I am making a little home made electronics project.

The plan is to turn on a Red LED when a SCART input is sending an RGB signal. Pin 16 on a SCART cable is 1-3V when the input is carrying an RGB signal (according to wikipedia - in my case it actually sends 4.5V so I figure this is quite a flexible spec).

The circuit I have designed works....but I don't know if it is safe, or done correctly! Please see the schematic below:


simulate this circuit – Schematic created using CircuitLab

I have 2 main questions:

  1. I don't seem to need the 1k Ohm resistor leading to the base of the transistor. I assume any voltage applied to the 2N3904 base will switch the circuit on...so this is just keeping a low current to prolong the life of the transistor?
  2. The circuit doesn't work unless I cannot Pin 21 of the SCART input up to the ground of the main circuit. I don't know if this is safe, or the correct way to make this circuit work, can you validate my decision here.
  • 2
    \$\begingroup\$ You can rotate parts with right mouse and drag wires and parts to make it look nicer \$\endgroup\$ Oct 5, 2016 at 0:31

1 Answer 1


1) The 1k ohm resistor IS necessary. The base - emitter junction of a transistor is a diode, and the voltage drop of a silicon diode is ~0.7 V. The transistor will allow as much current to pass as it need to to clamp that voltage to less than 0.7 V. This will interfere with the signal, as the RGB signal is indicated to the receiving device as a signal greater than 1V (1-3 V according to wikipedia: https://en.wikipedia.org/wiki/SCART). In order to keep interference to a minimum, I would probably use an analog buffer (https://en.wikipedia.org/wiki/Buffer_amplifier#Voltage_buffer_examples) instead of a transistor to turn on the LED, but if you find it doesn't interfere, there isn't much reason to change it.

2) If you find that pin 21 works as a ground for you, it shouldn't cause any damage. Looking at the pinout on that same wikipedia page, I would go with pin 18 instead. Pin 18 is specifically a ground for the signal on pin 16. Either way, a common ground is necessary for the signal to be able to turn on the transistor.

  • \$\begingroup\$ Hi, thanks for your answer. For point 1) I am not sure I fully understand your reasoning for why the 1k Ohm resistor is definitely required. Can you explain why it works without it, and why it is better to have it? For point 2) So you are saying use pin 18, makes sense, but it is definitely safe to connect a ground from another device, to the main ground on my main circuit? \$\endgroup\$ Oct 5, 2016 at 0:03
  • \$\begingroup\$ I am assuming the current flowing across the base is therefore 38mA if you consider the voltage drop in this case? So are you saying without that resistor, the current would be far higher, causing some form of damage? \$\endgroup\$ Oct 5, 2016 at 0:05
  • \$\begingroup\$ With no base resistor, the transitor will draw as much current as needed to keep the base-emitter voltage about 0.7 volts, sacrificing its own life if necessary. The base current doesn't need to be more than about 1/10 of the collector current (if that). \$\endgroup\$ Oct 5, 2016 at 0:24
  • \$\begingroup\$ Peter - thanks for weighing in here, but this still is not making complete sense in my head. I assume it is good practice to just keep the current going into the base as low as possible -- but I struggle to quite understand why this 0.7v is being involved, or adds this level of complexity. I think the current at the collector would be 16mA, and the current at the base would be 4mA with the 1k Ohm resistor, but if I took away the 1k Ohm resistor, what would that current be going to Base? How can I work that out as I don't know the resistance.... \$\endgroup\$ Oct 5, 2016 at 0:30
  • \$\begingroup\$ The 0.7 volts is just the nature of a silicon diode. The concern over the base current is because of power dissipation. Like Peter said, if the base draws too much current, the transistor will burn out. The circuit works without the resistor (probably) because the pin cant supply enough current for the transistor to burn out. This still interferes with the signal, which needs to get above 1 volt to be of use to the receiving device. Regarding 2) on your first comment, the grounds are probably internally connected anyway. I can't be sure, I am not familiar with SCART. \$\endgroup\$ Oct 5, 2016 at 0:37

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