![]() Here you might put a resistor between the base and emitter of the transistor to shunt leakage current as shown below (though I used a 3.3V zener in this case). You need to be able to ensure you design current flows in a DC circuit such as this to be immune to things such as component leakage currents. In this case your very low currents don't help. If you want to be able to design at even a very simplified level, you need to understand the characteristics of your components. So let's work out the input voltage based on a leakage current of the Zener of 20 uA at 1 V, and V(BE) for the transistor is 0.6 V.Ģ0 UA through the 10k resistor results in a voltage drop of 0.2 V, so the input voltage would be 0.2 + 1 + 0.6 = 1.8 V !!! ![]() Notice in the graph that they suggest that the transistor may be on between 6-10 uA for 1 mA collector current suggesting an Hfe of over 150 for the device they used, and this will vary over a range for the transistors you might purchase. Notice that the base current required to turn on the transistor could be less than the leakage current of the Zener!! ![]() Have a look at the datasheet (Table 3) to see the B/C transfer characteristics. So I'll ignore RL for the moment.įor the Transistor to be on the collector current would be around (ignoring VCE(sat)) 0.8 mA and with an Hfe of 50 that implies a base current of around 16 uA maximum. It's hard to know what the collector voltage is because you show a DC coupled restive divider of 4.3k and Rl (no value for RL). If the transistor were a 2N2222 then we might expect the minimum Hfe would be around 50. If the Zener were a BZX79-B/C2V7 the leakage current could be as high as 20 uA V. Let's deal with some practical components: The Zener characteristics look like this: Your calculations are incorrect for anything less than ideal components. 500ua with a 4.7v zener, Ic unchanged, just as we expected.Īgain, the point I have been trying to emphasize is that beta is a useful concept for linear applications and has no meaning in a switching application, as shown here. If anyone is interested, I can simulate the case with a 2.7v zener, :)Įdit2: due to popular demand, here is the same circuit, using a 2.7v zener. The blue line in the plot is Ib and it reaches 500ua at 10v input, with Ic at 800ua -> a beta of 1.6. the use of a 2.7v zener would have increased the Ib by 200ua in this case, with minimum changes in Ic. Here is a simulation I put together quickly - it mimics the circuit provided by the OP, with the exception that I used a 4.7v zener, not a 2.7v zener. I pointed out that beta is really a concept for linear applications (where the associated Vce is much higher than in a switching application here). Jack had concluded that the Ib would be 16ua max, based on an estimated beta of 50. I'm amending this in response to the answer provided by Jack Creasey below. A prudent design practice is to first assume that it is insignificant, go through your calculations, obtain Ib and then confirm that it is indeed significant. However, there can be cases where the voltage drop over R1 is significant. If you use typical values, you will find that that voltage drop is quite small vs. That current will cause a voltage drop over R1. You should calculate the Ic first, and then apply a Ib based on the assumed beta of the transistor. How will calculation changes if drop across R1 also need incorporate? and for digital applications, people tend to use "fully turn on" instead.Īs Vbe is not a constant, nor the reverse voltage of the zener, the figure you obtained is just a rough estimate, and should be adjusted further for specific applications if called for. for most analog applications, people use "turn on" to mean "starting to turn on". The calculation is generally correct, but dependent on your definition of "turn on".
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