Are transistor variability and thermal effects irrelevant for ac volt follower - Page 1
Forum section: electronics / beginners

author: Infraviolet
author_rank: Super Contributor
timestamp: September 17, 2023, 09:30:14 pm
content: I'm trying to consider how the properties within a transistor which are variable with device manufacture, ambient temperature, current through transistor... would affect the following circuit.My suspicion is they wouldn't, but I'm not sure if there's a variable property of transistors which I haven't considered.Consider the voltage follower circuit shown in the diagram, an AC waveform enters centred about some arbitrary, but fixed, level. The transistor acts as a follower and serves to buffer the current, the output waveform is the input waveform, minus about 0.7V, but now with a low source impedance, rather than whatever large impedance may have been present on whatever is supplying the signal to the transistor's base. The AC signal is then capacitively coupled so as to now be centred about some other arbitrary, but fixed, level, set in this exmaple by resistors 39 and 40.The transistor isn't amplifying voltages and hoping for known ratios of current, so I don't think Hfe variability matters. And because the output gets capacitively coupled to a new centre level I don't think changes to the 0.7V drop should matter either, if it drops more or less than that the offset is still removed by the capacitor.I'll assume that anything after the capacitor cannot possibly ever demand more than 50mA of current for even an instant, and the source impedance from anything inputting to this isn't going to be more than a few K ohms. Also, no input waveforms getting to within less than 1 volt of the 5V or gnd rails ever need be considered.Is there anything about this circuit that would be affected by variable properties of transistors? Or should this truly act as an almost perfect way to make a clone of the input signal's voltage, but with the ability to drive smaller loads without having voltage on the load being dragged down by the original input's source impedances.Thanks

author: TimFox
author_rank: Super Contributor
timestamp: September 17, 2023, 09:39:29 pm
content: The input impedance (seen by Sig In) is a function of beta (hfe), which is quite variable within a given part number from part to part, and increases with temperature.For a given driving impedance (output driving Sig In), similarly the output impedance driving the load depends on beta.For zero driving impedance and reasonable total load impedance (including the bias resistors, especially R38), the voltage gain (slightly less than one) depends onhib, which depends only (almost) on the bias current (for small signals).

author: Kim Christensen
author_rank: Super Contributor
timestamp: September 17, 2023, 09:43:43 pm
content: Because Hfe varies with temperature, so will the current gain. If the current gain varies, while load stays the same, then the input impedance will vary which will effect the loading on the source and thus the signal level at the input...No real magic here. No more magic than a well designed common emitter or common base amplifier mitigating variations in Hfe, etc.

author: Infraviolet
author_rank: Super Contributor
timestamp: September 17, 2023, 10:52:35 pm
content: If you're in a situation where (load current)/Hfe makes the base's input current small enough that the source impedance of the incoming signal gives only a negligible voltage drop then this would act perfectly?? If the ratio of the current you want to drive in to the load to the current you can draw from whatever is supplying Sig_in (without appreciably changing the voltage on sig_in) is less than the smallest Hfe possible with variation then it does act perfectly?? And this would still apply to large signals so long as they don't get within a volt or so of the rails??Effects of the current in R38 should cancel away ?? because this is an AC signal and where the centre sits is irrelevant, just so long as it doesn't stray far enough to get too near the rails.

author: TimFox
author_rank: Super Contributor
timestamp: September 17, 2023, 11:00:55 pm
content: Output impedance from the emitter to the load (including bias resistors) does not depend directly on beta, if the source impedance (to the base) is zero, but it is non-zero and roughly (26 mV)/IE(DC value) for small-signal amplification.Input current depends on beta, and on the output current. It is a quantitative question if that be important.

author: Kim Christensen
author_rank: Super Contributor
timestamp: September 17, 2023, 11:02:35 pm
content: Define "perfectly"...A change of output loading will always have some effect on the input. It'll never be zero.But maybe it's low enough to be considered "perfect" for your application.

author: TimFox
author_rank: Super Contributor
timestamp: September 17, 2023, 11:06:38 pm
content: Perfect implies zero effect, with a strong suggestion of hand waving.

author: Infraviolet
author_rank: Super Contributor
timestamp: September 17, 2023, 11:13:37 pm
content: Ok, lets consider perfection thresholds of:a)1%b)0.1%andc)0.025%For all the range of variability common within one model of transistor (but maybe from different manufacturers, to give a nasty worst case), and temperature ranges from -20C to +80C, current in the load ranging from nearly 0, up to as far as it can go before "perfection" thresholds get violated, the source driving sig_in to have internal resistance ranging somewhere from 0.5K to 5K. And consider signals which are NOT small, ranges of peak-to-centre of 0.5V from the centre-level, up to 1.5V from it.I should have thought better than to imagine true perfection for anything, anywhere, ever.Am I on the right lines thinking like this:Consider the AC current in the load, ignoring the DC level of current about which it is centredDivide this by Hfe (lowest case possible)If the resulting divided current would be small enough that the voltage drop caused by that much current across the internal resistance of sig_in's source is insignificant (as per thresholds) as compared to the voltage of the AC waveform's peaks, then the circuit is "perfect" enough.For any higher Hfe value the circuit would then be closer to perfection than in that calculation.Is there another varying parameter, beyond Hfe, which needs taking in to consideration. I saw your mention of h_ib, and am not sure what that parameter is.

author: tridac
author_rank: Regular Contributor
timestamp: September 17, 2023, 11:34:52 pm
content: You don't say what level of gain accuracy you need, nor the signal source impedance. An emitter follower configuration would be much more stable by adding a second transistor, pair of resistors divider feedback to the first stage, to stablise the gain and lower the output impedance. Would not use a simple emitter follower for anything other than the most trivial application, but still appropriate in some cases...

author: TimFox
author_rank: Super Contributor
timestamp: September 18, 2023, 01:03:13 am
content: Quote from: Infraviolet on September 17, 2023, 11:13:37 pmEven with very highhfe, the slope of theVBEvsIEcurve, which ishib(the input impedance of a common base stage) determines the highest gain for the emitter follower into a finite load impedance. Luckily, it is determined to high accuracy by only the emitter quiescent current. That parameter (in ohms) forms a voltage divider from an ideal output to the load impedance, reducing the voltage gain below unity.What elementary textbooks on transistor circuits do you have access to?

author: Infraviolet
author_rank: Super Contributor
timestamp: September 18, 2023, 03:02:27 am
content: I have Horowitz and Hill, but transistors used in analog fashions keep seeming to find ways to confuse me despite all the equations relating to them being really simple expressions.From what you've just said, H_ib acts as a resistance in series with the load (and with the emitter to ground resistor), and I can see how such a potential divide happens. This will mean the higher the resistance of the parallel combination of the load and that emitter resistor to ground the less H_ib will matter?And H_ib is one of the parameters which will vary with device manufacture and with ambient temperature?

author: TimFox
author_rank: Super Contributor
timestamp: September 18, 2023, 03:32:42 am
content: hibis the slope of Shockley's equation for the voltage across a diode as a function of the current through it: this should be in Horowitz and Hill.You can consider it to be the AC resistance around a quiescent bias point: dVBE/dIE.From one 2N3904 to another, the value is reasonably consistent and independent ofhfe= beta, but it is a function ofjunctiontemperature (the transistor knows nothing of the ambient temperature, only the temperature of its junction).IIRC, the parameter "26 mV" in the equation dVBE/dIE=hib= (26 mV)/IEis proportional to absolute temperature (in K), and therefore varies slowly with junction temperature in the neighborhood of 300 K.Yes, as you increase the total load resistance seen by the emitter (possibly using a constant-current source in the bias network), the less effect the output impedance has on the voltage gain.If you really need an accurate +1 V/V gain, why don't you use a simple op-amp follower (output fed back directly to the inverting input), which will give you a high input impedance, very low output impedance, extremely close to unity gain at reasonable frequencies, and a sensible output current capability?[edit: added temperature dependence of 26 mV]