Transistor CCS resistor value calculation.

[Diabolical Artificer]

I want to make this CCS, see pic. I know how to calculate R3, the emitter resistor sets the CCS current, so if the zener was 5v6, R3 would be 5v divided by I CCS, so for 5mA R3 would be 1k.

However R1 and R2 am not sure about. I think R1 is found by V+ divided by I zener (5mA say) +  BC337 emitter current divided by hFE (50uA)  +  MJE340 emitter current divided by hFE (160uA) Therefore if V- is - 80v, that'd be a 15.355 resistor for R1, or nearest value 15k. R2 is trickier, I make it that there would be about an 8v drop across R1, making Vb BC337 76v, beyond it's Vce.

So do I use another tranny instead of the 337 or have I cocked up the calculations?

Andy.

[ppppenguin]

Not sure that this cascode configuration gives you much benefit in the intended application. One important point is that the current in the resistor chain and zener must be much greater than the base currents of the transistors. Another way of putting this is that you must never rely on hFE when designing, except to make sure that the base current isn't going to give trouble at the lowest possible hFE.

The current is set by the zener and emitter resistor as you say. Won't be very stable with temperature but that may not matter. A current mirror would be better https://en.wikipedia.org/wiki/Current_mirror There are also many better current sources, easily found online. But as I said, it may not be very critical in your application.

The total of R1+R2 is given by the supply voltage minus the zener voltage divided by the zener current. You are then free to set the base voltage of the upper transistor to any convenient value by choosing the actual values of R1 and R2.

So for 80V HT and 5mA zener current the total of R1+R2 is about 75V/5mA =15K. The base currents are small compared to 5mA and so can be ignored. If you wanted 15V on the base of the top transistor (and hence just under that on the collector of the bottom one) there would be 10V across R2 and 65V across R1. Hence R1 must be 6.5 times greater than R2. So R2=2K and R1=13K would work. Since the zener current isn't critical and neither is the base voltage of the upper device there's a lot of latitude in the R1 and R2 values. Watch the power disspated in R1. 65V*5mA= 325mW so don't use a little 0.25W part.

[valves4amps]

Hi Jeffrey, the point of this cascade circuit is that the lower transistor with high gain may not tolerate the higher voltages that may exist across the whole network. So the upper transistor is closely controlled by the higher gain lower one.  Also the temperature coefficient of Zeners ranging from + to - is around zero at 5v. I agree totally with your summation of resistor values.

Dave

[Alistair D]

Instead of a zener use one or more red LEDs in series. The +ve temperature co-efficient of an LED closely matches the -ve co-efficient of the transistor's B-E junction. 

I have built a couple of CCSs using this method and they worked fine. I never tested the temperature stability to see if LEDs were better than zeners, the word of the gurus was good enough for me.

Al

[ppppenguin]

Dave, my point about the cascode is that the high voltage transistor might well be good enough on its own. The tempco of a 5V zener is close to zero but that of the base emitter junction of the transistor is not. Hence the tempco of the zener should be approximately matched to that of the transistor. Though this calculation also needs to account for the emitter resistor. If the drop across the emitter resistor is large compared to the change in Vbe that will improve the stability.

One easy way of getting good stability is a current mirror which ultimately relies on the matching between 2 transistors. Again emitter resistors improve the stability.

Of course all this may be academic. The actual current from the CCS may be fairly uncritical. Another factor is the output impedance at the frequencies of interest. Again we don't know how much this matters. Any CCS will be a lot better than a simple resistor.

Al, the tempco of the LEDs or zener needs to have the same polarity as the BE junction. Otherwise it will makes things worse, not better.

LEDs are a handy way to get a not very precise voltage drop. For example I've used them to get roughly 5V from an 8V supply. You get a free pilot light too

[Alistair D]

Jeffrey you are absolutely correct about the LED and transistors requiring the same polarity of co-efficient.  The years have been playing tricks on my brain. Red LEDs have -ve temp co-efficient similar to that of a silicon transistor VBE junction. NOT +ve as I stated earlier.

Al

[Diabolical Artificer]

The CCS is for a LTP, it also serves in a cathode follower. I've built the single tranny version just using the MJE340, but apparently this has a low Zout as the hFE of the 340 is low. So you plonk the smaller Q in to increase Zout. AFAIK Zout is basically Re x Q1 hFE x Q2 hFE - hoe (which I don't know how to find and isn't on datasheets) is low, so can apparantly be left out.

"the tempco of the LEDs or zener needs to have the same polarity as the BE junction. Otherwise it will makes things worse, not better. " Could you expand on this please Jeffrey, how is this calculated/solved?

As the application is for a valve circuit, precision isn't important. I keep meaning to have a play with the ring of two CCS, but am unsure how one figures out Zout and wouldn't a ring of two with added upper Q be better still?

Thanks for your help lads, Andy.

[ppppenguin]

I ask again, how much does the Zout matter in this application? if the single transistor isn't good enough then you need a better circuit such as this cascode. But that still raises the question: How good does the CCS need to be in this application? As you've said, precision isn't important in this valve circuit.

If temperature goes up and (for example) the zener voltage stays the same but the transistor Vbe goes down then the colelctor current will rise. The effect is reduced by the emitter resistor. With a 5V zener the effect will be fairly small as there will be around 4.5V across the resistor and the change in the Vbe will be small compared to the total voltage.

There are many different current source and current mirror topologies. Just look online and you'll find plenty. Some are really only for use on chips where you can match transistors accurately.

[Terry]

Temperature compensation can be critical in some situations.

For example, when the BB105 varicap diode was introduced for use in Varicap TV Tuners, it was important to ensure that the supply voltage was compensated to track the temperature coefficient of the diodes, otherwise tuning drift would be experienced until the temperature stabilised - particularly important in the valve sets of the time, of course. Eventually, Mullard introduced the TAA550 stabiliser which had the inverse temperature coefficient to that of the diodes but that didn't exist when we first started using them.

The solution was a string of five zener diodes - 3 x 6.8V and 2 x 3.3V - which had with the correct temperature coefficient and produced a supply of 27V (the maximum reverse voltage of a BB105 was 28V).

[Mark Hennessy]

Personally, it's far easier to measure Zout than to try to calculate it

I prefer the "ring of two" current source. It maintains the current using negative feedback, so generally performs better than these sorts of "hold a known voltage across a known emitter resistance" sources. Cascoding these types of sources won't make them inherently better - they just allow you to use different parts to get the voltage handling. Another approach is to use a Darlington instead.

Of course, the hFE has relatively little to do with the output current directly, but if you cock up the design by making R1 and R2 too high, then the base current might make the base voltage change, and that in turn reduces the emitter voltage, which in turn reduces the output current. The "stiffer" the base voltage is, the more constant the current will be. You still have the problem that the emitter current is not the same as the collector current (because Ie = Ic + Ib), but for the application, I can't see that being a real-world issue. It would be good to know why you're so worried about the tail current in this application, BTW...

Luckily not an issue here, but this circuit has poor "compliance" - meaning that the voltage can only get to within ~6V of the negative rail at best. When you add the cascode transistor, you reduce the compliance further, depending on the values of R1 and R2. Generally, you choose them such that the base of the MJE340 is only a couple of volts above the base of the BC337. So that answers one of your questions about the design of this - decide what the zener current should be (perhaps 10mA), add a bit for the base currents (perhaps a milliamp, but it's really not critical), then choose the R1 + R2 combination to provide this. Then, as you'd only like a couple of voltage across R2, that comes out at 180 ohms, give or take.

Back to the "ring of two" source - that can get to within a volt of the rail, typically. Choosing the two resistors is easy - one carries the current, so that is 0.6V / required current. The other needs to provide a milliamp or two, and the voltage across it is the rail minus 1.2V. Easy

Regarding temperature stability, I'd be astounded if that matters - and if it does, there are easier ways to fix it in a solid-state amp, at least. However, I've always understood that 5.6V zener diodes have a positive tempco of 2.1mV/C, and a "standard" PN junction has a tempco of -2.1mV per degree C. So-called "temperature-compensated" zener diodes consist of a 5.6V zener in series with a standard signal diodes, and such a combination can be implemented for a fraction of the cost using a 1N4148. Obviously, all numbers are "engineering" numbers, meaning they are approximate, but good enough.

In a current source, the zener doesn't generate much heat of its own and is likely to stabilise fairly quickly at power-on (ignoring the influence of the surrounding temperature rises). But the transistor is likely to run much warmer, so it's the change in Vbe that will contribute most to any change in current. The "ring of two" current source is much less affected by this, because the voltage across the sense resistance is compared to the Vbe of the control transistor, not the one handling the current.

Yet another problem is Early effect. With the high voltages present, this might raise its head. The "ring of two" is much more resistant to that, too.

If the current really matters, put an op-amp in there. This will give excellent results - limited only by the quality of the voltage reference at the input - but it is over-engineering, which is why it's rarely if ever done commercially. Again, why does the current source stability and impedance matter in this application?

Final thought on that - presumably the LTP is at the input to the amp? What is the common-mode voltage seen by this LTP?