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When I put 100 kΩ at the base the LED is off, but when I put 470 Ω the LED is on and it's even brighter than when I remove the resistor at the base.

Why is this happening?

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Here's the actual circuit. The transistor is a BC548.

enter image description here

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    \$\begingroup\$ Your circuit is not correct. I would guess that with the 470 Ohm resistor it is passing the current to the led. Why is the led in parallel with the transistor? \$\endgroup\$
    – Kartman
    Oct 27, 2022 at 4:57
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    \$\begingroup\$ If the circuit is really as drawn, with a 100K resistor in series with the LED and transistor, I'd be surprised if the LED shows any detectable light. \$\endgroup\$ Oct 27, 2022 at 5:01
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    \$\begingroup\$ Which transistor is it? And are you absolutely sure the resistance values and diagram are correct? Can you measure voltage over battery, resiators and LED (so all transistor pin voltages and currents are known) and report the values in the diagram? \$\endgroup\$
    – Justme
    Oct 27, 2022 at 5:24
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    \$\begingroup\$ @RomerBagasis Justme is directly on-target here. There is no way your diagram is correct if the behavior you describe is accurate. Either your description is wrong or else the diagram is wrong. Which is it? \$\endgroup\$
    – jonk
    Oct 27, 2022 at 5:25
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    \$\begingroup\$ @jonk Thank you and I appreciate the discussions, best wishes to you too. I see my idea about Vbc breakdown was not correct as the breadboard circuit does not match the drawing. The breadboard has a resistor between emitter and ground. The emitter may still have bad connection. \$\endgroup\$
    – Justme
    Oct 27, 2022 at 5:58

2 Answers 2

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You actually have implemented this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When R2 is 100K the current from R3 is shunted away and the voltage across D1 is lower than the Vf of the LED.

When R2 is 470Ω then the base-collector junction becomes forward biased and almost 10mA will flow through the LED.


When R2 is 470 ohms there is no transistor action and the behavior is similar to the below schematic (D2/D3 represent the transistor junctions). This is not a valid simulation for when R2 is 100K because there is transistor action in that case.

schematic

simulate this circuit

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  • \$\begingroup\$ Is this diagram different from the one I've drawn? What does shunted away mean? \$\endgroup\$ Oct 27, 2022 at 6:16
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    \$\begingroup\$ @RomerBagasis Indeed different. Please look closely. \$\endgroup\$
    – winny
    Oct 27, 2022 at 6:21
  • \$\begingroup\$ What does shunted away mean @winny \$\endgroup\$ Oct 27, 2022 at 6:22
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    \$\begingroup\$ It's slightly forward biased but the voltage across R1 is < 0.1V and the voltage at the collector of Q1 will be maybe 0.2V. With R2 = 470 ohms the voltage across R1 would like to increase to almost 4V, but the LED in conjunction with the forward biased B-C junction clamps it to the LED Vf + 1 diode junction, so a few mA flows through R1 and the rest (almost 10mA) flows through D1, causing it to light brightly. \$\endgroup\$ Oct 27, 2022 at 6:53
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    \$\begingroup\$ There is no transistor action and the transistor acts as two forward biased diodes conducting current from base to emitter (normal) and from base to collector (not normal). The latter is what passes through the LED. You can play with the simulator using the circuit I set up above. \$\endgroup\$ Oct 27, 2022 at 8:09
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The circuit is configured such that the transistor shunts current from the LED, and higher base current will reduce current and brightness of the LED. It is not a practical circuit for controlling an LED.

Here is my simulation of the actual circuit, showing LED currents for base resistors of 470 ohms, 100k, and 100 Meg (open):

LED Circuit

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    \$\begingroup\$ So why is the LED On when it's 470 ohms and off when it's 100k? \$\endgroup\$ Oct 27, 2022 at 5:07
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    \$\begingroup\$ I simulated the circuit and it does not act the way you describe. If the emitter is not connected to ground, current will flow through the base-collector diode junction and will have behavior similar to what you observe. \$\endgroup\$
    – PStechPaul
    Oct 27, 2022 at 5:24
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    \$\begingroup\$ @SpehroPefhany explained it in his answer. \$\endgroup\$
    – PStechPaul
    Oct 27, 2022 at 6:43
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    \$\begingroup\$ My previous comment was based on the original circuit you posted. That no longer applies. \$\endgroup\$
    – PStechPaul
    Oct 27, 2022 at 7:02
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    \$\begingroup\$ No. You have a 470 ohm resistor from emitter to GND, which your schematics do not show, \$\endgroup\$
    – PStechPaul
    Oct 27, 2022 at 7:06

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