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calculation
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John Woodgate
In message <kt8hbj+5vsb@...>, dated Tue, 30 Jul 2013, jean_claudeabeille <jean_claudeabeille@...> writes:
Is anyboby here who can explain me how to calculate Ib in this circuit : Draft2.asc. I found no formula(s) on the web while LTSpice knows. Thanks.You can't find a formula on the Web, even though there is one, because it's not a practical thing. The base current is exponentially dependent on the base voltage and the junction temperature. LTspice assumes values in the exponential equation, but the slightest change can make a huge difference to the current. Every individual transistor will give a different result. You simply don't use bipolar transistors with a fixed DC voltage between base and emitter. -- OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk Why is the stapler always empty just when you want it? John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK |
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--- In LTspice@..., "jean_claudeabeille" <jean_claudeabeille@...> wrote:
Hello, Let's assume an ideal transistor with base resistance Rb=0, current gain B=BF=constant, Early voltage VAF=infinity and the ideality factor NF=1. Ib = (Is/B)*exp(Vbe/(NF*VT) VT = k*T/q k=Boltzmann constant T = absolute temperature q = electron charge The default temperature in SPICE is 27 degree C. This gives VT=25.8641mV. If there is some Rb, e.g. 10Ohm, you can take is into account. Vbe = Vbe_extern - Ib*Rb Be aware that the current gain is a function of Ic and Vce too if more parameters are set in the SPICE model. I have uploaded an example using the basic equation. Fies > Temp > ib_transistor.asc Best regards, Helmut |
LTspice takes the "brute force" approach. It has all the network equations for the entire circuit, and it solves for every branch current, using iteration. That is, in effect, it makes a guess, then it evaluates it against the network equations, then makes another guess (not a wild guess but based on what the network equations say), evaluates again, and so on. After some number of these tries, it concludes it is very close to the "right" answer, and that's what it shows you. To be a "right" answer, the voltages and currents must be consistent with the network equations. When everything satisfies the network equations within certain tolerances, that's when it stops the iteration routine, and shows you the answer. Andy |
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It may be that the OP simply wants to determine what that current is (compute it). If that is the case, just run the simulation. When it is done, move the cursor over (almost) any component and you will be shown a graph of the current through that component. It is really easy.
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You do, however, need to be aware that there is assumed to be a "positive" current direction. With some components, such as resistors, are difficult to tell which is the "right" way. If you had to rotate the component and rotated it the wrong way, you might end up with a displayed current that is the negative of the real current. Jim Wagner Oregon Research Electronics ----- Original Message -----
From: "Andy" <Andrew.Ingraham@...> To: LTspice@... Sent: Tuesday, July 30, 2013 11:58:13 AM Subject: Re: [LTspice] calculation LTspice takes the "brute force" approach. It has all the network equations for the entire circuit, and it solves for every branch current, using iteration. That is, in effect, it makes a guess, then it evaluates it against the network equations, then makes another guess (not a wild guess but based on what the network equations say), evaluates again, and so on. After some number of these tries, it concludes it is very close to the "right" answer, and that's what it shows you. To be a "right" answer, the voltages and currents must be consistent with the network equations. When everything satisfies the network equations within certain tolerances, that's when it stops the iteration routine, and shows you the answer. Andy [Non-text portions of this message have been removed] |
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--- In LTspice@..., John Woodgate <jmw@...> wrote:
John, The equation Helmut gave is an excellent approximation for the relationship between Vbe and Ib. It is used in ever circuit with a BJT that anyone simulates; in other words, it is a fundamentally important relationship in BJT analysis. Rick |
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John Woodgate
In message <kt9dt1+fgsc@...>, dated Tue, 30 Jul 2013, sawreyrw <sawreyrw@...> writes:
The equation Helmut gave is an excellent approximation for the relationship between Vbe and Ib. It is used in ever circuit with a BJT that anyone simulates; in other words, it is a fundamentally important relationship in BJT analysis.Of course; didn't I write: 'The base current is exponentially dependent on the base voltage and the junction temperature.' As I tried to explain, you don't ever apply a DC voltage directly between base and emitter because in practice the base current (and emitter current) is unpredictable due to its extreme dependence on voltage and temperature, and which transistor you took out of the box. -- OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk Why is the stapler always empty just when you want it? John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK |
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--- In LTspice@..., John Woodgate <jmw@...> wrote:
John, Yes, but you wrote "You can't find a formula on the Web, even though there is one, because it's not a practical thing." This is absolutely wrong. Rick |
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John Woodgate
In message <kt9nnr+k9tf@...>, dated Wed, 31 Jul 2013, sawreyrw <sawreyrw@...> writes:
Yes, but you wrote "You can't find a formula on the Web, even though there is one, because it's not a practical thing." This is absolutely wrong.What is wrong with the statement? Do you find practical applications where a fixed DC voltage is applied, in the conducting direction, between base and emitter of a bipolar transistor? -- OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk Why is the stapler always empty just when you want it? John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK |
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Hi All
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The ubiquitous current mirror given the excellent matching of trannies in an IC. Other wise, you must test / match them well out of the junk box. Al D. On 07/31/2013 02:39 AM, John Woodgate wrote:
What is wrong with the statement? Do you find practical applications --
AC2CL I do not think there is any thrill that can go through the human heart like that felt by the inventor as he sees some creation of the brain unfolding to success... Such emotions make a man forget food, sleep, friends, love, everything. - Nikola Tesla |
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