By pressing the stop-button once the LED turns off and stays off. But if you remove R1 the LED only stays off if you are pressing the stop-button. Why does R1 affect this?
Electronic – the purpose of R1 in this circuit
transistors
Related Solutions
Here is a possible solution for you:
simulate this circuit – Schematic created using CircuitLab
The pin A0 from the microcontroller is a pin capable of both digital output and analog input. This pin is configured as input during start.
This circuit is wired so that the gate voltage of M1 (aka analog input A0 of the microcontroller) maps to this:
- \$0\mathrm{V}\$: circuit is turned off, or turning this entire circuit off.
- \$\frac{1}{2}V_{cc}\$: circuit is running, button not pressed.
- \$V_{cc}\$: Button is pressed.
Power on: The button SW1 is pressed down and +5V is present at the gate of M2, pushing M2 on and in turn push M1 on, turning this entire circuitry on. Microcontroller start and wait until the voltage of pin A0 drop to \$\frac{1}{2}V_{cc}\$ and start the program.
Power off: When pin A0 is pulled to +5V for several seconds and released, switch pin A0 into digital output and output a logic 0, turning M2 off and in turn cutting off M1, turning this circuit off.
There are a number of problems with your circuit.
- You have a 1 V power source. This will not turn on an LED, never mind two in series.
- Assuming that's a mistake and you have about 5 V supply (based on R5 and R7 being 100 Ω and a current of 10 mA) then when you close SW2 you will put 5 V across D3 and destroy it.
- It's not clear what purpose D1 and D2 serve.
- You have wired Q1 and Q2 as voltage followers. Logic circuits would normally use some kind of common emitter.
simulate this circuit – Schematic created using CircuitLab
Figure 1. A NOR and a NOT will do the job.
- Q1 and Q2 form a NOR gate. Turning either on will pull the common collector to ground. Note that the transistors will be driven into saturation so they could be within 0.2 V of ground whereas your arrangement could at the very best get to about 0.7 V below supply.
- When Q1 is turned in it will turn on Q3. Its collector will be pulled high. This will turn on Q2 and latch.
Figure 2. Minimalist version.
Figure 3. Analysis of your circuit. Note that there is 0 V to bias Q2. It can't turn on.
How can I get the base voltage higher than the emitter?
You can't. That's why logic gates aren't designed that way.
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Best Answer
T1 and T2 are connected in a classic 'thyristor' bistable latch.
If T1 is on, it biasses T2 on, which holds T1 on.
If T1 is off, it doesn't bias T2 which stays off, which doesn't bias T1 which stays off.
That's in theory, with perfect transistors. In practice, all transistors have some degree of leakage current. This is amplified by the other transistor, and an off circuit may turn on if there's sufficient gain.
Without R1, any current flowing through R3 is amplified by T2. R1 sets a threshhold current, below which T2 will stay off. That's about 700mV/10k = 70uA, which is waaaay above any likely leakage through T1.
However, there will also be leakage through T2, which will be amplified by T1. Assuming reasonable hFE, say <= 300, the max permissible T2 leakage now becomes 230nA, which is still easily met by good quality transistors at room temperature.
The LED does not prevent T2 turning on, as its threshhold voltage for significant conduction will be above 700mV.