Okay you wanted transistors.
Below is the LTspice simulation of a transistor amp whose output goes from 0V to near the supply voltage for a 0-5V input which uses feedback to give a stable gain of slightly over 10.
A three transistor complementary circuit is used to get sufficient drive for the desired output load with low quiescent current.
The gain is approximately R3/R1.

View attachment 91347

Maybe, this is my level of misunderstanding. If I am starting with 54V and I want the circuit to provide a maximum of 54V go feed the current across the slice such that it follows the voltage from near 0 to about 2.3uA (i.e., the actual voltage across the slice is unimportant, it is the current that is important). My battery current source is obviously not a problem so I think there is something fundamental about the operation of a transistor that I'm not getting. Sorry for this, as I said way above, my level of understanding of how electronic components work is probably interfering with me seeing the forrest because of all of the trees. It is nice to see, a design that could work that doesn't use an op-amp.

 

Actually, you are asking about building a clock, and we're just trying to translate accurately your concept of time from a clock-watcher's point of view into our concept of time from a clock-builder's point of view. Also, don't dis electronic component datasheets. Some need a more careful reading than others, but overall they get way too much scrutiny to be marketing toys.

So, if what you really want is an Arduino-controlled current source, let's go there. This is a different kind of circuit, but normally not a significant increase in overall circuit complexity compared to a voltage source. If you want linear programmability, with 0d = 0 mA and 4096d = 2.55 mA, that is 600 nA per step. That's a pretty small number. I'll have to chew on this a bit.

ak

 

More questions:

What is the largest voltage ever measured across a test slice?
What is the maximum current you want to pump through a slice?
Why is the CLD in the circuit? I suspect it is because someone said it is necessary, probably to protect a slice from accidental exposure to excessive current if Rp is set too low.

While at the gym I thought through another way to explain things so you can see what you have from our point of view. What you have is a simple series circuit with 4 elements and no external gorp hanging on and confusing things. Because of this, all 4 elements are commutative. You can arrange them in any order (paying attention to the diode polarity), and the current through the loop and the voltages across each element will not vary. Because of this, we can rearrange things to show more clearly what you have.

Rearranging things, you have a battery and diode that form the power supply, and two resistors in series. The battery and diode form a current-limited voltage source, and the two resistors form a simple voltage divider, what the audio people call an L-pad. Rp is the series leg and Rs is the shunt leg. Note that the s stands for slice, not shunt.

First, some definitions:

Vb = V-battery = 54 V
Vd = V-diode = the voltage across the CLD (current-limiting diode) = 0 to 2 V (ish)
Rp = R-pot = 0 to 10K Ohms
Rs = R-slice = ? to ??? Ohms depending on slice preparation and probe placement

Vout = the voltage at the cathode of the CLD, the "output voltage" of the power system.
Il - I-loop = the loop current that flows through all circuit elements

Basic Ohm's Law analysis tells us that if Rp is set to 0, any value of Rs greater than 20.8K will mean the diode is not in current limiting mode, and is just a voltage drop of somewhere between 0 V and 2 V depending on the actual current. With Rp at 10K, the max Rs for current limiting is 10.8K.

With medium to low values of Rp and Rs, the CLD will leap into action. It becomes a current-dependent variable resistor, decreasing Vout such that the total voltage across the two resistors is whatever it takes to maintain 2.5 mA through them. So for low values of Rp and Rs, Vout might be10 or 20 V. Because it is based on a FET, the limit value of a CLD is very poorly defined. +/-15% is a common value. If you want tight current limiting, a very simple transistor circuit will give much better regulation.

The final circuit element is you, the experimenter, acting as a feedback control loop. You take an input (voltmeter reading) process it against static and dynamic variables, and adjust the pot to bring the reading to a desired condition. Because that condition is based on a voltmeter, it is safe to say that you turn this collection of semiconductors and resistors into a voltage regulator or (after some arithmetic) current regulator.

Digest that, and I'll pile on more. Awaiting your answers to the questions.

ak

About 48V

About 2.3mA

You are correct about the CLD. It is just there for protection of the slice. Remember, it is controlling the current across the slice that I'm interested in. My average preparation requires about 45V to maintain 2.3mA across the slice. Because of the variability between slice resistances and the occasional occurrence of an outlier, the nature of using batteries in series is that I usually have 10V available that I don't need which could potentially allow about 4mA to flow across a 13K slice resistance. Do I really know if 4mA would hurt the slice? No, but using a CLD seemed the easiest way to avoid operator error (i.e., leaving the pot all the way closed). In an automated circuit that provides feedback to the controller, an operator error is unlikely if the code is operating correctly. So, if you are implying that the CLD ultimately doesn't need to be there, I agree,

I've already digested that. I don't think I would have a working circuit if I didn't. However, Vd is marginally less than 54V. I think that the one time I measured it was with fresh batteries (in which case Vb was 58.00somethingV), Vd was about 57V. So the CLD did drop the voltage a bit. I've thought about putting a voltage regulator on the battery and standardize to something like 53.5V. But as you point out, as the operator, I can hold the Il current where I want it by adjusting the pot. The only real requirement is that the voltage available to the slice is not lower than that needed to maintain 2.3mA (i.e., your Il) across the slice. All of this would hold true for the Arduino automated system as well.

 

Actually, you are asking about building a clock, and we're just trying to translate accurately your concept of time from a clock-watcher's point of view into our concept of time from a clock-builder's point of view. Also, don't dis electronic component datasheets. Some need a more careful reading than others, but overall they get way too much scrutiny to be marketing toys.

So, if what you really want is an Arduino-controlled current source, let's go there. This is a different kind of circuit, but normally not a significant increase in overall circuit complexity compared to a voltage source. If you want linear programmability, with 0d = 0 mA and 4096d = 2.55 mA, that is 600 nA per step. That's a pretty small number. I'll have to chew on this a bit.

ak

Thank you! In the meantime I'm going to find some reading on how transistors operate. There is still something I'm obviously not getting.

 

Thank you! In the meantime I'm going to find some reading on how transistors operate. There is still something I'm obviously not getting.

You're not getting that an emitter follower needs as much voltage on the base (+0.7V) as you want at the output.
So if you want 0-55V on the output, you need 0-55.7V on the transistor base.
The problem is, how are you going to generate that 0-55.7V?
That's what my circuit does. It converts 0-5V to 0-55V.
If you want something different, then you need a different circuit.

 

Note that #12, Wally, and I combine for about 140-150 years at this, so we're a little bit ahead of you. Take your time. I think you're coming along nicely.

To clear up something, "rail to rail" - A normal opamp has two power pins, traditionally intended for balanced + and - voltages, the two "rails". Nowadays more and more can run on a positive supply voltage and ground, but that still is two rails even though one of them is 0V. The term rail-to-tail refers to how close to the external supply voltages the input and output circuits can get while still operating linearly. For example, an LM741 (a venerable critter and the darling of this forum) is not rtr. If the inputs get to within 2 or 3 volts of either rail the amp stops amping normally. Also, even if the inputs are nicely centered between the rails, if due to circuit gain the output tries to get to within 2 or 3 volts of either rail it stops, clipping the output signal at a much lower peak-to-peak level than implied by the rail voltages available.

There is no single standard definition for rail-to-rail operation. Some newer opamps are rtr for the inputs only, output only, or both. Note that outputs never are truly rtr, in that they cannot swing down to 0.0V for a 0V negative rail. The details are in the datasheets. OTOH, many input stages that have a common mode range up to something like Vcc - 2 V (2 volts below the positive rail) have a negative input range than extends *below* the negative rail, a very cute trick from National Semiconductor in 1971.

ak

 

Remember, it is controlling the current across the slice that I'm interested in.

Did you know that we can do a current control circuit? You name the current and the circuit adjusts the voltage at analog speeds like, microseconds, continuously, infinitely variable to anyplace within the power supply voltage.

 

You didn't want to worry about what you imagined somebody might think about what might be a problem?
Not the best way to get a correct answer quickly.

I fail to see how not treating the slices as different fixed value resistors added anything. For all intents and purposes, that's what they are. I didn;t want the fact that I measure electrical neurophysiological signals to distract anybody.

 

Did you know that we can do a current control circuit? You name the current and the circuit adjusts the voltage at analog speeds like, microseconds, continuously, infinitely variable to anyplace within the power supply voltage.

Not really, but it didn't surprise me when you just let me know.

You're not getting that an emitter follower needs as much voltage on the base (+0.7V) as you want at the output.
So if you want 0-55V on the output, you need 0-55.7V on the transistor base.
The problem is, how are you going to generate that 0-55.7V?
That's what my circuit does. It converts 0-5V to 0-55V.
If you want something different, then you need a different circuit.

Thank you for that. But if you need to apply as much voltage to the base as you are supplying to the collector to let all of the voltage through the emitter, how does a transistor amplify anything?

 

.....................
Thank you for that. But if you need to apply as much voltage to the base as you are supplying to the collector to let all of the voltage through the emitter, how does a transistor amplify anything?

An emitter-follower (common-collector) circuit amplifies current, not voltage (giving a low impedance output).
The collector of a common-emitter circuit can amplify voltage across a collector resistor but has a high output impedance.
So that's why I combined two complementary common-emitter stages for gain with a low bias current, along with an emitter-follower stage for low output impedance.

Here's a short discussion of the three basic BJT transistor amplifier configurations.

Incidentally op amps are rather the "Swiss Army Knife" of analog components so you shouldn't be afraid to consider their use in any analog circuits you may want to build.
Often a circuit to do a particular analog function is simpler to design using op amps then it is using discrete transistors.
Once you understand how the "summing junction" works in an op amp circuit with negative feedback, you'll have it made. That was the "eureka" moment for me when I was learning op amps.

 

Communications people talk about sources and destinations; circuit people talk about sources, sinks, and loads. At its core, a transistor is a current-controlled current source. If you inject a small current into the base (that comes out the emitter), the physics of a transistor will adjust itself such that a larger, proportional current flows into the collector (and come out the emitter). The constant of proportionality is the gain, or beta of the device. So if a transistor has a gain of 83, and you inject 1 mA into the base, 83 mA will flow into the collector and 84 mA will flow out of the emitter, assuming the power supply and other circuit impedances can provide the currents involved. Mechanically, a transistor is two diodes plus some weird physics. The base-emitter junction is closest to a traditional diode, with the typical 0.6 V voltage drop across it (for a small signal silicon type). Based on your posts, your view of a transistor is what is called an emitter follower, one of several standard one-transistor circuit configurations. It is known for a constant DC shift across it, no voltage gain, but lotsa current gain. It often is a part of an output stage because many applications, like an audio power amplifier, need more output current than is available from the input signal.

But in your case you don't really need any current gain but you need a low-to-medium amount of voltage gain to take a 3-4 volt output from the DAC and turn it into a 50+ volt exciting signal for the slice. So while Q3 in Wally's circuit is in fact operating as an emitter follower, it is preceded by two voltage amplifier stages. An unfortunate characteristic of all transistors is that the gain is not a constant. It varies with temperature, the amount of base current, the amount of collector current, the voltage across the collector-emitter, and just about everything else you can imagine except the phase of the moon. That is why circuit topologies have been developed that make the overall circuit gain dependent on external circuit components like resistors and capacitors rather than on the instantaneous gain of the active devices. In Wally's circuit, R3 and R1 set the overall circuit gain independent of the gains of the three transistors. As long as the required overall circuit gain is less than the product of all of the internal gain stages, you can do all kinds of interesting, and stable, things.

One of those things is to turn what is usually a voltage-mode circuit, like a simple microphone preamp, into a current-mode circuit. A typical mic preamp has a gain of about 100, so if you put 5 mV into it you get 500 mV out of it. And, that 500 mV is a constant whether the load on the output is 100 Ohms, 1K, 100K, etc. Unless you overload the output with something too small like 10 ohms, the output will not vary with load resistance, the definition of a voltage source. One way to describe this circuit is that its transfer function is 100 V/V, 100 volts out per volt in.

But a different circuit can be built with a transfer function of current per volt. In your case you have a voltage-output DAC and want a current that is proportional to it. So to keep the math easy, lets say you want a maximum current of 2.56 mA. If the DAC has a max output of 2.56 V, you want a voltage-controlled current source with a transfer function of 1 mA/V; 2.56 volts produces 2.56 mA. With a 12 bit DAC this makes for simple programming: an output of 2560d is full scale; 1 LSB = 1 mV and produces 1 uA.

Good so far?

ak

 

An emitter-follower (common-collector) circuit amplifies current, not voltage (giving a low impedance output).
The collector of a common-emitter circuit can amplify voltage across a collector resistor but has a high output impedance.
So that's why I combined two complementary common-emitter stages for gain with a low bias current, along with an emitter-follower stage for low output impedance.

Here's a short discussion of the three basic BJT transistor amplifier configurations.

Incidentally op amps are rather the "Swiss Army Knife" of analog components so you shouldn't be afraid to consider their use in any analog circuits you may want to build.
Often a circuit to do a particular analog function is simpler to design using op amps then it is using discrete transistors.
Once you understand how the "summing junction" works in an op amp circuit with negative feedback, you'll have it made. That was the "eureka" moment for me when I was learning op amps.

Thank you oh so much for that. I greatly appreciate it.

 

Communications people talk about sources and destinations; circuit people talk about sources, sinks, and loads. At its core, a transistor is a current-controlled current source. If you inject a small current into the base (that comes out the emitter), the physics of a transistor will adjust itself such that a larger, proportional current flows into the collector (and come out the emitter). The constant of proportionality is the gain, or beta of the device. So if a transistor has a gain of 83, and you inject 1 mA into the base, 83 mA will flow into the collector and 84 mA will flow out of the emitter, assuming the power supply and other circuit impedances can provide the currents involved. Mechanically, a transistor is two diodes plus some weird physics. The base-emitter junction is closest to a traditional diode, with the typical 0.6 V voltage drop across it (for a small signal silicon type). Based on your posts, your view of a transistor is what is called an emitter follower, one of several standard one-transistor circuit configurations. It is known for a constant DC shift across it, no voltage gain, but lotsa current gain. It often is a part of an output stage because many applications, like an audio power amplifier, need more output current than is available from the input signal.

But in your case you don't really need any current gain but you need a low-to-medium amount of voltage gain to take a 3-4 volt output from the DAC and turn it into a 50+ volt exciting signal for the slice. So while Q3 in Wally's circuit is in fact operating as an emitter follower, it is preceded by two voltage amplifier stages. An unfortunate characteristic of all transistors is that the gain is not a constant. It varies with temperature, the amount of base current, the amount of collector current, the voltage across the collector-emitter, and just about everything else you can imagine except the phase of the moon. That is why circuit topologies have been developed that make the overall circuit gain dependent on external circuit components like resistors and capacitors rather than on the instantaneous gain of the active devices. In Wally's circuit, R3 and R1 set the overall circuit gain independent of the gains of the three transistors. As long as the required overall circuit gain is less than the product of all of the internal gain stages, you can do all kinds of interesting, and stable, things.

One of those things is to turn what is usually a voltage-mode circuit, like a simple microphone preamp, into a current-mode circuit. A typical mic preamp has a gain of about 100, so if you put 5 mV into it you get 500 mV out of it. And, that 500 mV is a constant whether the load on the output is 100 Ohms, 1K, 100K, etc. Unless you overload the output with something too small like 10 ohms, the output will not vary with load resistance, the definition of a voltage source. One way to describe this circuit is that its transfer function is 100 V/V, 100 volts out per volt in.

But a different circuit can be built with a transfer function of current per volt. In your case you have a voltage-output DAC and want a current that is proportional to it. So to keep the math easy, lets say you want a maximum current of 2.56 mA. If the DAC has a max output of 2.56 V, you want a voltage-controlled current source with a transfer function of 1 mA/V; 2.56 volts produces 2.56 mA. With a 12 bit DAC this makes for simple programming: an output of 2560d is full scale; 1 LSB = 1 mV and produces 1 uA.

Good so far?

ak

Yes! ak, you are really making this a learning experience and I really appreciate it. I was looking through my ancient books (I was getting into hardware design in the late 70s, just before I finished my doctorate) and I came upon the Op Amp Cookbook by Walter Jung). I recently found a simulator I'm comfortable with (docircuits) that is quite limited from the digital side, but is fine for doing linear stuff. I intend to experiment using some of the circuits in Jung's book. That's the way I learned TTL/CMOS logic 45 years ago, so maybe it will work again. Anyway, I have "Wally's?" circuit, but I've removed my stubborn hat and have decided to confront my fear of op-amps. So, while I'm trying to learn about op-amps, I would really like to get my lab stuff going. The Arduino Due arrived today, and I think I have everything I need to get going with this. So, again, any help would be greatly appreciated. You have been very patient with me. So, thank you in advance.