Sat Feb 22 10:51:58 CET 2014
BJT & MOSFET, switching high-voltage load from 3.3V uC
1. SWITCHING LOADS
Breaks down into two cases, symmetric from the pov. of the load:
- N) disconnect GND (load from V+)
- P) disconnect V+ (load to GND)
We want the transtor, either BJT or MOSFET, to be fully on/off. This
requires an N-type for case N) and a P-type for case P), preserving
symmetry of solution.
When control voltage is the same as the output voltage, the switched
behave as half of a CMOS inverter, with the other transistor replaced
by the load.
For BJT, simply add base resistor for current limiting, making sure it
is driven to saturation.
2. WRONG POLARITY
It's easy to set that the above works. So what goes wrong when we use
an N type instead of a P type for a load to GND?
An N type MOSFET won't be fully switched on. Since V_DS > V_GS - V_t,
it is in saturation (current source) mode.
For an N type BJT with a load to GND, there will always be at least
0.6V over the BE junction in the case the transistor is on, and if
it's off it is operated in reverse cutoff (correct?)
3. LOW DRIVE VOLTAGE
N/P symmetry gets broken when the control voltage is much smaller than
the output voltage. Say V+ is 24V, typical for industrial control
This is where interrupting the ground lead becomes easer than
interrupting the V+ lead. In the latter case, an extra inverter is
necessary to generate the control voltage near V+.
4. BJT vs MOSFET
* +BJT is easy to switch on (V_BE = 0.6V)
* -MOSFET has unpredictable V_t
* -BTJ introduces extra current in C->E path
* -BTJ has I_B
5. GATE SOURCE BREAKDOWN
The BS250P P-channel MOSFET has a G-S breakdown voltage of +- 20V.
This means that in a 24V circuit load to GND switching situation, care
needs to be taken to not pull the gate to GND, as this creates a -24V
G-S voltage. To solve this, a 100k/100k divider network can be used
at the gate input, switched to ground by a NPN.