1
\$\begingroup\$

I'm working with a BC637 transistor. I want to deliver a 48 V, 2.5 mA signal to my load (diagram below.) I have the load connected through the emitter pin (19.2k Ω resistor.)

I was under the impression the collector-emitter current was supposed to be controlled by the equation ß×Ib (base current.) While I was testing the circuit, I assumed ß would be about 100, and so I would get ~4.3 V / 200kΩ = ~ 2.15 mA. However, both in testing and simulating the circuit, I'm getting about 215 µA instead. I even took out the 200kΩ resistor and the current remained basically the same (230 µA.)

Since the base resistor had minimal effect on the output current, out of curiosity I calculated what the output current would be if I had just connected a 5V source to the 19.2k resistor. After accounting for the ~0.55V drop from the diode, the math roughly checks out: 4.447/19.2k = ~231 µA. I can't figure out why this would be happening though.

I'm a first year EE student with minimal previous electronics experience so I'm guessing there's something fundamental I've overlooked here. Any insight would be greatly appreciated.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ @asultan I may have missed it, but did you anywhere describe your load? \$\endgroup\$
    – jonk
    Commented Apr 11, 2022 at 20:42
  • \$\begingroup\$ It's an input pin on a golf cart speed controller with a built-in resistance of 19.2k Ω \$\endgroup\$
    – asultan
    Commented Apr 11, 2022 at 21:21

3 Answers 3

Reset to default
3
\$\begingroup\$

I was under the impression the collector-emitter current was supposed to be controlled by the equation \$\beta\ I_b\$ (base current.)

That is correct, but your problem is in thinking "controlled by" rather than "the relationship holds".

In this case, the emitter voltage is controlled by the base voltage -- it's one diode drop* below the base voltage. If you went and measured your base current in this circuit, you should find that it's very roughly 100 times lower than the emitter current ("very roughly" meaning somewhere between 50 and 250 -- read your data sheet for details, but in general transistor \$\beta\$ varies a lot over part variation, emitter current, collector voltage, temperature, etc., etc.)

* Usually taken as 0.6V, but it varies depending on all sorts of things.

\$\endgroup\$
3
\$\begingroup\$

You have an emitter follower circuit.

The voltage at the emitter will always be about 0.7 volts below the base voltage, regardless of the collector voltage.

\$\endgroup\$
1
  • \$\begingroup\$ Gotcha, didn't realize that. My "load" is a sort of current sink, so there's only one terminal into it. Is there a different circuit configuration which will let me send the 48 V 2.5 mA signal given that I can't put the load between the 48 V source and the collector pin? \$\endgroup\$
    – asultan
    Commented Apr 11, 2022 at 20:10
3
\$\begingroup\$

If you have to drive the load from the high side you can use the NPN transistor to drive a PNP output stage.

schematic

simulate this circuit – Schematic created using CircuitLab

Alternatively, given the relatively low output current to your load, an opto isolator would cut your component count considerably if whatever is driving the input has the capability to supply enough current.

\$\endgroup\$
4
  • \$\begingroup\$ Is it an issue that the base voltage of the Q1 goes to 48 V? That's what I'm seeing in my simulation. I thought 5 V was the maximum base-emitter voltage (from the 2N3906 datasheet) \$\endgroup\$
    – asultan
    Commented Apr 12, 2022 at 22:34
  • 2
    \$\begingroup\$ It's not an issue that the base goes to 48V as the emitter also goes there, and base-emitter voltage will never be too high in this circuit and 5V is the maximum for reverse biasing the junction anyway. But the issue is the maximum collector-emitter voltage which is 40V for the 2N3906. I've specified a more suitable part but you should really select a suitable one from what's available to you. \$\endgroup\$
    – Finbarr
    Commented Apr 12, 2022 at 23:54
  • 1
    \$\begingroup\$ For the benefit of the OP being new to this game: Q1 it's still the emitter to base voltage that's important which is still small, 48 minus a tiddly bit to 48. To work out how this works, imagine Q2 doesn't exist and is just a switch (R3 and R4 also gone). Then if the switch is open the base only sees R1 and gets pulled up to 48V so is off. Turn the switch on and it sees the whole divider, R1+R2 and it sees 46V (or whatever) from the divider (R1<R2), which is a volt or two between BE, all quite small voltage differences even though way above ground. Your load R5 is on the collector. \$\endgroup\$
    – Dannie
    Commented Apr 13, 2022 at 0:50
  • \$\begingroup\$ Of course! That makes sense - I forgot it was the emitter of the PNP connected to the 48 V source rather than the collector. Thanks! \$\endgroup\$
    – asultan
    Commented Apr 13, 2022 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.