You should NEVER have to calculate the gain of a transistor. You must always calculate a range of gains for the final circuit based on a worst case analysis of the passive components that determine the gain,

How do I calculate a range of gains? What is a worst case analysis? Here is a schematic


The battery is 8 volts. Resistor is 160 ohms. Motor is a load of 400 ohms. Resistors have a tolerance of 5%.

 

How do I calculate gain? Give me an example. Should I take temperature into account?

Here's a fragment of a tracking regulator circuit that just requires a minimum beta for the (discrete) darlington:

Only one opamp was required. I was this as a teaching exercise about how to use simple building blocks to implement a more complex circuit.

 

How do I calculate a range of gains? What is a worst case analysis? Here is a schematic
View attachment 238808

The battery is 8 volts. Resistor is 160 ohms. Motor is a load of 400 ohms. Resistors have a tolerance of 5%.

Transistor is tip110.

 

The battery is 8 volts. Resistor is 160 ohms. Motor is a load of 400 ohms. Resistors have a tolerance of 5%.

In this circuit, the transistor and resistor aren't needed.

 

So how do I design a transistor to be largely independent of temperature?

It depends on the device and application.
FETs have a sweet spot where the I-V curves at different temperatures cross.

 

View attachment 238811
In this circuit, the transistor and resistor aren't needed.

I still want to know how to calculate a range of gains and what a worst case analysis is.

 

I still want to know how to calculate a range of gains and what a worst case analysis is.

This circuit has a nominal gain of 10 and is independent of beta variations in the transistor:

 

I looked at the data sheet for tip110. According to 1 graph. The gain depends on temperature. It was a graph of collector current vs DC current gain. I saw three lines, 1 for 25 Celsius, 1 for -55 Celsius, 1 for 125 Celsius. What if the temperature is 10 Celsius, how do I calculate the gain?

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

The three graphs are giving you an idea based on AMBIENT. 25C = 77F. You derate, accordingly.



Note your hFE AND the test condition of 25C. How much current are you trying to run through it? Thermal Junction Temp is IMPORTANT.

Ib = Ic / hFE
Ib = 1 / 1000
Ib = 0.001 (aka 1mA)

 

So how should I calculate gain?

You don't.
You look at the datasheet and determine what the minimum and maximum gains would be at the temperatures and currents that you wish it to operate.
Then you analyse your circuit to make sure that it will function within its specifications at the extremes of gain you determined from the datasheet.
The minimum temperature is the minimum ambient temperature, but the maximum temperature will depend on both ambient temperature and heat dissipation.
The gain may vary from worst transistor at it coldest and maximum current to a good warm transistor at low current by a factor of 5 or more. Negative feedback keeps your circuit operating within its specification.

 

Your simple circuit has no gain because it is just a switch that is turned on. The TIP110 is saturated and the datasheet says for it to be saturated then its base current should be 1/250th of its collector current.

400mA/250= 1.6mA.
The max base voltage is about 1.5V when the collector current is 400mA and the base current is 1.6mA. Then Ohm's Law calculates the base resistor to be (8V - 1.5V)/1,6mA= 40.6k ohms. Your 160 ohms is 254 times too low.

@BobaMosfet: hFE is not used when a transistor is a saturated switch. The datasheet shows when saturated Ic= 2A and Ib= 8mA. 2000/8= 250.

 

Your simple circuit has no gain because it is just a switch that is turned on. The TIP110 is saturated and the datasheet says for it to be saturated then its base current should be 1/250th of its collector current.

400mA/250= 1.6mA.
The max base voltage is about 1.5V when the collector current is 400mA and the base current is 1.6mA. Then Ohm's Law calculates the base resistor to be (8V - 1.5V)/1,6mA= 40.6k ohms. Your 160 ohms is 254 times too low.

@BobaMosfet: hFE is not used when a transistor is a saturated switch. The datasheet shows when saturated Ic= 2A and Ib= 8mA. 2000/8= 250.

@Audioguru again Yes, I know that- I was only answering his specific request which was _how_ it's calculated. He needs to do a lot more learning about BJTs. That is why I pointed him at a basic electronics book which will explain all of it.

 

If you had, for example, a common emitter amplifier with voltage divider bias and a bypassed emitter resistor, then the voltage gain of the circuit would depend entirely on resistors, both actual and intrinsic, and NOTHING else. Period, done, end of story. For each resistor involved in the gain calculations you would use the minimum [and maximum] values allowed by the tolerance specification and you would compute a set of gains for the various combinations of values. Then you would look for the best result and the worst result. That would be a worst case analysis.