I have this block inside my circuit, which is supposed to drawn a current from the main line to the GND whenever there is this signal in the transistor base (simple diagram):

Current source

The main line can reach 50V, so I chose a BC546B which has a Vcbo of 80V and Vceo of 60V.

After that I discovered that the main line also could reach peaks of 150V and -50V sometimes. This would damage my BC546. I don't want to put a reverse diode in paralell with the transistor because I really need to block current in both ways. So I thought that maybe putting a series diode with the transistor may help me. So I did this simulation:


It seems to work. Note that when Vce>0 all the main line voltage is accross the transistor. When Vce is < 0 the main line voltage is accross the diode. Since a simple 4148 can hold up to 100V, I could choose a new transistor (MPSA42 for instance) that can hold up to 300V.

Since the diode voltage drop is not an issue because i'm dealing with current direclty, this could do the trick. Although the problem seems to be solved, how can I explain this behavior (that shows the scope)? How can I know where is the voltage drop when more than one PN junction is put in series?

**Corrected picture:

enter image description here

  • \$\begingroup\$ In the second schematic, the base of the transistor seems to be connected to 0V through a resistor. Is this intentional or a mistake? \$\endgroup\$
    – deadude
    Nov 9 '13 at 21:41
  • \$\begingroup\$ Yes, its just to make sure it is not conducting. This way I could reach maximum voltage across CE. \$\endgroup\$ Nov 10 '13 at 0:25

If you want to save the transistor put the diode in series with the collector and use a transistor capable of withstanding greater than 150V. With it drawn as you have it, the base-collector region will still see large negative voltages AND, to control the current in normal operation you need to keep R6 in the emitter without a diode.

  • \$\begingroup\$ My mistake. I've made it in a hurry. Check the corrected picture (i've edit the original post) if it's like you said. \$\endgroup\$ Nov 10 '13 at 0:28
  • \$\begingroup\$ R6 is missing but basically this should work. \$\endgroup\$
    – Andy aka
    Nov 10 '13 at 8:08
  • \$\begingroup\$ But my question still remains: why it works? I have 2 devices which should block signal (diode + transistor). How can I know which one will hold the voltage difference? \$\endgroup\$ Nov 10 '13 at 16:15
  • \$\begingroup\$ the reverse current of the diode is nano amps - the transistor will behave as a transistor still but hfe will be much lower and very little voltage will develop across the transistor. \$\endgroup\$
    – Andy aka
    Nov 10 '13 at 16:21
  • \$\begingroup\$ But note that what prevents a reverse current to exists (ideally) is a pn junction inside the transistor, just like in the diodes. But it I put 2 series diodes and a reverse biasing power supply like -50V, my simulation says that each of them will hold -25v. Altough if I change one of them to a transistor, almost all this reverse voltage will be accross the diode. How can I explain that? \$\endgroup\$ Nov 12 '13 at 19:58

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