Can somebody explain this simple circuit please?

[arashi256]

Hi new guy here - just building my first circuit and I'm looking at creating an LED flasher. I've come across a circuit like this: -

...which I don't understand. According to how I *think* it should work, I'm sure it should only work with non-polarised capacitors rather than the polarised ones used above. Is the diagram wrong or am I? I hope I'm wrong, actually, as as a beginner with a limited set of components, I only have polarised capacitors and no ICs (yet). Besides, simply flashing an LED shouldn't require any ICs, right?

Thanks for any help and hello all!

 

[duke37]

Look for information on a multivibrator, there should be lots of references out there.

Polarised capacitors can be used as long as the polarity is not reversed. They have the advantage that they are smaller and cheaper than non-polarised capacitors.

You can make a multivibrator or similar oscillator with an IC, it all depends on what you have available. You can get a flashing LED with the oscillator inside the LED

 

[arashi256]

I know what it is - I *did* do some research before-hand, which is how I came across the above diagram. I can't use ICs yet because I've not got to that yet - like I said, beginner. No point running before I can walk.

It's just that from this simulation, it seems as though the capacitors should be bi-directional or non-polarised to provide current to the base for the transistor Q1 and Q2 collectors to be able to pass current through the emitters to ground or an LED or whatever. After reading a beginner's article on capacitors, you can't reverse polarised caps without having to up the voltage significantly and break their voltage threshold which would destroy the capacitor anyway.

I can see how it works with non-polarised capacitors, I just don't understand how it could work with polarised ones. Which was what I was asking in the first place

 

[arashi256]

Unless I've got the direction of the capacitors or current direction wrong and current in that diagram is going from the "straight" plate to the "curved" plate, of course.

In which case, this is a very embarrassing first post

 

[duke37]

The collectors will swing from about 0.2V to 7V.
The negative side of the capacitor will swing 6.8V but the maximum voltage at the base will be 0.6V (the base conducts) so the swing will be from 0.6V to -6.2V.

Note that 6.2V negative on the base is close to the limit for normal silicon transistors.

The base current comes through the 100k resistors.

ICs are not necessarily more complicated to use than transistors. An oscillator can be made with a 4093 Schmitt trigger, one resistor and one capacitor.

I do not have Java, it messed up my e-mails, so I cannot see the simulation.

 

[(*steve*)]

It's just that from this simulation, it seems as though the capacitors should be bi-directional or non-polarised to provide current to the base for the transistor Q1 and Q2 collectors to be able to pass current through the emitters to ground or an LED or whatever. After reading a beginner's article on capacitors, you can't reverse polarised caps without having to up the voltage significantly and break their voltage threshold which would destroy the capacitor anyway.

Firstly, you can have reverse polarity on most polarised capacitors. The datasheets for these devices typically go into it if you're interested enough to look. However a relatively simple rule of thumb is that you can place a reverse voltage of about 10% of an aluminium capacitor's rated voltage across it. It will be leaky(er), but it won't kill it. Having said that, if you plan on doing this, check the specs for the device.

In this case, the base of the transistors will never fall below 0.7 volts. The other end of the capacitor *could* swing down to 0.2V, but up to (I think it was calculated as) 6.8V.

The polarity on the capacitor is not reversed until the voltage on the +ve end is lower than that on the -ve end.

In this circuit, the -ve end is fixed at approximately 0.7V and the other end gets pulled high and low. However the voltage across the capacitor doesn't fall immediately to 0.2V. It discharges slowly down to 0V before charging in the reverse direction.

Discharging a capacitor is NOT placing a reversed voltage on it.

Check the voltages at each end of the capacitor in that simulation and see if the capacitors do get a reverse voltage, and if so, by how much.

 

[KrisBlueNZ]

Yes, the capacitors will be charged slightly in the reverse direction, but only by about 0.5V. This doesn't cause a problem with a normal electrolytic that's rated at say 16V.
In that astable multivibrator circuit, when a transistor turns ON, its collector voltage falls rapidly to about 0.2V above the common rail. The capacitor attached to the collector will, at that time, have about 6.5V across it, with the correct polarity. So the negative end of the capacitor will swing down to about -6.3V, which as duke37 pointed out, is about the maximum negative base-emitter voltage that a typical bipolar NPN transistor can withstand, because the base-emitter junction behaves as a zener diode so the base will draw current; the transistor can be damaged if this current is too high.
When the negative end of the capacitor swings down to -6.3V (approx), the other transistor is turned OFF. Current through the connected 100K resistor steadily pulls that negative voltage (the base voltage of the other transistor) upwards, towards the ground rail, and past it. Once the voltage passes above +0.2V, the voltage across the capacitor has the wrong polarity, but this only lasts until the base voltage reaches the 0.7V base voltage threshold of the other transistor, at which time the circuit flips and the same behaviour starts with the other capacitor.
The small 0.5V reverse voltage will not damage a typical electrolytic capacitor that's rated at say 16V.

 

[TedA]

Yes, the capacitor might get a little reverse voltage. This might not be the best thing you could do to the capacitor. But, if this is just an experiment on a breadboard, nothing bad will happen before you get tired of staring at the blinking lights.

In fact, I suspect the cap may never see reverse voltage. When a given transistor is first turned on, it gets lots of forward base current, through the capacitor. But as the capacitor charges-up, this current drops, until most of the current comes from the 100k resistor to +9V. The 100k resistors will not supply enough base current to the transistor to hold it in good saturation. The collector voltage will float a bit higher as the transistor is starved for base current. This will happen before the other transistor's base voltage gets above 0V.
The degree to which this happens will depend on the individual transistors.

Even a possibility of reverse voltage on the timing caps can be eliminated by adding more parts, but what's the point?

Ted

 

[arashi256]

Okay, thanks all. Not sure I understood all of that, but it's a start

 

[Cannonball]

Remember the sole purpose of the caps is to turn the transistor off that it's negative lead is connected to until it discharges through the bias resistor for that transistor.

 

[davenn]

Remember the sole purpose of the caps is to turn the transistor off that it's negative lead is connected to until it discharges through the bias resistor for that transistor.

this thread is 5 yrs old .... try and avoid reviving very old threads ... thanks

thread closed