How can by-pass a power circuit?

Hello,

I have a power circuit that can regulate an input above 18VDC all the
way till 100VDC, the out being 12V at 2A. However the actual supply
line is in the range of 8V to 100VDC. So i would like to by-pass the
above "regulator" circuit when the input is below 18V. Can anyone help
me by suggesting a way to achieve this?

I was thinking about leaving my "regulator" ON all the time. And then
bypass the "regulator" and pass the input directly to the next stage
when there is no output from the above "regulator" (i.e., when supply
voltage is below 18V)

If anyone has implemented something similar or has any suggestions
that you could let me know, I would really appreciate it.

Thanks in advance,
SLK.

 

Hello,

a simple (but not the best) resolution will be in using a two-way
contact relay. Have a look at the coil data (often you get them from
catalogues in short form). There the minimum operating voltage should
be mentioned. If that hint does not help you, just try a 15V or 18V
relay.

The circuit of the relay's contacts are to be connected to the input
voltage, the output of your regulator, and the common pin leads to the
output of the "system".
To connect the coil, there has to be payed attention to the high input
voltage.
Normally a 18V relay is not able to withstand 100V. Therefore a
protection circuit (consisting of a resistor and a Zener diode) has to
be put in front of the coil.
The current through the coil to switch the relay is the parameter to
dimension the series resistor: at an input voltage Uin = 18V the
resistor has to be made such high that the relay just switches. W/o
data of the relay it's a matter of trial and error, otherwise
Rs = Rcoil * (18V - Ucoilmin) / Ucoilmin
where Ucoilmin is the lowest switching voltage of the relay.
The power dissipated by the resistor may not be omitted! It may be up
to several Watts (P = Umax * I). Umax = Uinmax - Ucoil
So use a relay with high restistance of the coil and low minimum
operating voltage!

A Zener diode is used to protect the coil against high
voltage/current.
It is paralleled to the relay coil.
Use a type with breakdown voltage Uz close to (i.e. lower as) the max.
coil voltage of the relay. Calculation of the dissipated power:
P = Uz * ((100V - Uz)/Rs) - (Uz/Rcoil))
As you probably will get a rather high dissipated power and high power
Zener diodes are rare and expensive, you may choose a low power diode
to drive a power bipolar Transistor mounted to a heat sink. E. g.
2N3055 for the Transistor, on heat sink approx. 5 K/W. The diode may
be a 1.3W type.
-------------------------------------------------
| | |
| ____ | |
+Uin -----|____|------------------------ | |
Rs | | | | |
| | | _____ |
| | | | Reg | |
____ | | |_____| |
| /\ ZD | | | |
/__\ | | | |
| | | | |
| | | |_ _|
| |C ___Rel /
| B / | / |======/
--------| T |_/___| /
\, | |
|E | |
| | ----------out
| |
| |
| |
| |
Gnd ------------------------------------