My question is: Does the transistor dissipate more power because of the freewheel diode? Ifso, why?
simulate this circuit – Schematic created using CircuitLab
Asked
Active
Viewed 520 times
4
-
How do you propose to use that circuit without a diode? It wouldn´t last too long. – F. Bloggs Jan 12 '17 at 17:10
-
It could be that when going from cut-off to saturation, a very little amount of additional current has to be spent to put the diode in inverse. However it is tiny and as Bloggs said, it is not an 'option', without the diode, the transistor will simply blow up. – Claudio Avi Chami Jan 12 '17 at 17:21
2 Answers
3
If we assume the diode has 2pF capacitance and 1nA leakage when reversed, then yes. The transistor will have a transient slightly larger current when switching on, due to charging the diode capacitance, this can be neglected as it only occurs once per operation. There is a continuing slightly larger current due to the diode leakage. You do the sums on the relative influence of the continuing extra 1nA, and decide whether it matters or not compared to the relay current.
The major difference comes when the transistor switches off. With the freewheel diode, the relay kickback (that is the continuing current driven by the energy stored in the relay coil inductance) is conducted through the diode, and the collector voltage rise limited to less than one volt above the rail. Without the diode, the kickback raises the collector voltage until the collector-base junction punches through, destroying the transistor.
Neil_UK
- 158,152
- 3
- 173
- 387
2
When Q1 is switched on, it passes the current taken by L1 and D1 has very little effect, drawing a relatively tiny leakage current.
When Q1 is switched off, the energy stored in L1 produces an inverse voltage that attempts to keep the inductor current flowing in the same direction. In your circuit, this collapsing field voltage is short-circuited by D1. Throughout all this, Q1 is switched off and dissipating no power.
TonyM
- 21,742
- 4
- 39
- 62