New to the site, and indeed fairly to electronics. I'm having a bit of trouble with this circuit I've made. It is an led indicator type circuit.

transistor switch indicator

The problem I'm having is that there is not enough power getting to the external load. Am I right in thinking that this is due to resistance of the transistor (losing about about 0.7V?).

I thought I'd try a MOSFET instead because I'd heard that they were lower power, but it seems there will always be a problem since no current can flow from gate to source, and the pulldown resistor gives the same problem.

enter image description here

Am I missing a really simple trick or am I trying to get something for nothing?

  • \$\begingroup\$ You are missing the power rating and colour of the LED and you should aim to choose a supply just slightly greater <3 pref <1V than the LED forward voltage \$\endgroup\$ Oct 3, 2016 at 20:05

2 Answers 2


It looks like you want an LED to light up when a load is connected, without affecting the voltage to the 1.8K load. The BJT is better than most any MOSFET in this application and yes, the Vbe junction will drop 600mV or more with 5mA flowing. Most MOSFETs need more than 600mV to turn on so even if you put a low value resistor in series with the load you will need to drop more voltage.

It would be easier and at least as efficient to just connect the LED and series resistor across the load, but maybe that is not convenient for some reason.

If you want to get to an even low voltage than 600mV it's easier to use a comparator, however the comparator will draw some current from the 9V even if the load is not connected.

The below circuit will drop 50mV with the load connected. It's possible to go lower again by using a better comparator- this one is very cheap but it has a Vos of almost 10mV worst-case, so I've set -25mV and +25mV roughly as the differential input voltages. It can drive a few mA to the LED if you need more use a different comparator or add a transistor.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ The LM393 seems like the best choice. It's cheap, readily available, and a discrete circuit can't come closer to doing what's needed. \$\endgroup\$
    – jonk
    Oct 3, 2016 at 19:40
  • \$\begingroup\$ I do worry that the 9V was the default CircuitLab value and not the real one, though. \$\endgroup\$
    – jonk
    Oct 3, 2016 at 19:46
  • \$\begingroup\$ @jonk Default is 1V so the OP definitely typed it in. Doesn't mean it is correct, of course. I take the 100R as being a place-holder. \$\endgroup\$ Oct 3, 2016 at 20:13
  • \$\begingroup\$ I take your point. It is the battery down at the bottom which defaults to 9V! I'd forgotten that the defaults are different. But cripes, I am getting to hate the default values. \$\endgroup\$
    – jonk
    Oct 3, 2016 at 20:28

Let's just do some basic math: With a base-emitter voltage of approx. 0.7V and a stable source (see below) of 9V, your "external load" gets 8.3V. If it really is resistive and has 1.8k, the resulting current will be $$I_{R2}=\frac{8.3\text{V}}{1.8\text{k}\Omega}=4.6\text{mA}$$ This is all you can get with this circuit. But even with an ideal transistor with zero base-emitter voltage, the condition $$I_{R2}\leq\frac{9\text{V}}{1.8\text{k}\Omega}=5\text{mA}$$ will hold, so this is the maximum current you can get into your load. Otherwise you would need to boost your 9V to some higher level first.

The MOSFET circuit does no good, as there is no gate current and, hence, you will only get a very tiny current through R3, as it has to flow through R1=1M. So the current will be about $$I_{R3}\approx\frac{9\text{V}}{1\text{M}\Omega}=9\mu\text{A}$$

In any case, you should also take care for your LED. The LTL-307EE is rated at 20mA, while the fully turned-on BC546 and the 100R will give something like $$I_{D1}\approx\frac{9\text{V}-2\text{V}-0.2\text{V}}{100\Omega}=68\text{mA}$$

Are you trying to build an indicator for some industrial 4-20mA thingy? Those are usually operated at 24V. Whatever the load is, be sure not to exceed the transistor's maximum base current. This value seems to be missing in the datasheet, but I'd not go above 10mA continuous for such a small-signal BJT.

*stable source: In case you are driving the circuit using a 9V block battery, please check its voltage while in use. Those have a rather high impedance, so your 9V source might easily drop to 8V or something depending on the load.

Regards, Philipp

  • 2
    \$\begingroup\$ I think it is a huge mistake to imagine that any schematic on EE.SE is accurately stating an LED type, if the LED is called "LTL-307EE." That's because you ALWAYS get that type when you drag out an LED symbol. Just like you ALWAYS get a \$100\:\Omega\$ valued resistor when you drag out a resistor. It could be an amazing coincidence, of course. \$\endgroup\$
    – jonk
    Oct 3, 2016 at 19:12
  • \$\begingroup\$ @jonk: Good point, I'm not familiar with CircuitLab or its defaults. Just commented on the image ;) Anyway, 20mA max. is still a reasonable limit for an indicator LED (5mA is enough for most of nowadays indicators anyway). \$\endgroup\$ Oct 3, 2016 at 19:15
  • \$\begingroup\$ I'm not even sure the 9V is correct. That's ALSO the default value you get when you pull out a voltage source. I wish the OP would have made things explicit. Or else I wish that CircuitLab would use nonsense values for default so that there would be no question about it. \$\endgroup\$
    – jonk
    Oct 3, 2016 at 19:56
  • \$\begingroup\$ @jonk I completely agree, having default values for voltages and resistors are even counterproductive. \$\endgroup\$
    – pipe
    Oct 3, 2016 at 20:13
  • \$\begingroup\$ @pipe Spehro was correct in noting that the V source is 1V by default. It's the battery source at the bottom of the icons that defaults to 9V. But yeah. The default values lead to confusion, if the OP doesn't take the time to specify that they actually specified the values. hehe. \$\endgroup\$
    – jonk
    Oct 3, 2016 at 20:30

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