View attachment 87431 Changing the duty cycle by definition changes the frequency, but modulating frequency does not neccesarily change the duty cycle. Am I missing something here?

Yes. Changing the duty cycle does NOT by definition change anything other than the duty cycle. It certainly does not have to change the frequency - that's how a pulse width modulator works, constant frequency, variable pulse width.

Your circuit is a shortcut, and you've run into what is lost. Without all of the voltage dividers and diodes, it is a modern version of the classic multivibrator circuit that is almost 96 years old. It basically is two monostables in series, and does in fact change frequency when you change the duty cycle because you are changing only 1/2 of the circuit. If you increase one collector resistance while decreasing the other, the duty cycle will change but the frequency will not. It is going to be very difficult to do compared to a standard PWM circuit that separates the frequency function from the modulation function.

ak

 

Yes. Changing the duty cycle does NOT by definition change anything other than the duty cycle. It certainly does not have to change the frequency - that's how a pulse width modulator works, constant frequency, variable pulse width.

Your circuit is a shortcut, and you've run into what is lost. Without all of the voltage dividers and diodes, it is a modern version of the classic multivibrator circuit that is almost 96 years old. It basically is two monostables in series, and does in fact change frequency when you change the duty cycle because you are changing only 1/2 of the circuit. If you increase one collector resistance while decreasing the other, the duty cycle will change but the frequency will not. It is going to be very difficult to do compared to a standard PWM circuit that separates the frequency function from the modulation function.

ak

Got it. I'd be interested in how to create a PWM modulating circuit, but since I'm no longer building a SMPS, I'll cross that bridge next time.

I have some new questions.

I don't have a bench-top power supply at the moment. From what I've read, it seems to be most people's first real project to build their own, so before I get into building the linear supply for the printer I thought I would do that.

From what I've seen on Amazon, your run of the mill bench-top power supply can output about 30V at 5A maximum. I can't see myself using more than that, but I could see myself using that current sometimes, but probably not a voltage higher than that. I've decided to aim for those specs.

My first thought was a bunch of adjustable voltage regulators in parallel for voltage control, and a bunch more voltage regulators in parallel for constant current adjustment. That's where I hit my first snag.

Putting voltage regulators in parallel is going to complicate the adjustment a bit. I found some 5A voltage regulators, but their max input voltage was 20V. 20 < 30, so no cigar there. I guess I'll just have to go parallel, unless someone knows of something else.

I ran into another problem. Most voltage regulators I've found seem to have a minimum voltage output, and a minimum load requirement.

I want to be able to go to 0V, so I can't use anything with a defined voltage reference it seems. I also don't want to load requirement.

I looked at the datasheets of a voltage regulator and it looks like an op-amp with feedback and the output connected to a BJT.

It seems like it would be easy enough to use two potentiometers for voltage, one for coarse and one for fine, and another two pots for controlled current adjustment. This poses a couple issues though.

I know that an ideal op-amp doesn't throughput current on the inputs...how reliable is this rule in the real world?

If the inputs don't draw any current at all and just detect voltage then everything is fine and dandy. I can use the pots as voltage dividers to the non-inverting input, and the output, regardless of current flowing through it, can be fed back into the non-inverting input.

If that's not the case, then I could have some serious issues with a circuit built on that principle. Since voltage regulators have maximum voltage, current, and power values it could be one of two things. Either the op-amp configuration I mentioned is flawed, or it's just limited by the BJT.

I have a feeling that it's the former.

What options do I have?

 

@WBahn,



I should have said more precisely, thats a 10:1 step down voltage ratio.
Example of LTSpice 10H:100mH transformer plot.
http://www.linear.com/solutions/5092

E.
The other Thread has been closed.

Oh, that wasn't the problem. It's just been so long that I had forgotten that coil inductance goes as the square of the number of turns. At some point it would be good for me to go back and re-derive all of these basic relationships.

 

A few responses...

A typical starter bench supply has 0-30V at 3A and fixed 5V at 3A, and things increase from there. Another common arrangement is 3 outputs, 0to +30, 0 to -30, and the fixed 5V. A nice extra feature is an optional tracking setting for the dual bipolar outputs. You adjust the + output and the - output follows along. As for max output current, it always is difficult to predict the future but is has very little effect on the actual circuit design.

You don't need two power stages in series. One power stage can respond to both the voltage and current control loops. Search for lab or bench power supply schematics and you'll get a ton of examples. Look at each one, and you'll see a pattern emerge on a common approach. Multiple independent regulators cobbled together for greater output current is more work and more risk than necessary. Grow a good regulator control circuit, then add output transistors in parallel to handle the worst case power dissipation. Note that with linear regulators you never get very good efficiency. So for a max output of 30 V / 5 A, the worst case is an output of 1 V at 5 A. That leaves 29 V plus the regulator headroom voltage at 5 A dissipated in the output device(s), something that could approach 200 W. Some power transistors are rated to dissipate 200 W, but you would need a high density heatsink and a ducted fan to cool it. Better to spread it out over 3 or 4 cheap parts like a 2N3055, with another 3055 as the darlington driver (common inventory).

There are circuit techniques to get a linear supply to regulate down to 0V output, but first make sure you really need that capability before committing to the increased circuit complexity. Opamps with an input common mode voltage range that extends below the negative rail can give you stable voltage control, and a high-side current sense circuit keeps the current sense resistor out of the ground return. Another way is to add a small charge pump circuit to get a little negative voltage output from the transformer secondary, opening up a much larger range of opamps to choose from. Either way, you will get better output regulation if the control circuits have their own small regulated supply to keep them independent of the output.

Transistorized bench power supplies have been around since the 60's. That is a huge knowledge base. Use it. Surf for the schematics, read the articles, ask questions, and learn more about how these kinds of things are done. The classic approach is with two opamps and some pass transistors, but it can be done with a bunch of LM317's, or one 317 and some power transistors wrapped around it for a little less performance but a lot less design work. Life is choice.

ak

 

A few responses...

A typical starter bench supply has 0-30V at 3A and fixed 5V at 3A, and things increase from there. Another common arrangement is 3 outputs, 0to +30, 0 to -30, and the fixed 5V. A nice extra feature is an optional tracking setting for the dual bipolar outputs. You adjust the + output and the - output follows along. As for max output current, it always is difficult to predict the future but is has very little effect on the actual circuit design.

You don't need two power stages in series. One power stage can respond to both the voltage and current control loops. Search for lab or bench power supply schematics and you'll get a ton of examples. Look at each one, and you'll see a pattern emerge on a common approach. Multiple independent regulators cobbled together for greater output current is more work and more risk than necessary. Grow a good regulator control circuit, then add output transistors in parallel to handle the worst case power dissipation. Note that with linear regulators you never get very good efficiency. So for a max output of 30 V / 5 A, the worst case is an output of 1 V at 5 A. That leaves 29 V plus the regulator headroom voltage at 5 A dissipated in the output device(s), something that could approach 200 W. Some power transistors are rated to dissipate 200 W, but you would need a high density heatsink and a ducted fan to cool it. Better to spread it out over 3 or 4 cheap parts like a 2N3055, with another 3055 as the darlington driver (common inventory).

There are circuit techniques to get a linear supply to regulate down to 0V output, but first make sure you really need that capability before committing to the increased circuit complexity. Opamps with an input common mode voltage range that extends below the negative rail can give you stable voltage control, and a high-side current sense circuit keeps the current sense resistor out of the ground return. Another way is to add a small charge pump circuit to get a little negative voltage output from the transformer secondary, opening up a much larger range of opamps to choose from. Either way, you will get better output regulation if the control circuits have their own small regulated supply to keep them independent of the output.

Transistorized bench power supplies have been around since the 60's. That is a huge knowledge base. Use it. Surf for the schematics, read the articles, ask questions, and learn more about how these kinds of things are done. The classic approach is with two opamps and some pass transistors, but it can be done with a bunch of LM317's, or one 317 and some power transistors wrapped around it for a little less performance but a lot less design work. Life is choice.

ak

I had a quad op-amp and some power BJTs lying around, so I tried a dummy circuit.

I had an old regulated 3A 14V PSU laying around, so I used that for power. This is my circuit...



I'm only getting ~2V on the output, and I don't understand why.

When I look at similar schematics, and linear voltage regulator schematics I notice something: the output is usually a darlington pair. I'm guessing this is my problem, but why do I need the darlington?

 

Because your voltage divider is connected to ground on both ends (r1/r2). Do you see anything else...?

 

Because your voltage divider is connected to ground on both ends (r1/r2).

Oops. In reality it's not. I just made an error on my spice model.

Will post a new one shortly.

 

Oops. In reality it's not. I just made an error on my spice model.

Will post a new one shortly.

Here we go. Actual bread-boarded circuit.



Also, using a MOSFET in place of the BJT yields supply voltage.

 

The TIP32 is a PNP transistor, and you show it as NPN. Another drafting error?

You don't say what the load is. A TIP31 (the NPN flavor) can have a gain of over 100 in some conditions, but less than 50 at 2 A collector current. A typical opamp cam make about 20 mA max output current, but you don't get the full output voltage swing at that high current. 20 mA x50 = 1 A, so if your load current is trying to be 2 A the circuit will pull out of regulation. A darlington transistor actually is a two-transistor integrated circuit (which Sidney Darlington invented ***10 years*** before the TI IC patent). At room temperature and 2 A Ic, a TIP110 has a gain of over 2000, so 2 A output current requires less than 1 mA of drive current from the opamp. The tradeoffs are that it is less efficient at high output voltages because there are two Vbe drops in the transistor instead of one, it is slower, and the gain curve is not very flat. None of which usually matters in a basic linear regulator design.

ak

 

The TIP32 is a PNP transistor, and you show it as NPN. Another drafting error?

You don't say what the load is. A TIP31 (the NPN flavor) can have a gain of over 100 in some conditions, but less than 50 at 2 A collector current. A typical opamp cam make about 20 mA max output current, but you don't get the full output voltage swing at that high current. 20 mA x50 = 1 A, so if your load current is trying to be 2 A the circuit will pull out of regulation. A darlington transistor actually is a two-transistor integrated circuit (which Sidney Darlington invented ***10 years*** before the TI IC patent). At room temperature and 2 A Ic, a TIP110 has a gain of over 2000, so 2 A output current requires less than 1 mA of drive current from the opamp. The tradeoffs are that it is less efficient at high output voltages because there are two Vbe drops in the transistor instead of one, it is slower, and the gain curve is not very flat. None of which usually matters in a basic linear regulator design.

ak

Actually, I got it to work. In just assumed that power NPNs had the same pin shceme, whichbthey don't. This is definitely an NPN transistor.

I had another issue though. I could end up with around 80W of power in the worst case scenario with this strategy.

I need to spread the power out over multiple transistors. Is thag done with transistors in parallel, or darlington pairs or?

 

Multiple power darlingtons in parallel, or multiple power transistors in parallel driven by one more power transistor as an emitter follower driver, a kind of hybrid darlington.

ak

 

Multiple power darlingtons in parallel, or multiple power transistors in parallel driven by one more power transistor as an emitter follower driver, a kind of hybrid darlington.

ak

Okay, so I can use one power NPN to drive say, three other transistors which will split the current...

I'm a bit fuzzy on a couple things.

Will they also split the voltage three ways?

Because parts are imperfect I assume that one of the NPNs will conduct more heavily than the rest...increasing it's temperature...further increasing its conductivity...so on and so on. I've read that this problem is solved by adding low value resistors at the emitter of each transistor. This doesn't make sense to me.

Where would I put the load then? It would essentially be equivalent to adding 3 resistors in parallel. If I used 1 ohm for each resistor each resistor's "real" value would be 1/3 ohm.

If I'm dividing current three ways, that means that the max will be 1A per resistor at 30V. I would need 30W resistors...

Plus if there were minute differences in resistance, I would have the same problem I originally had.

Also, I'm having trouble with my dummy circuit. The max output seems to be about 13/14 of the input voltage. I'm guessing this is the current drawn by my op-amp. How do I know what voltage to use in order to compensate for that?

 

Allenph: Okay, so I can use one power NPN to drive say, three other transistors which will split the current...

ak: Yes.

Allenph: Will they also split the voltage three ways?

ak: No. Think about it. The three output transistors are in parallel, so they all have the same voltage across them. Series items split voltages, parallel items split currents.

Allenph: Because parts are imperfect I assume that one of the NPNs will conduct more heavily than the rest...increasing it's temperature...further increasing its conductivity...so on and so on. I've read that this problem is solved by adding low value resistors at the emitter of each transistor. This doesn't make sense to me.

ak: It does to me. But seriously... One diode (such as a base-emitter junction) directly in parallel with another and with a slightly lower Vf can starve current away from it because of the sharp transition, or knee, between the conducting and non-conducting conditions of a diode. It basically is a voltage-dependent switch. Power transistors have higher Vbe than the standard 0.6V for small signal transistors, so lets start with Vbe = 0.8 V for conduction. For a perfect diode, at 0.799 V across the it there is no current through it, and at 0.801 V it tries to be a short circuit. So if two transistor junctions are directly in parallel and one transistor's Vbe is 50 mV less due to temperature differences, it is trying to short out the other diode. Now add a 1 ohm resistor in series with each emitter. The two parallel circuits no longer are just the base-emitter diode junctions, but a diode in series with a resistor. With 2 Amps being shared and a 50 mV Vf difference, each resistor will have approximately 1 V across it, with one resistor having 50 mV more than the other, a 5% difference compared to the 100% all-or-nothing without the ballasts. The tradeoff is the larger the emitter ballasts, the better the load sharing - but the lower the efficiency of the entire circuit due to the power dissipated in the ballast resistors. This is an issue only when the output voltage approaches the input voltage of the regulator. A *simplified* example: If the negative peaks of the bulk filter capacitor ripple voltage are at +31V and you want +30V out of the supply, you can drop 0.8V across the output transistors and still just barely be in regulation. But dropping 1.8 V across the transistors and their emitter ballast resistors means the output can go up to only 29.2 V before pulling out of regulation.

Allenph: Where would I put the load then? It would essentially be equivalent to adding 3 resistors in parallel. If I used 1 ohm for each resistor each resistor's "real" value would be 1/3 ohm.

ak: Yup. Before ballasts, the output (and the sense connection back to the regulator) was the junction of the three emitters. With ballasts the output is the junction of the three resistors. Each resistor is carrying approximately 1/n times the total output current.

Allenph: If I'm dividing current three ways, that means that the max will be 1A per resistor at 30V. I would need 30W resistors...

ak: Nope. That's what the power transistors do. For a constant 3 A output current, as the output voltage is turned down the collector-emitter voltage across the transistors increases. All voltage regulators essentially are a controlled variable resistor in series between the rectifier/capacitor and the output. We use a control circuit to emulate this resistance with power transistors. They are what varies as you adjust the output voltage. They also track the instantaneous value of the input, which is why there is no ripple on the output even though you might have 5 V peak-to-peak ripple on the main filter capacitor.

Allenph: Plus if there were minute differences in resistance, I would have the same problem I originally had.

ak: True, but if you go through the previous example and introduce a 5% difference between the two ballast resistor values, you'll see that the current difference between the two legs still is relatively small compared to a circuit with no ballasts. 1% or 2% wirewound power resistors are common in these circuits.

Allenph: Also, I'm having trouble with my dummy circuit. The max output seems to be about 13/14 of the input voltage. I'm guessing this is the current drawn by my op-amp. How do I know what voltage to use in order to compensate for that?

ak: Either you're splitting hairs unnecessarily, or you don't quite understand the difference between textbook semiconductor physics and real world devices. Before, I used an example of 0.8 V for Vbe. In fact it can be 1.5 V for very large devices, and it varies with both base and collector currents and from part to part. Plus there is a bulk resistance through the semiconductor material and the transistor package that adds some voltage drops. For a 90 W output supply the total power consumed by the regulator control functions probably is less than 1 W. Cost of doing business.

ak