + and
the L was -. Now I think the L will be + and the C -. So can I
'replace' the coil with a battery for the millisecond or two that the
field dissolves?

---
Yes, but you've got it backwards. You have to put the + of the
battery on the collector and the - on the supply
---

Anyway... that's the only way I can explain the use of the diode. When
the 72V is running through the coil it has no use. When it is switched
off, it will allow my new 'battery' with a + on the L side to short
circuit itself and that I guess takes care of the current generated...
Is this correct?

---
Kinda, but the + will be connected to the collector. What happens
is that when the coil voltage tries to rise past the supply plus the
drop across the diode, the diode clamps the transistor collector to
the supply plus one diode drop while the coil discharges through the
diode. So that'll be supply plus about 0.7V, which shouldn't bother
the transistor at all.
---

So if all this guessing of mine is correct then my question is... So
what? What will happen if the C of the BJT is more - than the E and B?

---
It won't go negative, it'll go hugely positive and let all of the
magic smoke out of the transistor.


--
JF