I'm an electronics newbie and I am puzzled by the behaviour of a circuit I have created. I have connected the output of one inverter to the input of another but the behaviour of the circuit is not what I expected. I've included a schematic and two photos of the circuit. The first photo shows the two inverters unconnected and the circuit behaves as I expect it to, that is, LED1 turns off when I press the tactile button. However, in photo 2 you can see that I have connected the output of the first inverter to the input of the second and LED1 is off, but it should be on. Can you explain what is happening?

Inverter Circuit Schematic Inverters Unconnected Inverters Connected

  • \$\begingroup\$ The base of Q1 is always low. Your SW1 does nothing! (Unless something is broken in your ground connections.) So, your schematic isn't showing your actual circuit. \$\endgroup\$ Feb 15, 2020 at 12:01
  • \$\begingroup\$ The base of Q1 goes high when I press SW1 and the LED goes out. I am a little new to drawing schematics. \$\endgroup\$ Feb 15, 2020 at 12:13
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    \$\begingroup\$ Okay so I have got the schematic wrong, however the actual unconnected inverter circuit works. With SW1 unpressed LED1 stays on, when it is pressed LED1 turns off, which is correct. When I connect the two inverters together LED1 goes off and stays off. If I press the button LED2 turns on. \$\endgroup\$ Feb 15, 2020 at 12:49
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    \$\begingroup\$ So, without us knowing exactly what your circuit does, we can't help you – please investigate what your schematic actually looks like, and whether all connections on your breadboard actually are reliable. (A lot of time has been sunk into solderless breadboard connections simply not being good connections.) \$\endgroup\$ Feb 15, 2020 at 12:55
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    \$\begingroup\$ yes, but the circuit you show in your schematic has little to do with what you've built and little to do with what it's designed for. So fix your schematic, it's not correctly representing what you've built. Helping you without that is impossible. Full stop. \$\endgroup\$ Feb 15, 2020 at 13:21

3 Answers 3


The problem here is that the anode (top) of LED1 is connected directly to the base of Q2. When Q1 is off current will flow from R3 through Q2 base to ground. As Q2 is a silicon transistor the voltage on its base will be clamped to about 0.7 V. A red LED needs about 1.8 V to light up so LED1 won't see enough voltage to light up. (Q2 will switch on and off correctly so LED2 will behave as expected.)

To correct this, link LED1 to Q2 with a resistor (e.g. 1 kΩ, depending on the value of Vcc) rather than a wire link. This will allow the voltages on LED1 and the base of Q2 to be different.

Edit: I've removed text about an earlier version of your schematic which no longer applies to the updated version.

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    \$\begingroup\$ I connected a 220 Ohm resistor from LED1 to Q2 and the circuit works fine. Thanks for your help. \$\endgroup\$ Feb 15, 2020 at 13:39
  • \$\begingroup\$ @MarioGianota: you could look at this problem as the 2nd inverter having near-zero input impedance, so it's impossible for the 220 ohm output impedance of the first inverter to drive it to the desired voltage range. \$\endgroup\$ Feb 16, 2020 at 0:41

The base of Q2 in your schematic will shunt the LED and prevent it from ever turning on. However the second inverter will function coupled to the first. So the output LED2 should function as expected.

I didn’t look at your breadboard implementation.


Schematic seems to be showing only the circuit of second bread board photo.

As per second bread board picture, ... LED1 is getting clamped by the VBE drop of Q2. So, LED1 would not glow when its anode is connected to the base of Q2 regardless of the switch position.

As per first bread board picture,... The switch is off, The green wire connecting LED1 anode and Q2 base is removed. So, LED1 is no longer clamped by the VBE of Q2. That's why LED1 is glowing. The LED2 is also glowing because Q2 base is left open thus Q2 is OFF.


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