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I cannot seem to make the schematic below work.

As you will plainly see, the idea is to use the PNP as a switching device for a PhotoMOS (TLP354x). By switching SW1, Q1's base goes to the ground through R2 and I have 12,6V at the emitter of Q1.

All is well so far, however, whatever I do, with or without R3 or R2, trying with different values etc... I always have a lower voltage (12V instead of 12,6V) at the base of Q1 which leads to it partially conducting and having a 3V voltage at Q1's emitter constantly.

What's preventing it from completely pulling up?

Should I go with a MOSFET instead to resolve this problem?

schematic

simulate this circuit – Schematic created using CircuitLab

Thanks for the help,

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    \$\begingroup\$ You do have Q1 connected the right way around? Emitter to 12.6V \$\endgroup\$
    – RoyC
    Apr 12 at 10:38
  • \$\begingroup\$ I'm assuming there must be some good reason why you can't just wire the switch in series with the LED and its current limiting resistor. \$\endgroup\$ Apr 12 at 10:42
  • \$\begingroup\$ @RoyC, yes absolutely, error on the schematic, editing it to reflect that. \$\endgroup\$ Apr 12 at 10:51
  • \$\begingroup\$ @PeterJennings Yes, problem is that being a LED, switching the photoMOS from the low side always leaves a 12V (-Vf) on the input, not acceptable for various reasons for us. \$\endgroup\$ Apr 12 at 10:52
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    \$\begingroup\$ The schematic was correct it isn't now. \$\endgroup\$
    – RoyC
    Apr 12 at 10:55

2 Answers 2

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You probably have q1 connected the wrong way round try turning it round so the emitter is connected to 12.6V.

You can solve this without using a MOSFET but for most switching applications a MOSFET would be my preference. This is because you do not require any steady state gate current to switch it.

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  • \$\begingroup\$ That did solve it. \$\endgroup\$ Apr 12 at 11:05
  • \$\begingroup\$ As a side note, is there any interest in going to a MOSFET for this type of application? \$\endgroup\$ Apr 12 at 11:05
  • \$\begingroup\$ Answer edited to address this issue. \$\endgroup\$
    – RoyC
    Apr 12 at 11:08
  • \$\begingroup\$ Actually, in this particular instance, the circuit does want to put a reasonable-sized wetting current through the switch (often a couple of mA) so a MOSFET here would not have that advantage over a BJT. With a MOSFET, the 10K resistor would have to be dropped to, say, 4K7 and thence the 3K3 lowered to, say, 470R so it works but there's still something between gate capacitance and the switch (with any ESD). But I do take your point. \$\endgroup\$
    – TonyM
    Apr 12 at 11:13
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The switch doesn't pull the base all the way down, it make a voltage divider 10/3,3 resulting at approx. 9 V at the base.

To solve this, remove R2, and move R1 and D1 up before the emitter (to avoid short circuit through Q1 when activating the switch) connecting the collector directly to ground.

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  • \$\begingroup\$ not sure that is relevant in the case of a BJT (needing current more than voltage). \$\endgroup\$ Apr 12 at 11:16
  • \$\begingroup\$ It's a BJT. The base can't be pulled to below 12V anyway, when supply is 12.6V. \$\endgroup\$
    – Justme
    Apr 12 at 11:44
  • \$\begingroup\$ As I understand it, you are proposing to replace the common-emitter configuration with a common-collector one? \$\endgroup\$ Apr 12 at 13:30

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