Transistor output resistance and thermal voltage

In summary: However for bjt's the collector current is very small and it's very important to have accurate r_0\;.
  • #1
tindel
12
0
A couple questions:

I'm in the process of making some small amplifiers and using bjt's in the small signal realm. I have used bjt's as switches for quite a while, so I am quite familiar with their basic operation. I was reviewing small signal analysis trying to refresh my memory about how to do the analysis when I came across a couple variables that I'm not sure how to incorporate them into my analysis in a practical sense.

First, thermal voltage of a pn (diode) junction - VT - which my book describes as a 'constant', which is a function of temperature. VT is defined as

VT= k*T / q

where
k is Boltzmann's constant = 1.38 x 10^-23 joules/kelvin,
T= temperature kelvin,
q= magnitude of electronic charge = 1.6 x 10^-19 coulomb

Is this true for all pn junctions - or does this change with doping, diode types, etc?


Second - small signal output resistance, ro - my book does an awful job explaining how to arrive at this value, practically speaking. They give the following equation for the output resistance ro = (VA + VCE) / IC. I have never seen ro or VA in a datasheet. I'm not sure what VA even is. I do understand that ro is a function of collector current due to vce, so I understand why it's used, I just don't understand when I have a circuit I'm analyzing, how to come up with a value of ro!

True to form - my old electronics book tells me how to do all the calculations, but doesn't explain how to arrive at the value in a practical sense. A quick google search also turned up nothing.
 
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  • #2
The term [itex]kT/q[/itex] is not really a parameter of the diode (or transistor) itself, but is actually just the thermal energy of the carriers in the semiconductor, expressed in electron volts. The electrical properties of diodes and transistors do depend upon this quantity, but also on many other parameters of the device. So in summary, yes this quantity is the same for all semiconductors at a given temperature, but the characteristics of those different devices can still be different due to other factors like physical dimensions, doping levels and the intrinsic semiconductor material itself.
 
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  • #3
tindel said:
A couple questions:

I'm in the process of making some small amplifiers and using bjt's in the small signal realm. I have used bjt's as switches for quite a while, so I am quite familiar with their basic operation. I was reviewing small signal analysis trying to refresh my memory about how to do the analysis when I came across a couple variables that I'm not sure how to incorporate them into my analysis in a practical sense.

First, thermal voltage of a pn (diode) junction - VT - which my book describes as a 'constant', which is a function of temperature. VT is defined as

VT= k*T / q

where
k is Boltzmann's constant = 1.38 x 10^-23 joules/kelvin,
T= temperature kelvin,
q= magnitude of electronic charge = 1.6 x 10^-19 coulomb

Is this true for all pn junctions - or does this change with doping, diode types, etc?

I cannot answer about the doping, but yes [itex]V_T=\frac {kT}{q}\;[/itex] holds for all transistor and diodes.

Second - small signal output resistance, ro - my book does an awful job explaining how to arrive at this value, practically speaking. They give the following equation for the output resistance ro = (VA + VCE) / IC. I have never seen ro or VA in a datasheet. I'm not sure what VA even is. I do understand that ro is a function of collector current due to vce, so I understand why it's used, I just don't understand when I have a circuit I'm analyzing, how to come up with a value of ro!

True to form - my old electronics book tells me how to do all the calculations, but doesn't explain how to arrive at the value in a practical sense. A quick google search also turned up nothing.

A lot of transistor data sheets provide collector curves for different Ib. To find the early voltage, extend the straight portion of the curve ( the saturation region) to the left side where Vce going to negative. Keep extending all and they all will meed at Ic = 0. Read off the Vce and that's the early voltage.

[itex] r_0\;[/itex] is defined as change of collector current to change of Vce. That is the slope of the collector curve when the transistor is in saturation region ( the straight part of the curve). As you can see, the slope is different for every Ib. Therefore [itex] r_0\;[/itex] is collector current dependent. Normally [itex] r_0\;[/itex] is not that important as you have collector resistor [itex] R_L\;[/itex] which is usually much lower. The total load resistance is [itex] r_0\;[/itex] parallel with [itex] R_L\;[/itex] which essentially just [itex] R_L\;[/itex] when it is a lot lower than the output resistance. It is important when you use it as a current source where you need very high impedance at the output. In IC, most circuit use active load, then it is more important as it affect the gain of the transistor if this is the load on the collector.
 
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  • #4
tindel Second - small signal output resistance said:
o[/SUB] - my book does an awful job explaining how to arrive at this value, practically speaking. They give the following equation for the output resistance ro = (VA + VCE) / IC. I have never seen ro or VA in a datasheet. I'm not sure what VA even is. I do understand that ro is a function of collector current due to vce, so I understand why it's used, I just don't understand when I have a circuit I'm analyzing, how to come up with a value of ro!
Apart from using the "Early voltage" (as explained by yungman above) many datasheets will specify the common emitter [itex]r_o[/itex] indirectly via the parameter [itex]h_{oe}[/itex]. This parameter is in fact the reciprocal of [itex]r_o[/itex].

Remember that in an actual circuit that the (external) collector resistor will be in parallel with [itex]r_o[/itex] in the small signal model. Very often this external [itex]R_C[/itex] is much lower than [itex]r_o[/itex] and so dominates the parallel combination (meaning that you can often ignore [itex]r_o[/itex] without introducing too much error in the analysis).
 
  • #5
Thanks uart and yungman - The data you gave on both the early and thermal voltages is very useful. I understand that ro is very large, and with small output resistances, relative to ro I understand that it is not much of a concern unless you have large output resistances.
 

1. What is transistor output resistance?

Transistor output resistance is the resistance that a transistor exhibits between its collector and emitter terminals when a small voltage is applied to the base terminal. It is a measure of how much the output current from the transistor changes in response to changes in the input voltage.

2. How is transistor output resistance determined?

Transistor output resistance is determined by measuring the change in output current for a given change in input voltage. It can also be calculated using the slope of the output characteristic curve of the transistor.

3. What is thermal voltage in a transistor?

Thermal voltage in a transistor is the voltage that is generated due to the diffusion of charge carriers (electrons and holes) across the base-emitter junction. It is typically around 26 millivolts at room temperature.

4. Why is thermal voltage important in transistors?

Thermal voltage is important in transistors because it affects the performance and behavior of the transistor. It can affect the biasing of the transistor and the amount of current that flows through it, which in turn can impact the overall functionality of the circuit.

5. How does temperature affect transistor output resistance and thermal voltage?

As temperature increases, both transistor output resistance and thermal voltage increase. This is because the movement of charge carriers increases with higher temperatures, leading to a higher output resistance. Additionally, thermal voltage increases with temperature, causing a decrease in the transistor's current gain. This can affect the transistor's overall performance and lead to potential overheating issues.

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