Mirror Currents

[eestudent123]

Hello,

I have a question regarding the circuit attached below. Assuming that bettas are infinite and that Is2=N*Is1 since the emitter area of Q2 is N times larger than the emitter area of Q1.

My intuition tells me that Iout=N*Iref, but I dont know how to put it in equations....
There is no xurrent flowing along Vdd, is thia correct?
Where should I start from?

Thank you,

 

[cowboybob]

There is no xurrent flowing along Vdd, is thia correct?

What leads you to that conclusion?

Just asking.

 

[Roff]

Post deleted.

 

[bountyhunter]

One way to look at it: since the bases are tied together, the diode on the left forces the voltage onto the P-N junction on the right.

Assuming the devices are the same (same doping, same geometry, same emitter size) the currents must be equal through those two diodes.

If you increase or decrease the emitter area (size) of the right transistor, the current DENSITY through that emitter remains the same which means the current has to increase or decrease directly proportional to the emitter area. That circuit is the basic building block of all linear amplifiers and integrated circuits. We used it as far back as the early 70's.

 

[Ian Rogers]

One way to look at it: since the bases are tied together, the diode on the left forces the voltage onto the P-N junction on the right.

I don't think they are diodes... they look like current path indication to me.

I take it xurrent.. should be current............. The current though Vdd will be the sum of the two paths... The idea of a current mirror is to use matching transistors to MIRROR the current. If you use dissimilar transistors it would be difficult to calculate the current through Q2... Using mathing transistors the current through Vdd will be double the current through Q1.

 

[bountyhunter]

The term current mirror does not mean the two devices MUST be the same, it means that what happens in one mirrors over to the other. In reality, we often used emitter sizes scaled up to ratio a higher current in the transistor leg. That way you don't have to waste a lot of current in the diode leg to get the current you want in the transistor side. If the transistors are identical, it's a 1:1 mirror. If not, it's a different ratio. We would normally put that info on the schematics whatever it was like 4:1 or whatever in the bias chains.

That is shown above in the schematic where the diode side is labeled "x1" and the other side is labeled "xN", showing that the right side can be a multiple of the left.

 

[Ian Rogers]

Duly noted.... It must be for theory only.... I must admit, I only use current mirrors in 4.. 20mA current loops

It would be difficult to ascertain current flowing into Vdd then...

 

[Roff]

Duly noted.... It must be for theory only.... I must admit, I only use current mirrors in 4.. 20mA current loops

It would be difficult to ascertain current flowing into Vdd then...

It is (N+1)*Iref.

 

[RCinFLA]

The xN means there are N identical devices in parallel.

If all the devices, including the one fed with a current source, are identical devices, and have exactly the same Vbe what do you think their collector currents will be in comparison to one another?

An important assumption is the temperature of all the devices is equal since the Vbe has a negative temperature coefficient.

 

[bountyhunter]

An important assumption is the temperature of all the devices is equal since the Vbe has a negative temperature coefficient.

CORRECT. In an IC, it's not an issue since the transistors are all on the same die. Using discretes, I sometimes would glue them together when they had to "track" thermally. Use plastic transistors and glue the flat faces together.

 

[lebevti]

i don't think they are meant to be used as discretes....that's strictly an IC circuit

 

[bountyhunter]

i don't think they are meant to be used as discretes....that's strictly an IC circuit

Glad I never knew that..... I used them by the ton in my designs. In discrete form you can match VBEs and/or use a little degeneration resistance to force them to balance. I only glue them together to force thermal tracking when they set bias currents and one dissipates more power than the other.

 

[Ian Rogers]

Glad I never knew that..... I used them by the ton in my designs. In discrete form you can match VBEs and/or use a little degeneration resistance to force them to balance. I only glue them together to force thermal tracking when they set bias currents and one dissipates more power than the other.

That was my point... about calculating the current.... Its ok for chips that you test in front of you... but you can't calculate the hfe of a transistor in advance... That's why class A amplifiers are so inefficient.

 

[Roff]

That was my point... about calculating the current.... Its ok for chips that you test in front of you... but you can't calculate the hfe of a transistor in advance... That's why class A amplifiers are so inefficient.

That is a non sequitur.

 

[Ian Rogers]

Why? The OP wants to know the equation that you posted... All I was saying is.. Transistors have, on their datasheets, a value from ---> to... You need to know the exact figure to calculate... Or am I reading this wrong... (won't be the first time)

 

[Roff]

Why? The OP wants to know the equation that you posted... All I was saying is.. Transistors have, on their datasheets, a value from ---> to... You need to know the exact figure to calculate... Or am I reading this wrong... (won't be the first time)

What I meant was, I didn't see what the efficiency of class A amplifiers has to do with hfe. Thinking on it some more... maybe you are referring to designing the base bias network to make it insensitive to variations in beta?

 

[bountyhunter]

Class A amps are inefficient because they run so much bias current through the device with voltage across it. That translates to power dissipation with no signal out = low efficiency. It is not determined by beta of transistor.

 

[Roff]

Class A amps are inefficient because they run so much bias current through the device with voltage across it. That translates to power dissipation with no signal out = low efficiency. It is not determined by beta of transistor.

Yeah, that was my reasoning too, but I thought perhaps Ian was, as I said, thinking about the bias circuitry on a simple common emitter stage, which would waste less current if the value of beta were known.

 

[Ian Rogers]

I'm sorry for the confusion again.... I always look on things from a design perspective... The reason I brought the class a amp into the debate was to specify that building a circuit around a single transistor.. You cant replace the transistor with another as the base bias is slightly different every time... This must be the case in this issue...