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Old 23rd Oct 2019, 9:29 pm   #1
Uncle Bulgaria
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Default BC548 4.8V drop?

Why would the transistor, with a 5.6V Zener diode reference on the base and 9.5V on the collector, have only 4.7V on the emitter with no load?

It's my understanding that with a signal on the base, the transistor should switch on and the emitter should have the 9.5V of the collector, less 0.7V.

Have I got a duff transistor or is there some underlying arcana I am not privy to?
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Old 23rd Oct 2019, 9:42 pm   #2
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Default Re: BC548 4.8V drop?

With 5.6V at the base, I'd expect it to have 5V at the emitter. Assuming there really is 5.6V at the base (just because there's a zener there, it doesn't necessarily mean it's being provided with enough current to maintain 5.6V across it), and also assuming it's not oscillating (emitter followers are prone to breaking out into oscillation, which can easily confuse a meter).

A circuit diagram would help confirm some of my assumptions and clarify the question. Also, how much current is leaving the emitter?
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Old 23rd Oct 2019, 9:42 pm   #3
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Default Re: BC548 4.8V drop?

It’s an npn transistor so in linear operation the emitter will be 0.7v lower than the base, provided the collector is more than about 1v higher than the base. Your setup is just that, the transistor is operating as a voltage regulator in effect.
Edit: good point above re zener may not have enough current but basic point remains.
To operate as a switch with the load on the emitter, connect the base to the collector via a resistor. This is not the best way to operate though, if you have a load connected to 0v.
Better to use a pnp, emitter to supply, load from collector to ground. Base via resistor to ground for ON, via resistor to emitter for OFF. This mode minimises voltage drop in the transistor to Vce(sat), about 0.1v.

Ken

Last edited by Ambientnoise; 23rd Oct 2019 at 9:59 pm.
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Old 23rd Oct 2019, 10:30 pm   #4
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Default Re: BC548 4.8V drop?

Here is the circuit diagram - please excuse the mouse-writing! It's a sub-question to my ongoing CR 240 tape deck repair - this whole transistor assembly was burned out from a short.

It should detect when the mains power is being used (the 6.2V is then connected to the zener and thus the base, less the two 2.2k resistors), and turn the lights on. On battery power, the 9.5V is diverted to the base by the switch momentarily, presumably to save the battery.

The lamps used are wire-ended 3V, 30mA and are impossible to find. I'm trying to replace them with LEDs, and have some 3V, 25mA ones installed as replacements. Without the zener, the lamps all light. With it I get the reduced voltage at the emitter and no lights. I'm trying to work out the current limiter resistor I need.

I haven't used zeners before so perhaps mhennessy's point is apt - the current is not enough to maintain the voltage, so the transistor is not actually saturated. I hope I'm using the right terms here, as my understanding of transistors is limited!
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Old 23rd Oct 2019, 10:34 pm   #5
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Default Re: BC548 4.8V drop?

I trust you're using current-limiting resistors in series with the LEDs?

If not, the LEDs will be acting like zeners, and "fighting" with the transistor and zener.

But then, they would still light.

If they are white LEDs, then 5V might not be enough to light two in series. So removing the zener, which allows the transistor's emitter voltage to float up, might explain (some of) the symptoms.
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Old 24th Oct 2019, 9:36 am   #6
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Default Re: BC548 4.8V drop?

If I understand the cct, on mains the transistor operates as a voltage regulator with 5v on the emitter feeding the series 3v lamps with 2.5v each, in effect. Sounds reasonable. In battery mode, there is no feed to the zener or transistor base so the transistor is cut off, no current flows collector-emitter, the bulbs are resistors so the emitter will be at 0v.
Did it work correctly with bulbs ?
With leds, 5v is marginal at best for two in series so you need the 4 in parallel , each with their own resistor selected for 5v and the current you want. I suggest you try 5mA through each led at first, the cct prob fed about 50mA to the bulbs.
I suggest you start with just one led and resistor to try things out, with the zener replaced of course.

Ken
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Old 24th Oct 2019, 10:04 am   #7
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Default Re: BC548 4.8V drop?

I thought it rang a bell:

https://www.vintage-radio.net/forum/...=157951&page=3

Lawrence.
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Old 24th Oct 2019, 11:54 am   #8
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Default Re: BC548 4.8V drop?

Quite so, Lawrence! You can see what being away all summer does to the brain. When I came back to it to replace the burned parts, the bit I'd remembered was that the lights should be on when mains is on, not that you'd kindly gone through it before!

I think you're the only one now who understands that thread's twists and turns. I hope to have a go at this tonight...
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Old 26th Oct 2019, 1:06 pm   #9
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Default Re: BC548 4.8V drop?

Sorted! The LEDs are now in parallel with individual resistors. It took some re-wiring of the series connections on some of the boards, but now they all light with ~20mA.

However, I'm still confused.

I thought that when a transistor's base has a voltage applied, the resistance between C & E drops to allow current flow. Why is the voltage on the emitter not the same as that on the collector in this case? I had been counting on more to run the LEDs in series. Why is 5V being expected on the emitter by applying the ~0.6V drop to the 5.6V on the base rather than the 9.5V on the collector?
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Old 26th Oct 2019, 1:55 pm   #10
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Default Re: BC548 4.8V drop?

The collector and emitter voltages are only the same when the transistor is saturated (fully-on). When not saturated, they will be different.

The base voltage is usually 0.6V higher than emitter. So if there's 5.6V on the base, then there has to be 5V on the emitter, irrespective of the voltage on the collector.

Obviously, that's simplifying grossly, and no-doubt others will be itching to jump in and tell me how silly I'm being. But before doing so, they should consider that simplification is sometimes needed when learning - we all started somewhere.

In the meantime, the more reading up you can do on how transistors, the better. Many texts are simply terrible - far to theoretical. I've always quite like the explanations in "The Art of Electronics" by Horowitz and Hill - they manage to avoid the worst of the maths and keep things nicely practical. PDF copies can usually be found online. You might enjoy this extract:

https://blog.adafruit.com/2015/06/17...ectronicsbook/
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Old 26th Oct 2019, 2:21 pm   #11
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Default Re: BC548 4.8V drop?

The circuit is an 'Emitter-follower'

The transistor turns on only as much as necessary to make the emitter voltage follow the base voltage, just around 0.65v lower.

You may be supplying 5.6v (with respect to chassis) to the base, but the transistor doesn't know that. It has no connection to the chassis for comparison. It can only respond to the voltages on its three electrodes.

You apply 5.6v to its base and it starts to pass current from base to emitter, like a silicon diode. But it also passes a much greater current from collector to emitter. the total of these currents flow through the LEDs and their current-setting resistors. This brings the emitter voltage up, reducing the base-emitter voltage and reducing the base current. The collector current scales down in proportion to the base current.


The circuit settles in equilibrium with the emitter current needed to drive the emitter load to get the emitter up to about 0.65v below the base.

If the load current changes, the transistor turns on a bit more or a bit less to handle the change and to keep Vbe about 0.65v

It's a feedback mechanism built in the transistor. There's no loop to point at on a schematic.

Valve cathode followers work in a similar fashion, but Vgk varies quite a bit more than Vbe.

DAvid
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Old 26th Oct 2019, 2:22 pm   #12
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Default Re: BC548 4.8V drop?

If it’s 20mA to each led, that’s 80mA total whereas I think the original total load was about 50mA. I think it maybe worth reducing to 50mA total, so 12mA per led as the transistor will be running hotter due to the 120mW or so increased dissipation. The leds will be a bit dimmer but should be ok. You should just feel how hot the transistor gets !

Ken
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Old 27th Oct 2019, 2:54 pm   #13
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Default Re: BC548 4.8V drop?

Thank you all.

mhennessy - with no background in this, the simpler the better. I feel I sometimes have the hang of simple valve circuits after years of playing with them, but not working with the theory daily means it does seep away. My ex-girlfriend's father is a real engineer and has lent me "The Art of Electronics" on occasion. It is a wonderful book, but I couldn't justify getting my own copy for the limited need I would have for it. I may have to reconsider if I'm doing more transistor repairs. That page linked is particularly helpful.

He also is fond of saying that one should never be afraid of asking the 'silly wee laddie' questions, as they're always the ones that need asking!

I have the Van Valkenburg, Nooger & Neville US electronics series, which does have transistors in the final book, but it's in storage at present. Those have been great for valve circuits for me!

You're right that I've found most texts available online far too theoretical - the Wikipedia page particularly does that jump from 'that seems reasonable' to 'eh?!' in one paragraph. I always find that happens on technical entries!

Radiowrangler - so it's current dependent, not voltage dependent? The Transistor Man and his resistor is in the active area, not saturation. The current from the base voltage allows more current to flow from the collector supply through the LEDs. What then causes saturation? Greater voltage on the base to approach Vc, allowing for your feedback mechanism?

Ambientnoise - that's a very good call. Thank you. I had forgotten Kirchoff. I will see what the brightness looks like with larger resistors.
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Old 27th Oct 2019, 3:49 pm   #14
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Default Re: BC548 4.8V drop?

The job of a transistor is to pass a current from the collector to the emitter.

That current depends on the voltage applied to the base-emitter junction.

In that regard, a transistor is very much like a triode, so if you have a good understanding of those, you're half way there.

Many people believe that it's the base current that matters, not the base-emitter voltage. So, the transistor might be a current-operated device, not a voltage-operated device (like a triode, or a FET).

In fact, you can use either model, depending on the situation. The current-operated model is perhaps easiest for your circuit, but the voltage-operated model is essential when looking at voltage amplifiers, for example.

Because the current-mode model is so deeply engrained in engineers, it's my experience that it pays to get the voltage-control model out there as soon as possible when teaching transistors. Some engineers refuse to believe that a bipolar transistor could possibly be a voltage-controlled device, but it really is. Like valves, transistors have gm

Saturation takes place when the collector is passing the maximum current it possibly can given the surrounding conditions. Further increases in base-emitter voltage (or current) therefore have no obvious effect, and the voltage across the collector and emitter is next to nothing (less than 0.3V typically, depending on the collector current). In terms of Transistor Man, he has turned his dial as far as he can, and has hit the end stop.

Saturation is good when switching loads on and off, but not what you want when amplifying a signal.
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Old 27th Oct 2019, 4:07 pm   #15
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Default Re: BC548 4.8V drop?

Quote:
Originally Posted by Uncle Bulgaria View Post
so it's current dependent, not voltage dependent? The Transistor Man and his resistor is in the active area, not saturation. The current from the base voltage allows more current to flow from the collector supply through the LEDs. What then causes saturation? Greater voltage on the base to approach Vc, allowing for your feedback mechanism?
Watch this:

https://www.youtube.com/watch?v=cuOFG_ISHno

Lawrence.
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Old 27th Oct 2019, 5:30 pm   #16
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Default Re: BC548 4.8V drop?

Quote:
Originally Posted by Uncle Bulgaria View Post
Thank you all.

Radiowrangler - so it's current dependent, not voltage dependent? The Transistor Man and his resistor is in the active area, not saturation. The current from the base voltage allows more current to flow from the collector supply through the LEDs. What then causes saturation? Greater voltage on the base to approach Vc, allowing for your feedback mechanism?
It's both current and voltage!

Think of a humble silicon diode carrying a small current:

Is the voltage across it causing the current to flow?
Or is the flowing current causing the voltage?

THe answer is both at once, and they're related by the diode equation, just like the voltage and current of a resistor are related by Ohm's law.

So transistors have been modelled both in terms of the input voltage and of the input current. They are different ways of expressing the same thing. People use whichever viewpoint they think will make what they are trying to do easier.

Valves are a bit simpler to get your mind around. Unless you are doing horrible things to them, the grid current is beggar all, so you can just forget it.

With transistors you can think of voltage or current controlling it, but you have to keep an eye on the other.

Horowitz & Hill's 'Transistor man' model is rather neat, but notice the man is measuring the current through a diode as his input. This is how they got both the input voltage and input current into their model.

All transistors try to be linear, like good little current amplifiers, but if you drive them hard, eventually they try to pull more collector current than the circuitry and supply feeding their collector can provide. The collector and emitter voltages have been pulled as close together as they can get. This is saturation. Note that it is at a collector current set by the circuit, not (any longer) by the transistor.

If the collector circuit was improved to be able to supply more current, maybe the transistor would resume linear operation, multiplying its base current. Maybe the power dissipation would be too much and it goes up in smoke. It depends on the values.

As a rule-of-thumb, put in five times as much base current as you'd expect from the device gain and the amount of collector current you want to pass if you want a transistor switched on hard into saturation.

David
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Old 27th Oct 2019, 6:52 pm   #17
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Default Re: BC548 4.8V drop?

Excuse me but where are the LEDs? The circuit looks like a simple emitter follower used as a voltage regulator to feed two parallel connected series pairs of 3V 30mA filament lamps. This represents a 'nominal' load of 6V @ 60mA, however, since the transistor base is restricted to a maximum of 5.6V by the zener, this load will have a maximum voltage of about 4.9V across it and will therefore draw a maximum of less than 60mA (depending on the lamp characteristics. The slide switch (S?002) being used as a bright/dim control.

With the switch in the current position, the base current is supplied from the two 2.2K resistors in series from a 6.2V supply, this results in a maximum current of about 14uA (microamps) available for the zener and the base of the transistor, hence the emitter current is severely restricted by the current gain of the transistor (Ie = Ib x hFE) and therefore the emitter voltage will be lower due to emitter resistance (Vbe > 0.7V due to the transistor not being saturated). The result is that the lamps are dimmed.
With the switch in the other position, the maximum current available for the base and zener is about 4.3mA, plenty for the transistor to be driven into saturation and the emitter voltage rises to about 4.9V (Vbe ~ 0.7V) and the lamps will shine brighter . . .

* Do the math with a maximum hFE for the BC548 of 800 (at 2mA Ie) and minimum of 110 (at 50mA Ie) *

From the OPs voltage measurements it would seem that the circuit is working correctly.
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Old 28th Oct 2019, 11:09 am   #18
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Default Re: BC548 4.8V drop?

Just to note regarding the annotated snippet schematic that's been posted.....Normally the supply voltage to the bias resistors is approx. 9 volts with battery or mains/ext. DC...Not 9 volts and 6.2 volts.

In internal battery mode the transistor is biased via a 2.2k from the internal (nominally 9v) battery supply when the push switch S1002 makes contact.

When powered from the mains unit or an external DC source the two 2.2k resistors are connected in series to the 9 volts that's supplied via a regulator by either the mains unit or an external DC source, ie: a vehicle battery or whatever.

EDIT: The full schematic can be gotten from the vintageshifi site.

Lawrence.

Last edited by ms660; 28th Oct 2019 at 11:19 am.
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Old 29th Oct 2019, 1:57 pm   #19
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Default Re: BC548 4.8V drop?

Lawrence - right as ever. I was befuddled by the 6.2V output that runs just above this assembly.

RF Burn - the LEDs have been added by me, as the bulbs are unobtanium. My lack of knowledge about transistors meant that I was expecting different measurements when calculating my dropper resistors. The switch is what allows the lights to be turned on momentarily when on battery power - otherwise the lights are always on because of the lower feed connecting to a contact on the power pack.
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