# Transistor Amplification and PUT questions

Discussion in 'General Electronics Chat' started by geratheg, Aug 12, 2014.

1. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
I've been reading and doing experiments from the Make: Electronics book and had a few questions. I'll number my questions in red, the ones with a letter are the ones for which I am looking for an answer:

1) If we had 6V supply and a 100 ohm resistor in series with a collector of the transistor that amplifies at a 24:1 ratio such as the 2N2222 and the emitter to ground (see attached diagram) and no other device is connected in series, even though the transistor amplifies current, the maximum current would be limited by the resistor like so: 6/100 = 0.06 A by ohms law and the transistor cannot amplify to any more current than 0.06 A in this case.
1a) Is this correct?
1b) It's a little confusing to a beginner like myself to think about amplification because first I thought to myself "Does the transistor make current? Is that how it amplifies it and where do the limits come from?"

2) One of the experiments uses a 2N6027 PUT and a 2N2222 with a 0.0047uF capacitor and a couple of resistors (see attached diagram). The speaker barely makes any sounds when the 2N2222 transistor is eliminated.
So in the diagram, lets remove the 1K resistor and the 2N2222 transistor (Q3). Now wire 100 ohms to the bottom of the 2N6027 where the 1K was removed from, and the speaker in series with that and to ground.
To me it feels like this should work because the PUT isn't supposed to have much resistance inside it once it lets current through (what I learned), and the speaker is basically now wired to the capacitor, which acts as if it's the power source and the speaker isn't going to get much current from the main power source due to the large 470K resistor.

2a) The capacitor is theoretically supposed to be able to deliver infinite current if it was shorted, correct?
2b) Now, it appears that just above 2.7V is around what the capacitor is charged to before being discharged, is this correct? Or is it 3.3V?
2c) My other question: Is the capacitor not able to deliver enough current or is it that it does not store enough voltage for the speaker to make enough sound (it's very very quiet)? Basically, why do we need to add a 2N2222 transistor to make the speaker easily audible? Is it possible to make the speaker louder using just the Programmable Unijunction Transistor?
Does the PUT transistor have a maximum current that it can handle and thus is limited to delivering a limited amount of current? Is this why the speaker is very quiet?

Thank you for all the help!

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• ###### Speaker Schematic.jpg
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Last edited: Aug 12, 2014
2. ### ericgibbs Senior Member

Jan 29, 2010
2,418
369
hi,
Do you follow this OK.?
E

3. ### MrChips Moderator

Oct 2, 2009
12,248
3,283
Yes.

The transistor does not make current. Think of all circuits as a current loop.
Current is determined by Ohm's Law as you have rightly stated. The current I is determined by the voltage source V and the resistance R around the loop.

By Ohm's Law, I = V/R

If V = 6V and R = 100Ω, the maximum current I = 6V/100Ω = 60mA

Any other components in the loop such as the transistor and speaker will only make the current lower than this maximum.

Maybe. The current delivered will be determined by the resistance of the short. However, because the capacitor will rapidly loose its charge, the voltage will fall and hence the current will only last for a very short period of time dependent on the value of the capacitor.

The capacitor can potentially be charged to 6V through the 470kΩ resistor.
However, when the anode voltage rises above the 2.7V reference plus the 0.6V threshold, i.e. 3.3V, the PUT will avalanche and start conducting. Hence the voltage on the capacitor never exceeds 3.3V

The maximum current is determined by the circuit topology, not by the PUT itself. With 470kΩ and 33kΩ resistors in the circuit, the maximum current is about 6V/33kΩ which is less than 0.2mA, not enough to power the loudspeaker.

The 2N2222 NPN transistor is used as a switch to allow the larger 60mA current to flow to the speaker. Think of a transistor amplifier as a control valve that can regulate a larger flow of current using a much lower control current.

4. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
Thank you for all the responses.

I thought the 27K and 33K resistor wasn't there to limit current, but simply to provide the reference voltage. I also thought the speaker would get its sound not from current passing through the 470K resistor, but from the capacitor itself when it discharges to the speaker. Shouldn't that provide enough current for the speaker to be easily audible? If the capacitor discharges many times a second into the speaker, shouldn't I hear a sound from that?
Or is the capacitor just too small to provide sufficient power to the speaker?

Last edited: Aug 12, 2014
5. ### MrChips Moderator

Oct 2, 2009
12,248
3,283
The 27kΩ and 33kΩ resistors do set the reference voltage at 2.7V.
They also limit how much current is able to go down the resistor chain, equal to 100μA.

The 470kΩ resistor and 0.0047μF capacitor sets the frequency of the oscillation.

As a first order estimate, the RC time-constant is 0.47M x 0.0047μ = 2ms which would give a frequency of about 500Hz.

Now let's do a charge calculation.

The charge Q on the capacitor Q = C x V

C = 0.0047μF
V = 3.3V
Q = 0.0047 x 3.3 μC = 0.155 μC

As a rough estimate, assume that this charge lasted for 2ms
then the discharge current I = Q/t = 0.155μ/2m = 77μA

77μA is hardly going to be heard on the loudspeaker.

To get an audible click you would have to increase the capacitance by a factor of 500 which of course will reduce the frequency by a similar amount.

6. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
This is mainly what I was looking for. Thank you.

I didn't know current can be calculated this way. That was very helpful.

7. ### MrChips Moderator

Oct 2, 2009
12,248
3,283
Please note that the above is only a rough estimate assuming that the current is constant.

If the current I is constant for t seconds, the charge delivered to a capacitor is by the formula Q = I x t.

When the current is not constant, the charge or voltage has to be determined using differential equations.

8. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
Yes that would be about the average current according to my physics formulas, but the current would of course continue decreasing if it's actual current since this is a capacitor, which is more complex and doesn't use the basic formula you mentioned.

Last edited: Aug 12, 2014