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Newbie help with reading circuit please

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marchache

New Member
Hi,

I picked up "Getting Started in Electronics" by Forrest Mims and I'm having trouble understanding how a particular circuit works.

It's the "light flasher" circuit on pg 104 of the third edition. I'm hoping someone out there has the book and can help. If not, am I allowed to scan and post the circuit ?

Anyway, I think it's just a basic lack of understanding of circuits in general and capacitors and transistors in particular but I'm not sure I understand how this circuit work. let me take a stab at it:

First I assume the input voltage is DC.

My guess is that Q2 is allowing current flow as soon as power is supplied to the circuit and hence the light is on.

The capacitor begins to charge (rate being controlled by R1) and once it hits .6V Q1 allows current flow.

Does this in effect bypass the light (turning it off) because the resistance is now less through Q1 than through Q2 and the light ? The capacitor would then be grounded and discharge, closing the base at Q1 and starting the cycle over ?

I guess part of what is confusing me if I'm wright is why use Q2 at all if it is always on ?

I'm probably way off the mark but if some kind sole could help me out that would be cool.

Thanks
Marc
 

Johnson777717

New Member
Does your circuit look something like this? (See re-drawn circuit below)
If so, here's the explanation:

C1(1uF) charges through R1(100K) and R2(1M pot) until the charge on C1 (1uF) is high enough to turn on Q1. This causes Q2 to switch on. The LED is now connected to + 4.5 volts DC through Q2 and R4(100), so the LED lights up. Then, C1 discharges through the base-emitter of Q1. This causes Q1 to switch off, which then switches Q2 off. Thus, the connection from the LED to ground is broken and the LED turns off. The cycle then continues as the capacitor charges.--Taken in effect from "Basic Electronics, Transistors And Integrated Circuits, Workbook 1" by Forrest M Mims III

so we have:
1. Capacitor charges C1 (1uF)
2. Q1 switches on due to the proper amount of voltage.
3. Q2 switches on due to Q1 turn on.
4. LED lights up because the connection between +4.5VDC and ground is made through Q2.
5. The LED uses up all of the power stored in C1 (C1 discharges)
6. Once C1 does not have enough voltage, Q1 shuts off.
7. Once Q1 shuts off, Q2 will shut off, which breaks the connection between the +4.5VDC and ground.
8. Thus the LED turns off.
9. The circuit is now closed, which allows C1 to charge again.
10. Once C1 is charged, the cycle continues. (Back to step 1).

Schematic redrawn with "Express Schematic"
 

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marchache

New Member
Thanks so much. The circuit is very close, close enough for the light to start to go on...Sorry couldn't resist the pun..

A couple of follow on questions if I may.

Does a capacitor allow current to flow while it is charging ? I thought it did but if that were the case wouldn't a connection be made from 4.5V through the LED, C1, the 1K resistor, the R2 and R1 to ground turning the light on ?

Once Q2 is on providing a connection, why doesn't the LED draw power from the 4.5V source instead of the capacitor ? Does it have something to do with the bias of the LED ?

Thanks again, sorry for the simple questions. I'd really like to get my head around this so I can start building on my own.
Marc
 

ljcox

Well-Known Member
marchache said:
Thanks so much. The circuit is very close, close enough for the light to start to go on...Sorry couldn't resist the pun..

A couple of follow on questions if I may.

Does a capacitor allow current to flow while it is charging ? I thought it did but if that were the case wouldn't a connection be made from 4.5V through the LED, C1, the 1K resistor, the R2 and R1 to ground turning the light on ?

Once Q2 is on providing a connection, why doesn't the LED draw power from the 4.5V source instead of the capacitor ? Does it have something to do with the bias of the LED ?

Thanks again, sorry for the simple questions. I'd really like to get my head around this so I can start building on my own.
Marc
Marc,
Yes a current flows into capacitors when charging. That is how they build up charge.

The LED will not light due to the charging current since the current is too small. It has to flow through the 100 Ohm resistor, C1, the 1K resistor, R2 and R1 to ground as you said. The total resistance is equal to 100 Ohm + 1k + 100k + the pot resistance so it is at least 110k. With a 4.5 Volt source and 111k means a current of 4.5/111k = 40.5 uA. So the current will be less than this depending on what the pot is set to.

A LED needs several milliamps to make it light. So it certainly won't light to the charging current.

Once Q2 is on, the LED does draw power from the 4.5V source. The Q2 collector current flows through the 100 Ohm resistor & the LED.
 

marchache

New Member
Excellent, thanks for the explanation.

Last question (I think). How does C1 discharge if there is no load placed on it by the LED ? Johnson777717 said something about discharging through the base emitter pair of Q1 but since Q1 is on doesn't the current flow through the collector emitter ?

Thanks
Marc
 

ljcox

Well-Known Member
marchache said:
Excellent, thanks for the explanation.

Last question (I think). How does C1 discharge if there is no load placed on it by the LED ? Johnson777717 said something about discharging through the base emitter pair of Q1 but since Q1 is on doesn't the current flow through the collector emitter ?

Thanks
Marc
I had not studied the circuit in any detail before my last post.

When the power is first turned on, C1 is uncharged and so it starts to charge via the 100 Ohm, LED, 1k, pot and 100k.

When the voltage between the base and emitter of Q1 rises to around 0.5 ~ 06. Volt, Q1 collector current starts to flow and so Q2 receives base current. The collector voltage of Q2 thus starts to fall, this increases the Q1 base current (because of the charge on C1 increase the base/emitter voltage of Q1) hence its collector current rises further. Thus Q1 and Q2 are rapidly turned on by the regenerative action (positive feedback). Q2 saturates and the LED glows.

When Q2 saturates, its collector voltage will be about 0.1V. So the right hand plate of C1 is now at this voltage.

Since the capacitor was charged to about 0.6 V when the switching occurred, the left hand plate of C1 will be at about -0.5V, ie. 0.1 - 0.6 = -0.5V (the capacitor's charge cannot change instantaneously).

The base of Q1 will be at about 4.5 - 0.7 = 3.8V. Thus C1 starts charging (through the emitter/base of Q1) towards this potential. As the voltage across C1 rises, the current through it decreases and consequently, the base current deceases and the collector current of Q1 also decreases in proportion. When the Q1 collector current decreases to a small value, Q2 starts to come out of saturation and so the potential at the right hand plate of C1 starts to rise and, of course, the potential of the left hand plate also rises at the same rate. This starts a rapid regenerative action since it increases the base emitter voltage of Q1 thus driving it into cutoff. Hence the collector falls to zero and so Q2 also turns off.

Since the left hand plate of C1 has a positive charge of about 3V (with respect to its left hand plate), the base/emitter junction of Q1 is reverse biased thus holding it in cutoff.

The circuit is now back to its initial state with Q1, Q2 and the LED off. So C1 again starts to charge via the 100 Ohm, LED, 1k, pot and 100k.

You will notice that I have not mentioned the discharge of C1. This because charge and discharge are essentially the same. Current going into or out of a capacitor changes its state of charge and hence the voltage across it changes in proportion.

Capacitors charge towards the applied voltage (ie. the voltage you would measure if the capacitor was not there). In the description above, the capacitor changes from a positive charge to a negative one and then back to positive on each cycle. Thus at certain instants, its voltage (and charge) is zero. At these instants, the capacitor is (technically) discharged, but only at those instants.
 

ljcox

Well-Known Member
marchache said:
Excellent, thanks for the explanation.

Last question (I think). How does C1 discharge if there is no load placed on it by the LED ? Johnson777717 said something about discharging through the base emitter pair of Q1 but since Q1 is on doesn't the current flow through the collector emitter ?

Thanks
Marc
In order for there to be collector current, there must be some base current
 
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