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I'm just starting on electronics, trying to build a few circuits but can't understand properly how things work.

I got this circuit and put on a breadboard, but the power source I have is 5V instead of 9, so I had to recalculate the resistors to make it work, but it doesn't. enter image description here

I just don't understand how this circuit works and the source website is quite bad on explaining why they are connecting the components in that way.

How and why this works?

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  • \$\begingroup\$ You can't blindly rescale the resistors because the voltage drop across the LED and B-E pins of the transistor don't scale with the voltage. They remain relatively fixed. B-E stays at about 0.7V. I don't know what the LED stays at. Probably somewhere between 1.5-2.5V \$\endgroup\$ – DKNguyen Feb 3 at 23:18
  • \$\begingroup\$ Welcome to EE.SE. Please upload the circuit into your post but keep a link to the article to credit the authors. That way all your readers don't have to follow a link to understand your question and your question will still make sense if the link dies. \$\endgroup\$ – Transistor Feb 3 at 23:19
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    \$\begingroup\$ Please share your final schematic after adjusting resistor values. \$\endgroup\$ – The Photon Feb 4 at 1:39
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    \$\begingroup\$ "I had to recalculate the resistors to make it work," - What resistor values did you use? "...but it doesn't." - What does it do (eg. LED stays lit up, LED doesn't light up, etc.)? \$\endgroup\$ – Bruce Abbott Feb 4 at 6:07
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    \$\begingroup\$ This looks like a design by someone who knows almost everything about how to make money on electronics to really naive people but actually knows very little about designing electronics. I can count 10 things done poorly and the biggest is no spec on what it is supposed to do besides damage the transistor switch. However it does start a timer to shutoff the LED if the button is pressed long than 70% or T=C1*R2 but shuts off immediately with contact bounce or release. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 at 16:58
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Assume the switch is open. Supply voltage Vcc.

When power is applied, R2 supplies enough current to the transistor base to turn Q1 on, illuminating the LED. R1 also charges the capacitor quickly through the transistor base (time constant of less than 1/4 second - 470uF * 470 ohms). LED current is about (Vcc - Vf -0.1)/R3. Base current after the capacitor charges is (Vcc - 0.7)/R2.

When the switch is closed, the positive terminal of C1 is grounded. The capacitor is charged to (Vcc - 0.7V) so the base of the transistor is now at -Vcc+0.7V, which turns the LED off. The capacitor begins to charge towards +.7V with a time constant of 470uF * 22K ohms or about 10 seconds).

When it reaches about +.7V the LED turns on again. The capacitor from this point forward has a reverse polarity of about 0.7V across it.

So let's see if the numbers work with the original resistor values. Let's assume a red LED with Vf = 1.8V and Vcc = 5V.

LED current of 3mA. That should be okay for a modern LED.

Base current of (5-0.7)/22K = 200uA. So Ic/Ib = 15. That's fine too (for an LED drive).

Worst-case base voltage: -4.3V That's fine (absolute maximum is -5V).

So it should work just fine from +5V. The off time will not be exactly 10 seconds, but it should be somewhere roughly that neighborhood.

If you, for some reason, scaled the resistors by (say) 5/9, the circuit would probably still work, however the time would be reduced.


If you're paying attention you may have noticed a couple of things. In the original configuration the base would go to -8.3V. That's well above the absolute maximum -5V and could damage the transistor. It's "probably" okay for a hobby circuit, but only by a slim margin. If the base breaks down, as it almost certainly would with, say, a 12V supply, the transistor will be destroyed the instant the switch is closed.

Secondly, the polarized aluminum electrolytic capacitor gets a -0.7V charge. That's generally okay, contrary to popular belief, but not much more.

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    \$\begingroup\$ Thanks Spehro, that was very clarifying. As I'm just starting, I already have thousands of other questions to ask but I'll get the basics ready first so I can ask more relevant questions. I just started reading the book "Practical Electronics for Inventors" and I'm loving it. \$\endgroup\$ – Rafael S. Menezes Feb 7 at 4:18
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Spehro explained the circuit well, I will discuss troubleshooting steps. Even experienced engineers don't always design and build circuits that work the first time. Troubleshooting is a vital skill that you need to learn.

For this circuit, I would simplify it until the LED lights up.

Remove the cap to make a simpler circuit. Most likely, the LED will still be off.

Probe Q1-B with a DMM, it should be about 0.7V.

Probe Q1-C with a DMM, it should be < 0.4 V.

If these are good, remove Q1 and tie D1 anode to ground. The LED should light up.

If the LED doesn't light, it is probably in backwards.

If the LED does light when connected directly, your transistor is probably wired wrong. (or, you fried it). Re-install the transistor and check the wiring carefully.

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@Spehro Pefhany's explanations are, as usual, perfect but specific. I will consider, as usual, the "philosophy" behind circuit. Both approaches are useful and necessary to form a complete picture of the circuit... although, in this material world, the implementation is valued more than the idea...

The clever idea here is not to charge or discharge but to recharge the capacitor... and this is a crucial advantage of this configuration. Why?

If we make a timer either by charging or discharging the capacitor through the base-emitter junction of the transistor, it will switch at the end of the exponential curve where it has very small slope... and it will not switch sharply. In the first case, the capacitor voltage will change from zero to +VCC; in the second case - from +VCC to zero.

Here, the capacitor is reversely charged to -VCC; so its voltage tries to change from -VCC to +VCC (two times greater range). The base-emitter junction is forward biased and does not affect the voltage. But when approaching the ground voltage (more precisely, 0.7 V), the voltage rapidly changes... the base-emitter voltage sharply turns on and fixes the capacitor voltage.

That is why the same idea is used in BJT multivibrators. So, we can say this timer is a 1/2 multivibrator.

I have realized this idea not only from my work with students but also from my personal experience when I tried to make simple timers for any purpose, especially in cars.

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    \$\begingroup\$ How is charging different from recharging a capacitor? Does the capacitor know that it was charged before and it behaves differently? \$\endgroup\$ – Elliot Alderson Feb 4 at 14:54
  • \$\begingroup\$ Nice question... My understanding of the three cases is: 'Charging' is to connect a discharged capacitor to a source. 'Discharging' is to connect a charged capacitor to a load. 'Recharging' is to connect a reverse charged capacitor to a source. The difference is in the range (in the last case it is twice bigger) and in the rate of change around ground (in the last case it is bigger). \$\endgroup\$ – Circuit fantasist Feb 4 at 19:04
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Theory of Operation

It is a called a 1-shot when something is enabled then times out for some period. In this configuration, you have to hold down the switch to disable the LED switch for some desired 1-shot period.

It must be pressed longer than about 60% of "Tau"= τ = R2*C an estimate of the exponential ramp time of the negative voltage applied to the transistor Vbe to disable the switch.

Q1 acts as a switch because R2 applies more than 5% of the collector current into the base which causes Vce voltage to drop << 1V. So when going from OFF to ON collector starts as a high resistance current source Ic/Ib=hFE then goes into a low resistance conductor between collector and emitter when Vce approaches 0V where the transistor changes its characteristics.

"Forward biasing" the base emitter, Vbe with voltage >0.6V or collector current >1mA such that the V=IR + Vf LED drop from the supply causes the tranistor to "saturate" as a switch.

So initially with the switch open using a 5V supply below, I can demonstrate with values to blink off the LED whenever the switch is held down more than 60% R2*C1. But it will turn ON immediately if released quickly.

R2=33k, C1=10uF T= 330ms so we expect ~200ms blink OFF time. Your circuit had 470uF*22kΩ =τ=10s but Transistors are designed with high gain such that unfortunately, the Veb reverse voltage is almost always -5V max, as a safe limit before stress-damage will occur. This must be avoided to prevent wounding the device or a permanent short-circuit.

Simulation

enter image description here

Now for longer times a FET or a CMOS logic Inverter works best.

Here the switch is edge sensitive and power on also triggers the switch ON for 10 seconds.

enter image description here

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  • \$\begingroup\$ I just can't understand why you have increased the resistance of R1 to 10k. It is advisable to restore the capacitor charge quickly after opening the switch. BTW we can replace it with a push button if connect another transistor switch (driven by Q1's collector voltage) in parallel to the button. I have done this by another contact of the relay in place of the LED. \$\endgroup\$ – Circuit fantasist Feb 4 at 19:30
  • \$\begingroup\$ Tony, How can R1 reduce the voltage across the capacitor? It is charged to VCC regardless of R1; only the time depends on it... \$\endgroup\$ – Circuit fantasist Feb 4 at 20:10
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    \$\begingroup\$ sorry I was thinking of another variation that I deleted with R4 between R2 and base \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 at 20:12
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    \$\begingroup\$ Perhaps a more interesting demonstration is a 10sec 1shot with a turn ON time that is edge sensitive using a FET or logic gate. tinyurl.com/tmec2q5 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 4 at 20:41
  • \$\begingroup\$ Nice simulation... that is a good confirmation of my explanations about the slope of the exponent at the end of the process. As a result, the LED does not turn off sharply but smoothly. Of course, the best remedy is the positive feedback... \$\endgroup\$ – Circuit fantasist Feb 4 at 20:56

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