Current Mirrors

[davenn]

Hi guys

OK here's something I cannot get my head around
This circuit is being discussed on another forum ( NOT MY THREAD) and the explanations to my questions about it

The guy is trying to figure out why its not working properly ( it drives a set of LED's via resistors( not shown above))

I commented, maybe unwisely....

there's another obvious fault in the circuit

the left hand transistor has it's base and collector legs shorted out
.... both legs going to the bottom of R12

which was responded with ...

That transistor is connected fine. It's a diode connected device set to control a current mirror.

I then responded ...

how do you figure that ?

so the way its wired the transistor would ALWAYS be on in full saturation so what's the point of having it there ?

because the junction of R12, C6 is going directly to the other transistors
that transistor is being bypassed anyway
but because the transistor is fully on, the C6/R12 junction is effectively always grounded

just wanna know

a response of ....

R11 and R12 set the collector current of the first transistor, right at the defined edge of saturation.
The Vbe generated is then paralleled with all others so they have the same collector currents.
It is a ground referenced current mirror with multiple parallel output sinks.

and ...

The C6/R12 junction is not grounded at DC. It is ground from the point of view of noise which is what you want. It is just the input device of a simple current mirror.

so please help me understand this current mirror thingy and how that lefthand transistor is operating in that configuration .... I just don't get it

thanks lots

Dave

 

[(*steve*)]

These transistors are all closely matched. Therefore, at a given Vbe, they will have the same Ic.

The left transistor has an Ic that is effectively set by the resistors. In this case Vce will be the same as Vbe. Sure it *looks* like a diode, but it's acting like a transistor. If you were able to measure the current you'd find most is flowing through the collector.

The capacitor reduces noise and this Veb is applied to all the other transistors setting their Ic within a fairly close tolerance.

edit: to understand this you need to realise that transistors are actually controlled by the voltage on the base, not by the current. We normally say it is controlled by the base current because it's a lot more efficient and we can vary the current over several orders of magnitude to give the very tiny changes in base voltage which actually do the trick.

 

[Arouse1973]

The diode connected transistor is often used when better temperature response is needed and the voltage drop doesn't rise as much versus current.

@Steve
At last I have found someone that agrees with me, transistors being voltage controlled devices.

Adam

 

[(*steve*)]

@Steve
At last I have found someone that agrees with me, transistors being voltage controlled devices.

Remember the first rule of fight club.

The first rule of voltage control of transistors is "Never mention that transistors are voltage controlled".

There are 2 reasons:

1) it encourages people to attach a voltage source to the base

2) it isn't usually useful for doing practical calculations

Maybe there are other configurations where this voltage control gives an important insight into understanding the circuit operation, but the only one I can think of is the current mirror.

 

[davenn]

Thanks for your responses to my queries Steve
I have been familiar with the term current mirror for many years but have never had need to work with or understand their uses

Going to have to do some reading to catch up on some theory
Will ask for clarification on anything that doesn't sink in

cheers
Dave

 

[BlackMelon]

I have a question about the current mirror......

When the voltage two voltage source (48V and 12V in the picture) are not equal... Why do the current flow through R1 and R2 are not equal too?

Thank you

PS: I have measured the current in Proteus8 and I was very curious about this.... (sometimes proteus is unreliable so I don't want to trust the program much)

 

[KrisBlueNZ]

The currents are not equal because in that circuit, the current flow through R2 and Q1 is limited by R2 and the 12V supply it is connected to.

When a current mirror is operating normally, the current in the output path (Q1's collector) is limited by Q1. Q1 conducts (from collector to emitter) an amount of current that's (roughly) equal to the current coming in from R1.

If Q1 is driving a simple resistive load (i.e. R2) (which is not the way current mirrors are normally used), the current flowing into Q1's collector causes a proportional voltage drop across R2.

But the amount of current that Q1 can cause to flow through R2 is limited by R2 and the supply voltage connected to it.

Let's plug in some approximate numbers into your circuit. The input current will be:
I = V / R
= (48 - 0.7) / 1000
= 47.3 mA.

Already, that's quite a high current for a current mirror. But I'll continue anyway.

Now, to cause 47.3 mA to flow in R2 requires 47.3V across R2, because of Ohm's Law again. R2 is a 1000 ohm resistor; the only way to make 47.3 mA flow through it is to apply 47.3V across it.

Q1 can't do that, because it can only apply a maximum of about 11.7V across the resistor. That number comes from the 12V supply voltage minus the collector-emitter saturation voltage (V_CEsat) of the transistor, which is typically around 0.2~0.3V.

So Q1 will saturate - it will pull its collector nearly all the way down to 0V and its collector current will be a lot lower than its base current multiplied by its current gain - and the current mirror's output current will be limited to about 11.7 mA.

A simple way to fix this would be to reduce R2 to, say, 200 ohms, or short it out completely. Q1 will limit its collector current according to the current mirror's input current so it won't be damaged if you operate it into a short circuit.

It will get pretty hot in this case because you're using quite a high input current; dissipation in Q1 will be about 0.5W if R2 is replaced with a short circuit.

 

[BlackMelon]

Thanks a lot KrisBlueNZ, You have explained very very clearly!!

I'm still curious that how do you know that the Q1 is in saturation mode with 12V source? (Its collector and base aren't shorted like Q2 which is obviously in saturation for sure) Will the Base current of Q1 be enough to make it saturation?

In the circuit, if I change the 12V source to much above required voltage (47.3V) that you've suggested (Let's have 100V).... The current flowing through R2 will always equal to the current flowing through the R1 and the voltage across R2 will be held at 47.3V, right?

 

[LvW]

Quote Arouse1973: "At last I have found someone that agrees with me, transistors being voltage controlled devices. "
You have found another one: That´s me!

Quote Steve: The first rule of voltage control of transistors is "Never mention that transistors are voltage controlled".
There are 2 reasons:
1) it encourages people to attach a voltage source to the base
2) it isn't usually useful for doing practical calculations

1.) In more than 20 years experience in teaching electronics to newcomers I didn´t meet one single person who was tempted to connect a real voltage source directly to the base.
I think, each beginner will have - at least a short - look into a book or a similar text which in 99.9% of all cases presents a proper biasing scheme for the BJT.
2.) Don´t we use Vbe=(0.65...07)V for calculating the base voltage divider and/or the emitter resitor Re?

 

[KrisBlueNZ]

Thanks a lot KrisBlueNZ, You have explained very very clearly!!

Oh, thanks. I thought I was rambling a bit.

I'm still curious that how do you know that the Q1 is in saturation mode with 12V source? (Its collector and base aren't shorted like Q2 which is obviously in saturation for sure) Will the Base current of Q1 be enough to make it saturation?

Actually, Q2 is NOT in saturation. That's actually impossible in a current mirror! You need to understand what saturation actually IS. The following description is for NPN transistors; just reverse the polarities for PNPs. Some of the details aren't exactly correct; this is just a simplified overview.

The Wikipedia article (https://en.wikipedia.org/wiki/Bipolar_junction_transistor#Regions_of_operation) defines saturation as when the collector voltage is closer to the emitter voltage than the base is, or when V_CE < V_BE. It can also be defined as I_C / I_B << H_FE; in other words, the base current multiplied by the transistor's current gain is much more than the actual collector current.

Saturation is a property not just of the transistor, but of the circuit it's used in. Take a transistor, ground its emitter, and connect its collector through a resistor to a positive supply rail. Inject an adjustable current into the base.

As you increase the base current, the collector current will increase (roughly) proportionally, according to the current gain. The transistor is operating in its linear region.

As the collector current increases, it will cause an increasing voltage drop across the collector resistor, leaving less voltage across the transistor; in other words, V_CE will decrease. At a certain collector current, determined by the power supply voltage and the collector resistor, the transistor will start to "run out of voltage" in its collector circuit, and start to enter the saturation region.

If you keep increasing the base voltage, the collector voltage will keep dropping, but more and more gradually, until it reaches a point where even a large increase in base current won't cause the collector voltage to drop noticeably. In this state, the transistor is saturated, because you can't get any significant increase in collector current from it by just increasing the base current.

If you reduced the resistance of the collector resistor, or increased the power supply voltage, this could take the transistor out of saturation, and back into its linear region, if it allows the collector current to increase enough that it is again controlled in proportion to the base current; that it is equal to I_B x H_FE. That's what I mean about saturation being dependent on the circuit as well as the transistor.

For a typical transistor at lowish current (milliamps), V_CEsat is around 0.2~0.3V. That is less than V_BE. I don't understand the physics, but it could be that the base-collector junction (which behaves similarly to the base-emitter junction) starts to become forward-biased when V_CE goes that low. So some of the base current flows that way instead of directly to the emitter. Perhaps someone else can explain the details.

For the benefit of LvW and Adam: I think that the current mirror is one case where it can be useful to consider a transistor as voltage-driven. We've had this discussion already; I wasn't satisfied with the (lack of) answers to my questions, and I don't consider it a useful discussion. Please feel free to rewrite my explanation of saturation on the basis of a transistor being a voltage-driven device if you think it would add any clarity. (I don't think it would.)

In the circuit, if I change the 12V source to much above required voltage (47.3V) that you've suggested (Let's have 100V).... The current flowing through R2 will always equal to the current flowing through the R1 and the voltage across R2 will be held at 47.3V, right?

Yes, assuming the transistors are perfectly matched.

 

[LvW]

For the benefit of LvW and Adam: I think that the current mirror is one case where it can be useful to consider a transistor as voltage-driven. We've had this discussion already; I wasn't satisfied with the (lack of) answers to my questions, and I don't consider it a useful discussion. Please feel free to rewrite my explanation of saturation on the basis of a transistor being a voltage-driven device if you think it would add any clarity. (I don't think it would.)

Kris, I remember we had such a discussion some time ago. Hence, I do not intend to start again. However, I must admit that I cannot remember which of your questions remained unanswered.
Nevertheless, I cannot resist to present a short task description of the well-known base voltage divider. Furthermore, this descriptions leads directly to the corresponding formulas for computing the resistors (and says everything about the control mechanism):
Based on the fundamental principle of a loaded resistive voltage divider, both resistors are calculated to generate a voltage of ......Volts at the base node. For a correct calculation it is, of course, necessary to take the current into the load (into the base node) into consideration.

 

[BlackMelon]

Kris, In the case of current mirror, can I simply determine that the Q2 will never be in saturation by looking at its voltage on 3 pins only?
I mean the base voltage will never be higher than the collector so it can't satisfy the saturation condition (Vb>Ve>Vc)

If not, will I have to draw a load-line of the circuit to find the relation between load resistance, voltage source, and collector current and map the graph with BJT characteristic graph to find its mode of operation?

 

[KrisBlueNZ]

Kris, In the case of current mirror, can I simply determine that the Q2 will never be in saturation by looking at its voltage on 3 pins only?
I mean the base voltage will never be higher than the collector so it can't satisfy the saturation condition (Vb>Ve>Vc)

Saturation is (V_B > V_C > V_E). And that's why the input transistor can never be in saturation - because V_B = V_C.

There is only one current path into that transistor - the input of the current mirror, which comes from the driving circuit - R1 in that case. Some of this current goes into the base; most of it goes into the collector. No matter how much current flows into that input, the relationship between I_B and I_C is equal to H_FE. (H_FE varies significantly depending on I_C, but that's a different matter.)

Basically, the input transistor cannot saturate because the base and collector current paths aren't separate. Increasing the input current will increase I_B but the transistor can always draw more current through its collector; the collector current isn't limited by the collector circuit the way it can be with the output transistor (because of the limited current available from a limited supply voltage and a collector resistor).

If not, will I have to draw a load-line of the circuit to find the relation between load resistance, voltage source, and collector current and map the graph with BJT characteristic graph to find its mode of operation?

No, you don't need to do that. It can't saturate, in principle. Also the fact that the collector-base junction can never become forward-biased is a clue.