Yeah, as for me, the project that taught me the most was building a 4-bit calculator using an Alteration Quartus PLD. It could add, subtract, multiply, divide, AND, OR, XOR, etc. I could Karnaugh map the s#!% out of things... but that's digital electronics - this is analog, and I only got through Circuits II.



I guess I didn't post the article where I found this circuit. It lists out a lot of info:

http://www.edn.com/design/analog/4314544/Integrator-ramps-up-down-holds-output-level (article)
http://m.eet.com/media/1126960/12071-figure.pdf (actual circuit image)

I'm using the recommended LMC6484 which is rail-to-rail. The thing that gets me is that the voltage WILL ramp up linearly to max when the motor is removed, but when the motor is added, it only goes to 1.8V or so (measured from ground to base/output of the ramp circuit). It's as if I've somehow added a resistor to create a voltage divider. Shouldn't the collector and emitter be isolated from the base? Why is this impacting the circuit like this? That's why I tried randomly throwing diodes at the problem, but no joy.

Here is a possibility:
Check fig 4 in the op amp data sheet. http://www.ti.com/lit/ds/symlink/lmc6484.pdf
It can only supply about 10ma at 3.7 volts.
Then if you look at fig. 5 of the transistor data sheet. http://www.onsemi.com/pub_link/Collateral/2N3055-D.PDF
You might try shorting out the resistor from the op amp to the base to see if it speeds up. If not I think it's a real problem.
We can either try to fix it or do it another way.
How much current can your 12 volt supply provide??

 

Here is a possibility:
Check fig 4 in the op amp data sheet. http://www.ti.com/lit/ds/symlink/lmc6484.pdf
It can only supply about 10ma at 3.7 volts.
Then if you look at fig. 5 of the transistor data sheet. http://www.onsemi.com/pub_link/Collateral/2N3055-D.PDF
You might try shorting out the resistor from the op amp to the base to see if it speeds up. If not I think it's a real problem.
We can either try to fix it or do it another way.
How much current can your 12 volt supply provide??

This is the boost converter I bought (I know, I went cheap):
http://pages.ebay.com/link/?nav=item.view&id=331616617766&alt=web

It says 5-12V, but in the description, the minimum input voltage is 1.5V, so it should work with the 3.7V battery (albeit with a lower current). The timer that is sharing the boost circuit only uses 20mA, so that's a negligible drain. I'm not home right now, but I can test the current when I get back. I'll also try to use my bench-top power supply to see if that makes a difference - you could be right. I believe I've tried shorting out that resistor, too, but with no huge difference. I'll double-check and post back in the morning.

 

This is the boost converter I bought (I know, I went cheap):
http://pages.ebay.com/link/?nav=item.view&id=331616617766&alt=web

It says 5-12V, but in the description, the minimum input voltage is 1.5V, so it should work with the 3.7V battery (albeit with a lower current). The timer that is sharing the boost circuit only uses 20mA, so that's a negligible drain. I'm not home right now, but I can test the current when I get back. I'll also try to use my bench-top power supply to see if that makes a difference - you could be right. I believe I've tried shorting out that resistor, too, but with no huge difference. I'll double-check and post back in the morning.

To bad, it is not big enough to drive the motor. Can you measure the current of the motor when it is running?

 

To bad, it is not big enough to drive the motor. Can you measure the current of the motor when it is running?

No, no, no - the 12V boost isn't powering the motor. The 3.7V battery is powering the boost circuit; the boost circuit is powering the op amp; the op amp is exciting the base of the transistor; the motor is connected directly to the battery and to the collector of the transistor (emitter to ground). Check my second diagram again:

http://goo.gl/3Lq83z

I know the boost circuit won't power the motor. I'm powering the ramp with the boost to switch the transistor and power the motor with the battery. Make a little more sense...?

 

No, no, no - the 12V boost isn't powering the motor. The 3.7V battery is powering the boost circuit; the boost circuit is powering the op amp; the op amp is exciting the base of the transistor; the motor is connected directly to the battery and to the collector of the transistor (emitter to ground). Check my second diagram again:

http://goo.gl/3Lq83z

I know the boost circuit won't power the motor. I'm powering the ramp with the boost to switch the transistor and power the motor with the battery. Make a little more sense...?

Yes, I understand. The problem is the op amp can only supply about 10 ma. To saturate (fully turn on the transistor) the base current needs to be about 1/10 of the collector current. So it can only supply about .1 amps to the motor before it comes out of saturation. My thought was to use 12 volts for the motor as well. In the case where the transistor is running in the linear region it's gain is much higher - maybe 70 instead of 10.
Anyway knowing the run current of the motor would be very helpful so we can figure out how hard it needs to be driven. I'm trying not to throw the circuit away and starting over.

 

@summersab I just noticed the post where the motor current is over 2 amps at 5 volts. Ouch!
The second circuit won't really ramp - the first will. The first is a switch. You just can't drive it hard enough to make it switch very well.
The first circuit is an emitter follower so it will follow the ramp, but you can't drive that one hard enough either..
The best solution might be to build a triangle generator and use the digital pwm output and a FET to drive the motor.

 

I just noticed the post where the motor current is over 2 amps at 5 volts. Ouch!
.

If that is the one I read, this is based on the armature resistance, if so, this is a totally unrealistic means of detecting the operating current of the motor, especially where ramped start is used.
OP quotes the motor as 'very small'!
Max.

 

If that is the one I read, this is based on the armature resistance, if so, this is a totally unrealistic means of detecting the operating current of the motor, especially where ramped start is used.
OP quotes the motor as 'very small'!
Max.

Strange but ---

Lastly, I connected the motor to an old bench-top power supply that I have, and provided that it is working right (fingers crossed), the motor pulled 2.2A at 5V. So, 2.27ohm? Still seems really low...

So, I'm off to do pwm. Shame it is such a cute circuit.

 

If that is the current rotating at no load then the actual resistance would normally be much lower than that?
Max.

 

If that is the one I read, this is based on the armature resistance, if so, this is a totally unrealistic means of detecting the operating current of the motor, especially where ramped start is used.
OP quotes the motor as 'very small'!
Max.

Max is right. My mistake, and I didn't measure the motor's properties correctly. It's an encased motor, so taking it apart is a little difficult (I can do it if that's absolutely necessary). My bench-top power supply is an old analog beast from the 1960s that I got from my alma mater, I'm not thinking it showed current properly, so I got my good multimeter out. I measured the current draw to be 500mA at 4.2V, so that's 8.4Ω.* Note that the motor is only rated for about 5-6V, though, so I don't want to go shoving 12V at it as previously suggested.

*Well, that's nice - ALT+234 in Windows or CTRL+SHIFT+U2126 in Linux in my case.

@summersab I just noticed the post where the motor current is over 2 amps at 5 volts. Ouch!
The second circuit won't really ramp - the first will. The first is a switch. You just can't drive it hard enough to make it switch very well.
The first circuit is an emitter follower so it will follow the ramp, but you can't drive that one hard enough either..
The best solution might be to build a triangle generator and use the digital pwm output and a FET to drive the motor.

I'm not sure if I follow. Which two circuits - the goo.gl links? Both circuits I posted are basically the same and they ramp just fine without the motor attached. Further, the output of the op-amp ramps as expected as does the collector of the transistor, so it behaves as designed. However, when the motor is connected, that all goes to hell - the op-amp output drops, and the collector behaves accordingly.

Maybe I don't understand a PWM properly, so let me know if this is right. The basic idea is to charge a capacitor with pulses. The faster the pulses, the more charge the capacitor has. This can then be used to control the speed of a motor. If so, my scenario is different (yet related). A pure PWM controller doesn't ramp but rather is used to control a motor's fixed speed via a potentiometer. I want the circuit to ramp and hold.

Since this might have to do with the base current, would this be a case for a Darlington like a TIP120? Thanks for all the attentive help, guys - I really appreciate it.

 

Max is right. My mistake, and I didn't measure the motor's properties correctly. I got my good multimeter out. I measured the current draw to be 500mA at 4.2V, so that's 8.4Ω.* Note that the motor is only rated for about 5-6V, though, so I don't want to go shoving 12V at it as previously suggested.
.

That is more like it, however as I previously mentioned, you cannot determine the armature resistance this way, for e.g., if you did this test rotating and off load, then place various small loads on the motor and see the current change.
Max.

 

I'm not sure if I follow. Which two circuits - the goo.gl links? Both circuits I posted are basically the same and they ramp just fine without the motor attached. Further, the output of the op-amp ramps as expected as does the collector of the transistor, so it behaves as designed. However, when the motor is connected, that all goes to hell - the op-amp output drops, and the collector behaves accordingly.

The circuit in post 1 has the motor in the emitter. This is where you want it so it can ramp.
Post # 17 has it in the collector. It cannot ramp correctly there.

Maybe I don't understand a PWM properly, so let me know if this is right. The basic idea is to charge a capacitor with pulses. The faster the pulses, the more charge the capacitor has. This can then be used to control the speed of a motor. If so, my scenario is different (yet related). A pure PWM controller doesn't ramp but rather is used to control a motor's fixed speed via a potentiometer. I want the circuit to ramp and hold.

PWM... Well maybe google it. But it can be ramped as well. Here is a circuit: It inverts the ramp, but you can swap leads on the pot for that.

Since this might have to do with the base current, would this be a case for a Darlington like a TIP120? Thanks for all the attentive help, guys - I really appreciate it.

A darlington built into a transistor won't work because both collectors are tied to the same voltage. But here is a circuit that is similar.

The drawback - if you look closely is that the ramp will be shortened. Don't know if that's a problem for you or not.
Take note the op amp runs on 12 volts now, not 5 like in the other schematics.

 

That is more like it, however as I previously mentioned, you cannot determine the armature resistance this way, for e.g., if you did this test rotating and off load, then place various small loads on the motor and see the current change.
Max.

Alright, Max, you win. I took the casing off, held the armature still, and it hovers around 1.5A at 4.2V, so 2.8Ω. Should I try other voltages?

The circuit in post 1 has the motor in the emitter. This is where you want it so it can ramp.
Post # 17 has it in the collector. It cannot ramp correctly there.


PWM... Well maybe google it. But it can be ramped as well. Here is a circuit: It inverts the ramp, but you can swap leads on the pot for that.


A darlington built into a transistor won't work because both collectors are tied to the same voltage. But here is a circuit that is similar.
The drawback - if you look closely is that the ramp will be shortened. Don't know if that's a problem for you or not.
Take note the op amp runs on 12 volts now, not 5 like in the other schematics.

Ron, I'm gonna mull over what you just posted, but I just took a bunch of readings, and I think I figured at least something out. In the circuit, there is a resistor that can be adjusted to change the speed of the ramp. From the site I got it from, it is labeled as R2; in my diagram, it is the 50kΩ that feeds the inverting input of the third op amp. I took measurements using R2=1MΩ and R2=0Ω both with and without the motor connected. Note that I have the motor connected to the collector and the emitter to ground for these measurements. The supply to the op amp is 12V while the supply to the motor is 4.2V (directly connected to the battery). Voltages are measured from ground.

No motor (resistor makes no difference):
B: 0.7V
C: 2mV
Op amp current: 11.3mA

1MΩ + motor:
B: 0.66V
C: 2.4V
Op amp current: 1.8mA

0Ω + motor:
B: 0.64V
C: 50mV
Op amp current: 11.3mA

So, when R2 is increased (i.e. I'm trying to slow the ramp), it decreases the current available from the op-amp to the transistor base and prevents the motor from running at full. Only when this is shorted does the motor get full power. That kinda defeats the purpose of this ramp-and-hold circuit. I'm sure I probably just came to conclusions that were mentioned in previous posts (like post 25, I believe), but I likely got lost in the nomenclature and missed it until figuring it out on my own. Is there any way to fix this? Gonna look at the diagrams you posted in more detail, now...

 

Alright, Max, you win. I took the casing off, held the armature still, and it hovers around 1.5A at 4.2V, so 2.8Ω. Should I try other voltages?
.

No real need, it would only see the current based on armature resistance when/if full voltage applied initially or at stall.
Max.

 

Alright, Max, you win. I took the casing off, held the armature still, and it hovers around 1.5A at 4.2V, so 2.8Ω. Should I try other voltages?



Ron, I'm gonna mull over what you just posted, but I just took a bunch of readings, and I think I figured at least something out. In the circuit, there is a resistor that can be adjusted to change the speed of the ramp. From the site I got it from, it is labeled as R2; in my diagram, it is the 50kΩ that feeds the inverting input of the third op amp. I took measurements using R2=1MΩ and R2=0Ω both with and without the motor connected. Note that I have the motor connected to the collector and the emitter to ground for these measurements. The supply to the op amp is 12V while the supply to the motor is 4.2V (directly connected to the battery). Voltages are measured from ground.

No motor (resistor makes no difference):
B: 0.7V
C: 2mV
Op amp current: 11.3mA

1MΩ + motor:
B: 0.66V
C: 2.4V
Op amp current: 1.8mA

0Ω + motor:
B: 0.64V
C: 50mV
Op amp current: 11.3mA

So, when R2 is increased (i.e. I'm trying to slow the ramp), it decreases the current available from the op-amp to the transistor base and prevents the motor from running at full. Only when this is shorted does the motor get full power. That kinda defeats the purpose of this ramp-and-hold circuit. I'm sure I probably just came to conclusions that were mentioned in previous posts (like post 25, I believe), but I likely got lost in the nomenclature and missed it until figuring it out on my own. Is there any way to fix this? Gonna look at the diagrams you posted in more detail, now...

We have to go back to 20 questions again.
If the emitter is grounded how high can the voltage on the base go.
So if the output of the op amp wants to go to 12 volts how much current would it need to supply.
How much can it supply?
If you can answer those questions you will know why the grounded emitter kills the circuit.

 

We have to go back to 20 questions again.
If the emitter is grounded how high can the voltage on the base go.
So if the output of the op amp wants to go to 12 volts how much current would it need to supply.
How much can it supply?
If you can answer those questions you will know why the grounded emitter kills the circuit.

20 questions, I.E. me really sucking at this, but here we go . . .

Okay, so from the op amp spec sheet, figure 4, the op amp can supply just under 100mA at 12V (just tested that by shorting the output resistor - I'm suspecting I picked the wrong value...?) Figure 5 of the transistor spec sheet shows that I have infinite Vce at under 10mA, but Vce at ~90mA is under 0.2V, and it looks like I need at least ~50mA from the op amp to pull it that low. The motor needs ~0.5A to run, you said the base needs to be ~1/10 of the collector, so the base needs to be at least 5mA (which . . . don't we have that? I feel like I'm talking in circles, now). I have zero idea how to answer your first question. One of the "on conditions" for the transistor says Vbe (Ic=4.0A, Vce=4.0V) = 1.5. So . . . if the base is held low, Vce gets pulled low instead of 4V. Vbe can only go to 1.5V, but what is this 4V?

I'm officially lost.

 

I am getting lost here... are you trying to ramp voltage, current or speed?
Or perhaps some combination of those, speed with a current limit perhaps.

When the thread started the problem wasn't complex and now it appears to be both complex and not well defined.

My point is you may want to take a step back and define the problem.
I don't disagree with the academic points RE armature current and how to manage it with a linear circuit but I also don't see the point, unless this is just that an academic exersize.

Your solution requires two essential elements an one/two optional ones
Controlling power to the motor, based on some control parameter
managing the perimeter value.

You absolutely could use an analogue system control of the motor current but you probably don't want to.
You could attempt to use this analogue control circuit to self regulate in some way but again you probably don't want to.

Using PWM to provide power, however you manage it, to the motor is easy and will work with a range of motors. This isolates the power switching from the control system, on is just on and off is just off with #% duty cycle being just that #%

If all you want is a linear ramp with no feedback then ramp the PWM ffrom 0% to #% and you are done.
If you want a current control ramp then measure the current, but not with the motor, and control the PWM to maintain some current setpoint.
now ramp the set point and you are done.
Voltage control is the same as current apart from the fact that you look at motor voltage as opposed to current.

If what you actually want is speed then measure it.... You can estimate it by measuring both speed and current but the speed/curent relationship is NOT linear at any load so this is a little tricky to do without an MCU and either a lookup table of complex equation.
However if you go there you still want to control the power with pulses and have a speed set point which you ramp.

With respect I suggest you draw a block diagram of what you want to achieve and then design discrete elements that together form your overall circuit.

My personal take on this thread is thatr it ahs lost sight of the goal... controlling the motor and seems to be focused on making an inappropriate control topology do stuff that is inherently hard to predict/stabilize.
I am not saying the transistor theory being discussed is wrong, per se, just that it is a poor choice of control that is massively complicating an ostensibly simple task... but that's just me, O and every manufacturer of motor/power control modules, small and large.

Al
(Sorry I overloaded you earlier)

 

By the way, a FET in its linear region, partially on, is a good controller of current and is voltage controlled with a very small gate current.
You can almost treat them as a voltage controlled resistor. You could certainly drive one with the output of an opamp provided it was capable of supplying/sinking 10-12V for 'common FET's and 5V if you used a low gate voltage version.

FET's for small power applications are not hard to use. The gate, which controls it behaves like a small capacitor and and the source/drain impedance is proportional to the gate voltage.

Al

 

@Dyslexicbloke I agree to a point that yes, this thread has become ridiculously long in order to solve a pretty simple problem. One thing that I think has made it a bit longer than it could be is the desire of some of the forum members to help me by helping myself (something that I do have an appreciation for despite its frustrations). I agree that feeding me an answer just keeps me coming back with more questions . . . but dang, I want an answer.

The Project
With that said, the problem is really, really simple. You know those Sonic Boom alarm clocks with a bed vibration motor? Well, they're great save for a few limitations. First, they come on at full power. You do not know terror until you have one of those things launch you from sleep to consciousness. Second, their travel/battery version is woefully underpowered, and as someone who travels, I'm basically making one that works a little better.

Here's what I've got. I bought a 12V programmable timer/relay, a 3-6V vibrating motor, and a beefy Sanyo 14500 lithium battery. Since the timer relay requires 12V and a minimal 20aA to latch, I didn't think I needed a 11.1V lithium pack just to for that, so I grabbed a cheap 12V boost circuit. It works just fine to power the timer and latch the relay with just one lithium cell.

So, the timer is powered by the battery via a 12V boost. When the timer goes off, the relay activates and supplies power straight from the battery to the motor. That's easy. The missing part? Some sort of circuit to make the motor ramp up from off to the full power supplied by the battery (in roughly 30 seconds or so; no need to be variable - fixed time is just fine).

The Solutions
I looked at PWMs and actually built one, but either I didn't build it right or I don't understand them correctly. A user on another forum explained them like this. Suppose the motor is connected to the power via a pushbutton. To turn on the motor, you push (or "pulse") the button repeatedly. The faster you push the button (i.e. increase the frequency of the pulses), the faster the motor goes. My understanding is that a PWM's pulse frequency is usually controlled by a pot, and adjusting this pot lets you control the speed of the motor. That's great for manually adjusting the speed of a motor, but I'm looking for a circuit where the speed automatically increases from off to full power, not one where I have to fiddle with a pot to set the speed (obviously, that's impractical if I'm asleep).

So, I built this circuit (click here for the diagram). The first paragraph defines the circuit perfectly and seems to solve my problem:

Much searching through online “cookbook” circuits turned up no means of ramping an op-amp integrator to hold at a preset constant voltage level. This Design Idea describes a single-supply op-amp circuit that outputs a rising or falling linear-voltage ramp in response to a step change of a positive dc-input voltage of 0V to VCC

Perfect!

Arthur's Limited/Flawed Understanding of Electronics
However . . . what do I connect the op-amp to so I can provide power to the motor? A transistor, I thought, but obviously, my understanding of transistors is pretty limited and/or flawed.

• Resistors? Easy - they reduce current flow.
• Capacitors? Got that - they store electrical energy kinda like a battery, but they have a lower potential and can provide higher current (I remember building Leyden jars in college physics).
• Relays? Simple - supply a voltage with a small current, and they activate a switch that turns on and off a higher-powered circuit.

Then we get to transistors. Uh . . . these are like relays(ish) where the more you "excite" the base terminal with a current(?), the more (or less) power will flow between the collector and emitter. So . . . kinda like a variable relay, right? FETs . . . are also transistors, but they (somehow) behave differently and exist because of some patent that expired so Bell Labs could invent them. It's language like this where you totally lose me:

A PNP transistor consists of a layer of N-doped semiconductor between two layers of P-doped material. A small current leaving the base is amplified in the collector output. That is, a PNP transistor is "on" when its base is pulled low relative to the emitter. In a PNP transistor, emitter-base region is forward biased. so electric field and carriers will be generated. They should flow towards the base junction, but the base part is very thin and has low conductivity. the reverse biased collector base part has generated holes. so due to the electric field, carriers or electrons get pulled by the holes.

Um . . . monkey push the button. Monkey want variable relay? Monkey like relay - relay easy. The only thing that is "doped" here is me. My brain is officially saturated (*snort* see what I just did there?)

 

@Dyslexicbloke I agree to a point that yes, this thread has become ridiculously long in order to solve a pretty simple problem. One thing that I think has made it a bit longer than it could be is the desire of some of the forum members to help me by helping myself (something that I do have an appreciation for despite its frustrations). I agree that feeding me an answer just keeps me coming back with more questions . . . but dang, I want an answer.

The Project
With that said, the problem is really, really simple. You know those Sonic Boom alarm clocks with a bed vibration motor? Well, they're great save for a few limitations. First, they come on at full power. You do not know terror until you have one of those things launch you from sleep to consciousness. Second, their travel/battery version is woefully underpowered, and as someone who travels, I'm basically making one that works a little better.

Here's what I've got. I bought a 12V programmable timer/relay, a 3-6V vibrating motor, and a beefy Sanyo 14500 lithium battery. Since the timer relay requires 12V and a minimal 20aA to latch, I didn't think I needed a 11.1V lithium pack just to for that, so I grabbed a cheap 12V boost circuit. It works just fine to power the timer and latch the relay with just one lithium cell.

So, the timer is powered by the battery via a 12V boost. When the timer goes off, the relay activates and supplies power straight from the battery to the motor. That's easy. The missing part? Some sort of circuit to make the motor ramp up from off to the full power supplied by the battery (in roughly 30 seconds or so; no need to be variable - fixed time is just fine).

The Solutions
I looked at PWMs and actually built one, but either I didn't build it right or I don't understand them correctly. A user on another forum explained them like this. Suppose the motor is connected to the power via a pushbutton. To turn on the motor, you push (or "pulse") the button repeatedly. The faster you push the button (i.e. increase the frequency of the pulses), the faster the motor goes. My understanding is that a PWM's pulse frequency is usually controlled by a pot, and adjusting this pot lets you control the speed of the motor. That's great for manually adjusting the speed of a motor, but I'm looking for a circuit where the speed automatically increases from off to full power, not one where I have to fiddle with a pot to set the speed (obviously, that's impractical if I'm asleep).

So, I built this circuit (click here for the diagram). The first paragraph defines the circuit perfectly and seems to solve my problem:

Perfect!

Arthur's Limited/Flawed Understanding of Electronics
However . . . what do I connect the op-amp to so I can provide power to the motor? A transistor, I thought, but obviously, my understanding of transistors is pretty limited and/or flawed.

• Resistors? Easy - they reduce current flow.
• Capacitors? Got that - they store electrical energy kinda like a battery, but they have a lower potential and can provide higher current (I remember building Leyden jars in college physics).
• Relays? Simple - supply a voltage with a small current, and they activate a switch that turns on and off a higher-powered circuit.

Then we get to transistors. Uh . . . these are like relays(ish) where the more you "excite" the base terminal with a current(?), the more (or less) power will flow between the collector and emitter. So . . . kinda like a variable relay, right? FETs . . . are also transistors, but they (somehow) behave differently and exist because of some patent that expired so Bell Labs could invent them. It's language like this where you totally lose me:

Um . . . monkey push the button. Monkey want variable relay? Monkey like relay - relay easy. The only thing that is "doped" here is me. My brain is officially saturated (*snort* see what I just did there?)

Use the 2nd circuit in post 32.
Q1 and R9 are the only 2 new components. The rest is just so I could simulate your circuit.