Consider the Monostable.....

[UTMonkey]

Hi All,

I am studying the humble transistor monostable.

According to my trusty(?) electronics book, transistor Q1 that has its base fed by the 50k x 10uF cap (see attached) is switched on bringing Q1 collector voltage low.

Putting these bits into my simulator it does what the book says (which is a first).

I get everything about what this circuit is doing EXCEPT why is it that Q1 powers on first? Surely the base voltage is reached at a slower rate because of the RxC, compared to the base of Q2 which is fed just from the 50k resistor?

Thanks in advance

p.s. yes I know a 555 monostable is easier.

 

[on1aag]

Hi UTMonkey,

It's because C1 charges through R1 and the base-emitter junction
of Q1.

on1aag.

 

[UTMonkey]

Thank you, but why not the 50K resistor to the base of Q2?

 

[ericgibbs]

Thank you, but why not the 50K resistor to the base of Q2?

Because while the cap is charging, the transistor Q1's collector is low, so there is no voltage to drive any base current into Q2 via the 50K.[R3]

 

[UTMonkey]

Thanks Eric. But surely at some point the cap starts at 0 and takes "some" time to get to 0.6.

And whilst it tries to do that Q2 is getting its full base voltage?

Am I missing something?

 

[ericgibbs]

Thanks Eric. But surely at some point the cap starts at 0 and takes "some" time to get to 0.6.

And whilst it tries to do that Q2 is getting its full base voltage?

Am I missing something?

hi,
Think in terms of current rather voltages, the cap charge current via the base is instantaneous.

EDIT: pix added

 

[UTMonkey]

Ahhh, I get you.

But surely voltage becomes a factor if it is to exceed the 0.6 of Q1?

 

[ericgibbs]

Ahhh, I get you.

But surely voltage becomes a factor if it is to exceed the 0.6 of Q1?

hi,
At switch ON, due to the charge current into the cap, the base/emitter voltage is instantly at 0.6/0.7V, so the transistor ON.

When the cap is fully charged the 50K supplies base current keeping the transistor ON, so you still have 0.6/7v between base/emitter.

When the switch is closed, Q1 is switched OFF and Q2 switched ON, this connects the +5v charge on the cap to the 0V line.
The end of the cap will now be - (5-.6v), that is the base of Q1 will be reverese biased at about -4.3V.

This reverse voltage will keep Q1 OFF until the cap discharges via the 50K [this is the Mono timing period], as soon as the Q2 base gets to +0.6V it will turn Q1 ON again.... and wait for the next sw press.

 

[UTMonkey]

Many Thanks Eric,

I think I may be in danger of falling into one of the "Transistor is a Voltage\Current controlled device (delete where appropiate)" camps.

I understand the action but I am approaching it in terms of voltage.

Is this "so" wrong?

 

[ericgibbs]

Many Thanks Eric,

I think I may be in danger of falling into one of the "Transistor is a Voltage\Current controlled device (delete where appropiate)" camps.

I understand the action but I am approaching it in terms of voltage.

Is this "so" wrong?

In using transistors, I find that considering it to be current controlled device, makes understanding transistor operation easier.

Extract from Wikipedia:
[edit] Bipolar junction transistor
The bipolar junction transistor (BJT) was the first type of transistor to be mass-produced. Bipolar transistors are so named because they conduct by using both majority and minority carriers. The three terminals of the BJT are named emitter, base and collector. Two p-n junctions exist inside a BJT: the base/emitter junction and base/collector junction. "The [BJT] is useful in amplifiers because the currents at the emitter and collector are controllable by the relatively small base current."[4] In an NPN transistor operating in the active region, the emitter-base junction is forward biased, and electrons are injected into the base region. Because the base is narrow, most of these electrons will diffuse into the reverse-biased base-collector junction and be swept into the collector; perhaps one-hundredth of the electrons will recombine in the base, which is the dominant mechanism in the base current. By controlling the number of electrons that can leave the base, the number of electrons entering the collector can be controlled.[4]

Unlike the FET, the BJT is a low–input-impedance device. As the base–emitter voltage (Vbe) is increased the base–emitter current and hence the collector–emitter current (Ice) increase exponentially (, where K is a constant). Because of this exponential relationship, the BJT has a higher transconductance than the FET.

Bipolar transistors can be made to conduct by light, since absorption of photons in the base region generates a photocurrent that acts as a base current; the collector current is approximately beta times the photocurrent. Devices designed for this purpose have a transparent window in the package and are called phototransistors.

 

[crashsite]

I understand the action but I am approaching it in terms of voltage.
QUOTE]

While I acknowledge that transistors are "current devices", I also tend to think of circuit actions in terms of voltage. The "engineer types", who tend to think mathematically, seem to typically like the current model with current flowing into and out of nodes. But, current is difficult to measure except by its voltage effects so "technician types" (like me...who are lousy at math) often tend to bypass the "current modeling" and go right to the voltages.

But, to your original question, look at the current source for Q1. It consists of a 50k resistor AND also a capacitor in series with a 500 Ohm resistor. Thus, for the first instant, at power on, it gets a heavier (and more importantly, a quicker) dose of base current.

Q2 is also getting base current but must contend with various stray capacitances that must charge through 50k to turn it on so, it's Q2 that gets the delayed turn-on. By that time, Q1 is already on and the rest is history...

In practice, you also have to account for another option which your simulator probably wont deal with. What if the supply voltage is applied slowly? In astable multivibrators this can actually bring up both transistors in the ON state and the circuit wont oscillate. In the monostable, that is an illegal condition but, what can happen if the wrong transistor comes on first is the possibility of getting a spurious signal that can disrupt other circuits the monostable connects to as it settles into its legal status.