Transistor Confusion?

ok in the first example, the 12bulb comes on. (from 12v to gnd)

in the 2nd example, i pass 12v to the collector, 5v to the base, that connected in series with a bulb and then to ground?..

but the bulb lights dimly, is the output only going to be 5v? and where’s the current being drawn from? the 12v rail or the 5v rail?

(

is the output only going to be 5v? and where's the current being drawn from? the 12v rail or the 5v rail?

  1. Yes.
  2. 12V

I see..

interesting, the current is drawn from the 12v rail, but only a max of 5v..

Would that be a safe way to boost current from a 2nd power source intentionally using say 3 amps of current from the 12v rail but keeping it a constant 5v? (max supply of 100ma, eg a to power a USB device?).. it sounds like this method could be used to charge batteries? from any 12v source providing the current is there....

Thanks...

eg…

whatever the resistor value, infact not even an LED< but any device on 5v.

point is, that’s safe to do? draw a 5v supply but take the amps from the 12v rail?..

it’s a little weird for me to get my head around lol… (collector 12v, emitter 5v, base 5v) if 5v is the “Switch” then the 5v allows emitter and collector to flow? but only up to the voltage
on the base pin?..

ok i get it, just weirded out by it (powering devices from 12v using nothing more than a low amp 5v rail? ?) weird.

With an NPN transistor, you normally want the Emitter closer to ground potential for maximum transfer. And you ALWAYS want to limit transistor Base current with a resistor to prevent abusing the transistor or what is controlling it…

lol i'd use a resistor (always do).. i was thinking in theory (to power a device which required more amps than i have, but using this method i could get a regulated 5v (current from the higher voltage rail) out power what i want... surely this way would be better than a 5 watt zener diode? (rated at 5v) vs a Tip31 and the above method?.. powering a 1amp 5v usb device from 12v via a resistor and 5v on the base pin, i still have issues with this (yes yes with a resistor to limit the current....)

pwillard:
With an NPN transistor, you normally want the Emitter closer to ground potential for maximum transfer. And you ALWAYS want to limit transistor Base current with a resistor to prevent abusing the transistor or what is controlling it...

in the circuit you posted, you're switching on via the 5v base, but the collector/emitter sit on 12v/0v

what if, you only had 10ma to play with at 5v

but, you have a 2nd 12v rail of 3amps.

collector to 12v, emitter to your relay (with reverse diode protection) , when the 5v switches on the transistor, your relay would not fire off, because it's 12volts
but, let's supposing it was rated 5v and requires 100ma to switch the relay on...

using the above method, the relay would switch on as it consumes 100ma from the 12v rail but supplys only 5v energizing your relay.... I don't get the physics,,, i guess i'll just have to accept i understand the principal not so much how it works .

cjdelphi,
Get on-board and use the standard symbols. :slight_smile:

Anyway, what you’re attempting to describe, with the output from the emitter is what’s known as a power amplifier, common-collector configuration (or “emitter follower”) - it’s a pass transistor, the voltage regulator circuit depicted below.

The thing to bear in mind is that the difference between V_in and V_out is across the transistor, that’s VCE. Your transistor’s rated PD >> your output current * VCE.

For a BJT transistor to pass current, there should be approximately 0.7V between B and E (called Vbe).

In your circuit (with collector to +12v):Now if your base is at 5v (Vb), then in order to pass current, the transistors emitter will be 0.7v below that, so Vb - 0.7V = Ve.

That means at the emitter you will have 5-0.7 = 4.3V, as long as the collector voltage is larger or equal to the base voltage. If it is less, the transistor is turned off.

If you have it the other way around (emitter to gnd):
As the emitter is to GND, the Ve = 0v. This means that in order for the transistor to conduct, Vb = Ve+0.7 = 0 + 0.7 = 0.7V. That means you need to have 0.7v at Vb. Not 5v, not 12v, but 0.7V.

Due to the way a BJT works (it is a CURRENT amplifier), then if you let a certain amount of current flow through the base, you let a larger (but proportional) amount of current to flow from collector to emitter. As such we first work out how much current needs to flow through the collector. This depends on what load you are driving. Say 100mA.
The transistor has a gain called Beta (aka HFE), which is Ic/Ib (collector current/base current). This can be found in the datasheet, but lets say for our case it is B=100. Than means we then need a base current of Ib = Ic/B = 100mA/100 = 1mA.

We are using a 5v logic, which means we need to use a resistor to drop the 5v down to the 0.7v required across the gate, so:
Vr = 5v - Vb = 5-0.7 = 4.3V.
We also know we need a current through the gate, and hence through the resistor, of 1mA, so:
R = V/I = 4.3/0.001 = 4.3kOhm.

If you go down to your local shop, you are unlikely to find a 4.3k resistor. Furthermore, it is unlikely that the current in the load will be exactly constant. Fortunately, the transistor can happily run whereby the current flowing from collector to emitter is no longer being limited by the base, but by the load.

Say we use a 3.9k resistor. Working backwards, we get Ib = V/R = 4.3/3900 = 1.1mA. Notice that we have increased the current through the base of the transistor slightly? What does that mean for the load. Well, Ic = Ib * B = 1.1mA * 100 = 110mA. But wait I hear you cry, the load only draws 100mA... well, the transistor can conduct 110mA, but it doesn't have to, thats the wonder of a semiconductor. As long as the 'allowed' current is greater or equal to the 'required' current, all is well.

If for example we go the other way and choose a 4.7k resistor, we get Ib = V/R = 4.3/4700 = 0.91mA. This means we have Ic = Ib*B = 0.91mA * 100 = 91mA. Now we do have a problem. The load needs 100mA, but the transistor can only conduct 91mA. The transistor conducts the 91mA and no more by adjusting the voltage between collector and emitter.

One thing I haven't mentioned is if you put a logic 0 on the transistor. Well you would have 0v on the base, which means Vbe = 0, which means no current can flow - it is turned off :smiley:

So, In summary:
In your first scenario, you have limited the voltage of the load to 4.3v because of the 0.7v Vbe requirement. This is not what you want as all that will happen is the transistor will likely get hot as it dissipates more power than the load (what a waste). Say if your load was 100mA, the transistor would drop 12-4.3 = 7.7V, and thus dissipate 7.7*100mA = 0.77W. This is a lot. If you tried for 2A, you would dissipate 15.4W which would almost certainly blow it up.

In the second scenario, you now have proper control over the load. The load can dissipate almost 12v depending on the transistors resistance. You have an upper limit on current flow which is controlled by your base resistor which helps protect both load and transistor.

thanks for the info guys, i'll re-read it a few more times but something just clicked in my head which i never really fully understood..

but now i do - cheers.

Tom very nice explanation, except for one little thing - what you call "saturated" isn't!! For BJTs saturation is when the base current is higher than needed and Vce drops to a low value (Vsat in fact). I think you confuse with FETs where "saturation" means something else.

MarkT:
Tom very nice explanation, except for one little thing - what you call "saturated" isn't!! For BJTs saturation is when the base current is higher than needed and Vce drops to a low value (Vsat in fact). I think you confuse with FETs where "saturation" means something else.

Yeah, must of got mixed up. Never mind, I have edited that out :). Thanks.

Thanks for that explanation Tom, I've done a 'cut and paste' and will refer to it when I'm back on another project later in the year!