I'm using a separate low-voltage source to apply a small current to a transistor base which will then turn on a higher-voltage current to power an LED. I guess this is more like two separate circuits instead of a parallel circuit powered by one battery. I wasn't sure how to connect the negative terminal of the low-voltage source, so I tried three different ways, illustrated below. All three worked, lighting the LED. Does it matter which I use and if so why?

Setup 1 - negative terminal to higher-voltage battery's ground:


simulate this circuit – Schematic created using CircuitLab

Setup 2 - negative terminal to transistor emitter:


simulate this circuit

Setup 3 - negative terminal to transistor collector:


simulate this circuit

  • \$\begingroup\$ Setup 1 and setup 2 are identical. You can find a lot of these circuits by googling as well. \$\endgroup\$
    – Andy aka
    Feb 4, 2022 at 16:29
  • \$\begingroup\$ Setup 1 and setup 2 are electrically identical, just drawn differently. With the third setup, you will have no control of the transistor - it will always be passing as much current as it can. \$\endgroup\$ Feb 4, 2022 at 17:01
  • \$\begingroup\$ consider what happens if your battery doesn't output exctly 2.5 V? \$\endgroup\$
    – Neil_UK
    Feb 4, 2022 at 17:30
  • \$\begingroup\$ Peter Bennett: where you say "you will have no control of the transistor - it will always be passing as much current as it can", can you elaborate on that? How does Setup 1 control the transistor but Setup 3 doesn't? Thanks! \$\endgroup\$ Feb 5, 2022 at 13:38

1 Answer 1


It is normal to use ideal symbols in logic diagrams yet the real circuits have equiv. series R or ESR in the bulk resistance of every junction, caps, batteries, and diodes (even current meters). Most of the time, these (Level 0 logic schematics) with RLC parasitics are inserted mentally by those with experience to see the effects.

Let me show you a level 1 example. Imagine each level going to higher complexity with Miller Capacitance and trace inductance with complex impedance and variable Ic vs Vbe curves. The ESR values are just ballpark estimates on my side, but when a current flows thru a diode, you expect the exponential behaviour for voltage, it will be also increased by the series resistance and voltage drop V/R=I internal to each part. On batteries it is relatively easy to measure this and on transistors it can be looked up in datasheets as Vce(sat)/Ic {= Rce which is not shown}.. There are also thermal effects, so all specs in tables state the standard test temp of 25'C which is easy to control above room temp of say 22'C.

enter image description here

Now my schematic was not simulated so the values are wrong, but used your results. So with Vbat- Vce - Vf(LED) = error, do some KVL to verify the results. What if the 2.5V battery was ESR=10 ohms ? what would you expect?

Notice the ratio Ic/Ib = 22/1.2 is about 20:1 which is a reasonable value for a saturated switch, but which part would you compute is really limiting the current? What is a closer value for ESR1?

  • \$\begingroup\$ Thanks for the great explanation. Is it important whether the "tiny battery" negative is connected at the Collector vs at the Emitter? Both turn on the LED, and they result in slightly different voltage and currents (due to the various ESR impacts as you explain). But is it an okay design to connect the "tiny battery" negative to the Collector or the Emitter, or is one preferable? \$\endgroup\$ Feb 5, 2022 at 13:37
  • \$\begingroup\$ Polarity of Vbe matters which controls NPN Ic max. Here Vbe ~ 0.6 to 0.7 Normally you don't rely on an extra battery to drive the base, but just a current from a pullup base resistor to V+ in order to control base current about 5 to 10% of Ic= LED current. so Ib=1mA for Ic=20mA ratio is OK.. tinyurl.com/ybkvb5ox < click switch \$\endgroup\$ Feb 6, 2022 at 2:10
  • \$\begingroup\$ Then I can improve model with battery ESR=? change with mouse thumbwheel or edit. and see the results tinyurl.com/ycldux6k \$\endgroup\$ Feb 6, 2022 at 2:14
  • \$\begingroup\$ then with hFE=20 tinyurl.com/y76hub9m \$\endgroup\$ Feb 6, 2022 at 2:20
  • 1
    \$\begingroup\$ Thanks a million for the explanations and especially the schematics. Very helpful! \$\endgroup\$ Feb 6, 2022 at 21:04

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