B1135 transistor amplifier

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• 11C90 650 MHz Prescalers (Rev. A)
• 2SB1135 Transistor TO-220F B1135
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• 2SB1135 B1135 Transistor 2SB891 B891 2SB1010 B1010
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11C90 650 MHz Prescalers (Rev. A)

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That current will cause a voltage drop over R1. If you use typical values, you will find that that voltage drop is quite small vs. However, there can be cases where the voltage drop over R1 is significant. 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. I'm amending this in response to the answer provided by Jack Creasey below. Jack had concluded that the Ib would be 16ua max, based on an estimated beta of 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.

Here is a simulation I put together quickly - it mimics the circuit provided by the OP, with the exception that I used a 4. Your calculations are incorrect for anything less than ideal components. The Zener characteristics look like this:. If the transistor were a 2N then we might expect the minimum Hfe would be around It's hard to know what the collector voltage is because you show a DC coupled restive divider of 4.

So I'll ignore RL for the moment. For the Transistor to be on the collector current would be around ignoring VCE sat 0. Notice that the base current required to turn on the transistor could be less than the leakage current of the Zener!! Notice in the graph that they suggest that the transistor may be on between uA for 1 mA collector current suggesting an Hfe of over for the device they used, and this will vary over a range for the transistors you might purchase.

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. Far from what you might expect. If you want to be able to design at even a very simplified level, you need to understand the characteristics of your components.

In this case your very low currents don't help. 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. 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. So where is the transistor on?

I eventually learned to use "transconductance" to design amplifiers, and I use emitter resistors to set the currents. After lotta painful design failures as a student. Sign up to join this community. The best answers are voted up and rise to the top.

Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. What is the minimum input voltage to turn on the Transistor? Ask Question. Asked 4 years, 9 months ago. Active 4 years, 8 months ago. Viewed 10k times. Questions Can someone verify the calculation? How will calculation changes if drop across R1 also need incorporate?

At that point current starts to flow, so ideal world at 3. Now think about what the voltage drop will be in both cases. If you doubt that, assemble the parts and test it. Show 3 more comments. Active Oldest Votes. Ib tops out at about ua, vs. Add a comment. Jack Creasey Jack Creasey Look at the simulation and you can see that the switch slowly turns on as Ib increases, just as you'd expect.

With an analog drive such as this you expect analog results. You could change the input to a square wave to overdrive the base but you still need to be out of the leakage current range. At minimum you need all the zener leakage current diverted. So if Vbe is 0. The question was however, "What is the minimum voltage to turn on the transistor" I think I answered that, though there is clearly no definitive number that could be calculated without having the detailed characteristics of both the zener and transistor available.

With 58 milliVolts change in Vbe for each decade, we only need milliVolts change in Vbe. Lets take a simple example: The pn diode because the B-E junction behaves like a diode. Some people consider the diode to be "on" if the forward bias is 0. But this is a a raugh estimate and b nothing else than a definition.

There will be a current even for a voltage of 0. And the same applies, in principle, to ther BJT. We normally assume 0. So - is there a "physical" theshold? Sign up or log in Sign up using Google. Sign up using Facebook.

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2SB1135 Transistor TO-220F B1135

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2.4G Optical [ KOOLmouse Wireless ] Valentine [ ] Mouse Computer Mice the most fashionable

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2SB1135 B1135 Transistor 2SB891 B891 2SB1010 B1010

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