Finding hFE of a transistor without a meter that does it?

[Chris Goodson]

I understand that hFE=IC/IB, I mainly just don't know what values of resistors I should use to get the best results.  Could someone recommend me a circuit?

[brett]

Hi.  RG Keen has an excelent article; "Picking transistors for FF Clones" over at geofex.com

If you're dealing with modern silicon transistors you can ignore leakage.  Just multiply the reading by 100 and that's your Hfe.

cheers

[gez]

I understand that hFE=IC/IB, I mainly just don't know what values of resistors I should use to get the best results.  Could someone recommend me a circuit?

What type of circuit?  Do you mean a common-emitter type thing?  Divider bias makes for a more consistent circuit and you don't have to worry about individual gain of each transistor (if it's set up right that is).

Tell us what you want to build and I'm sure you'll be given loads of examples.

[Chris Goodson]

Common-emitter would be fine.  I just want a way of knowing what the gain is of different transistors so I can play around with matching them and such.

[gez]

You can go by typical gain figures quoted in data sheets and choose resistor values to suit these figures.  Then check where everything biases up with a meter and you can calculate the individual gain of the device (assuming you’re using a base resistor method for the bias) and make adjustments accordingly.

If you’re designing stuff and don’t want all the hassle of measuring the gain of each trannie, then a simpler approach is to use a trim pot somewhere in the design and adjust the bias using a volt meter.  This will account for variations in gain.  

However, divider bias is the simplest way to go.  You block out a proportion of the supply voltage and allocate it to the emitter resistor.  Different designers choose different amounts, but 1V is as good a figure as any.  Next choose your collector/emitter current (there’s usually little difference).  Again, this varies but .5mA is pretty common.  If you have 1V across the emitter resistor and .5mA flowing through it, using Ohm’s law you need a resistor value of 1/.0005 =  2K.

How do you guarantee that you get 1V across the emitter resistor?  Hold the base steady at 1.7V with a divider.  0.7V is accounted for by the forward drop across the base/emitter junction leaving you with 1V across the emitter resistor.  Negative feedback prevents the emitter from rising any higher; if current increases (due to thermal conditions etc)  more voltage develops across the emitter resistor which means less voltage across the emitter/base junction, resulting in less base current which means less collector current and less voltage across the emitter resistor - the emitter voltage is kept stable and you don’t have to worry about the individual gain of each trannie.

To choose divider values check the trannie’s data sheet for typical gains.  Whatever the lowest gain quoted is, lets say 400, divide the collector current by it - in this case .0005/400 = .00125mA.  This is the maximum base current that the devise could draw, so to avoid loading, the divider needs to have at least 20 times (some designers use 10) this current flowing through it i.e. 0 .025mA.  If you have a 9V supply then the max resistance that the divider could have is 9/.000025 = 360 000 Ohms.  We want 1.7V across the lower resistor so we need a resistor of value (360 000/9 ) X 1.7 =  68k.  This means that the other resistor would be 360k - 68k =  292k (nearest value 300k).

Now you need a collector resistor that will bias the collector at around half the supply, i.e. 4.5V.  The collector current is more or less equal to the the emitter current (.5mA) so using ohms law the resistor value needed is 4.5/.0005 = 9k (nearest value 9k1).

This will give you a gain of about 4.5.  For higher gain decouple/partially decouple the emitter resistor with a cap.