Timeline for Choosing Collector Current for Transistor Amplifer
Current License: CC BY-SA 4.0
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| Jun 11, 2022 at 5:09 | answer | added | Simon Fitch | timeline score: 1 | |
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| Oct 22, 2019 at 10:38 | comment | added | D.A.S. | Edmund That is exactly right set Ib = Ic/hFE is the answer but has a tolerance often of 50% due to hFE. But but regulating Ve by regulating Vb=Vbe+Ve(=IeRe) The value of 1mA is arbitrary and dependson on what Rc impedance you want to drive and what Vcc is avail. like 10k or 50 ohms demands a much different Ic | |
| Oct 22, 2019 at 3:23 | comment | added | jonk | @EdmundBlackadder If you gave me five different amplifier specifications, I might start by selecting the quiescent collector current in only one of them; while in another case selecting the voltage rail first, instead, and letting the quiescent collector current fall out of other considerations. Different goals are prioritized differently in different cases. So sometimes? The quiescent current falls out of other more important decisions, instead, and I don't even get anywhere close to deciding that until later on. | |
| Oct 22, 2019 at 3:14 | comment | added | jonk | @EdmundBlackadder Different BJTs are designed to cover different ranges of collector current. The datasheet will show the range the manufacturer feels comfortable with, when you read it. And the datasheet doesn't tell you what quiescent collector current to use. Your specifications lead you, directly or indirectly, to that choice. Your specifications won't be found on the datasheet. Datasheets serve a different purpose. And yes, the load you are driving will help inform you of what you might select as a quiescent current. But it doesn't determine it, solely. | |
| Oct 22, 2019 at 2:35 | comment | added | Edmund Blackadder | @TonyStewartSunnyskyguyEE75 But how do I ensure that Ic is 1mA. From my understand and correct me if I am wrong, I have set up a potential divider (following common emitter config.) to 1.) ensure that the transistor is always on and 2.) to set the Ib. And my thinking behind this is that if I set Ib to Ic/hFE, then I can ensure that Ic is what I want it to be. Help me understand how to set the value for Ic. | |
| Oct 22, 2019 at 2:35 | comment | added | Edmund Blackadder | @TonyStewartSunnyskyguyEE75 Also in this first part we Ic*Rc = V+/2 so that we get max symmetrical swing. Now if we assume Ic = 1mA and V+ = 18V, then Rc = 9k ohms. This ensurse that there is a voltage drop of 9V across Rc assuming that Ic = 1mA. | |
| Oct 22, 2019 at 2:34 | comment | added | Edmund Blackadder | @TonyStewartSunnyskyguyEE75 I am mainly following your steps for my design, and as you said let Ic*Rc = V+/2 which I use to find Rc but this requires knowing Ic. How do I choose this value of Ic. | |
| Oct 22, 2019 at 2:26 | comment | added | Edmund Blackadder | @jonk What do you mean by a range of collector currents over which it is useful. Is this from the data sheet. Also the specs given are a gain of 10 and supply voltage of 18V. But I'm not really looking for a solution, I am just trying to understand how to use the data sheet to choose a value for Icq. | |
| Oct 22, 2019 at 2:22 | comment | added | Edmund Blackadder | @Andyaka I also, wouldn't firstly choose Icq. But these are steps we have to follow, so I just trying to figure out the justification behind this assumption. | |
| Oct 17, 2019 at 3:47 | answer | added | analogsystemsrf | timeline score: 0 | |
| Oct 16, 2019 at 20:01 | comment | added | jonk | If you are stuck using a specific BJT (because that's already been decided for you), then you can find the range of collector currents over which it is useful (typically, this is two to three orders of magnitude range -- so you have many options, still.) It's possible you have a specific battery system (button battery or 9 V battery) and this may, taking into account the entire circuit functions, also suggest some limiting boundaries. (A teacher could demand the value.) But if you were to write down a list of specs, I think others may help you re-orient yourself with respect to this question. | |
| Oct 16, 2019 at 12:03 | comment | added | D.A.S. | let Ic*Rc = V+/2 then let Ib = Ic/ hFE (100 ish). Vb will be <600 mV with a current limiting Bias from Vref with R divider Ohm’Law then AC couple In faster than input swing V/ T(=RC ) slew rate. Add Re and choose gain = Rc/Re | |
| Oct 16, 2019 at 11:49 | comment | added | Bimpelrekkie | That 1mA is just an example / starting point. The most optimum biasing current depends on the transistor and what you want to do with it. For example, high frequency transistors often have a biasing current value that will result in the highest bandwidth. Another example: for an Audio stage driving a speaker of 8 ohms, a 1 mA biasing current is extremely low considering that more than 1 A can flow in normal operation. | |
| Oct 16, 2019 at 11:46 | comment | added | Andy aka | One thing I would definitely not do first when designing a transistor amplifier is: Choose a quiescent current of 1mA. That's my cunning plan and don't assume I'm Baldrick LOL. | |
| Oct 16, 2019 at 11:44 | history | asked | Edmund Blackadder | CC BY-SA 4.0 |