cd4066 output current?

[Nigel Goodwin]

ok, had momentary brain fart... was thinking the pnp base resistor in series between two transistors.

Like this (R3 is opto current limit resistor):

Sorry, but that's complete nonsense

Check my PIC 'extras' for an example of high side switching using NPN/PNP.



But the simple emitter follower suggested earlier (post #5) would do all that's required, as you're only powering a grounded LED.

 

[crosslakeguy]

Operating Power
At a minimum, optocouplers require current to bias the LED and some form of bias on the output side. The total input plus output current varies widely, depending on the type of optocoupler. When forward biased, the optocoupler LED is low-impedance, and device power consumption increases with LED forward current, which can range from 1 mA to over 15 mA. In some cases, the LED may require an external driver, further decreasing system efficiency while increasing BOM complexity and cost. The optocoupler output impedance can be low or high depending on its architecture. Most low-cost optocouplers have a simple transistor output that is high impedance when LED forward current is at zero and relatively lower impedance when LED forward current is in its specified operating range. Other (usually higher speed) optocouplers have an active photo coupler and output driver that requires an external bias voltage. Such devices have low output impedance but at the expense of increased total operating current, which can range from 15 mA to over 40 mA
this is from silicon labs.....

the engineers that manufactured and implemented these high current optos surely also designed a non discreet component to drive them at a reasonable current level? ....... The PIC is happy at 25ma but what about 45ma, Is there an IC that can operate multiple I/O's at the increase in current using non discreet electronics?

 

[Nigel Goodwin]

the engineers that manufactured and implemented these high current optos surely also designed a non discreet component to drive them at a reasonable current level? ....... The PIC is happy at 25ma but what about 45ma, Is there an IC that can operate multiple I/O's at the increase in current using non discreet electronics?

You can always parallel two pins together to give double capacity

But as for 'design', if you design things correctly then such problems don't occur - as for a suitable IC, the much mentioned ULN2003 would do it if you didn't ground the opto-couplers (and any sensible design wouldn't cripple itself by not offering the option).

And yet again, simple emitter followers would do the existing job anyway, and you can (or certainly could) get multiple transistors in IC type packaging.

 

[Mike odom]

Sorry, but that's complete nonsense

Check my PIC 'extras' for an example of high side switching using NPN/PNP.



But the simple emitter follower suggested earlier (post #5) would do all that's required, as you're only powering a grounded LED.

Actually, your post is utter nonsense. We were talking about a way to give a positive output without the .7V loss. We both understand you don't have to worry about the .7V drop using an NPN to turn on a PNP because you are driving an LED that's grounded. That's what my post #6 commented on that using the NPN to drive the PNP would drop .7V, and Crutchow mentioned an NPN follower and PNP follower wouldn't.

You can always parallel two pins together to give double capacity

very poor design indeed

But as for 'design', if you design things correctly then such problems don't occur - as for a suitable IC, the much mentioned ULN2003 would do it if you didn't ground the opto-couplers (and any sensible design wouldn't cripple itself by not offering the option).

He's trying to interface to an existing board, so the optos are already grounded.

And yet again, simple emitter followers would do the existing job anyway, and you can (or certainly could) get multiple transistors in IC type packaging.

And then you suggested the dual follower from post #5 would do all that's required. So show me the schematic of this follower pair, if mine isn't correct... and by the way, mine would work nicely, giving a hi out for a hi in, positive logic like he wanted. What is your better suggestion (for the follower pair)?

Please read the entire thread before picking apart a post 'out of context'...

 

[Mike odom]

Sorry, but that's complete nonsense

Check my PIC 'extras' for an example of high side switching using NPN/PNP.

P.S. I saw nothing in your tutorial of any additonal value, that any ENGR101 student hasn't already learned.

 

[Mike odom]

Aren't the transistors in the wrong places? Doesn't the Vb of the PNP have to be at least .7v less than the emitter?

er, no.
When the logic input is low, the NPN is off. R2 then pulls the base of the PNP to ground, turning it on and the emitter will be at .7V, turning the opto off.

When the logic input is high, the NPN is on, pulling the base of the PNP to +V - 0.7V, turning it off. R3 then supplies current to the opto, turning it on.

 

[Nigel Goodwin]

Actually, your post is utter nonsense. We were talking about a way to give a positive output without the .7V loss. We both understand you don't have to worry about the .7V drop using an NPN to turn on a PNP because you are driving an LED that's grounded. That's what my post #6 commented on that using the NPN to drive the PNP would drop .7V, and Crutchow mentioned an NPN follower and PNP follower wouldn't.

Sorry, but the BMP circuit you posted (and please don't use massive BMP's, use GIF or PNG) is exceptionally poor, and very current hungry (using more current when switched OFF than when switched ON).

The correct method (as linked in my tutorials) produces far better switching and uses no current when switched OFF - not that it's 'my' circuit or anything, it's the standard common way of doing it (for good reason).

He's trying to interface to an existing board, so the optos are already grounded.

And as I said, it's a very poor design - why cripple a design in that way?

And then you suggested the dual follower from post #5 would do all that's required. So show me the schematic of this follower pair, if mine isn't correct... and by the way, mine would work nicely, giving a hi out for a hi in, positive logic like he wanted. What is your better suggestion (for the follower pair)?

The suggestion from post #5 is a simple emitter follower, and is all that's required, the 0.7V loss is meaningless for driving an LED - I've no idea what you mean by 'dual follower'?

The example high side drivers in my tutorial would do it without the loss, as the link already posted.

Please read the entire thread before picking apart a post 'out of context'...

I have done.

I never took it out of context, it was an exceptionally poor 'solution'.

 

[crosslakeguy]

Hey guys, don't get your shorts in a bunch....... This was a very good learning thing for me..... You know whats a POOR design..... This crappy Chinese controller...... I have decided to modify whatever is wrong with it or throw it away...... because It is flawed does not work with higher current either!!!! going remove the improperly connected optos. and go right to the chip, which is pulled low internally, and accepts 5v inputs...... all common grounding anyways, and somewhere along the line they have the inputs pulled high......Its a simple design.....I don't know how they could have screwed it up that badly...... thought it was lack of current, but looking much, much more messed up than that!!! arg...

 

[4pyros]

Hey guys, don't get your shorts in a bunch....... This was a very good learning thing for me..... You know whats a POOR design..... This crappy Chinese controller...... I have decided to modify whatever is wrong with it or throw it away...... because It is flawed does not work with higher current either!!!! going remove the improperly connected optos. and go right to the chip, which is pulled low internally, and accepts 5v inputs...... all common grounding anyways, and somewhere along the line they have the inputs pulled high......Its a simple design.....I don't know how they could have screwed it up that badly...... thought it was lack of current, but looking much, much more messed up than that!!! arg...

Do you have a schematic of the cheep chinese motor driver board?

 

[rc3po]

ok, had momentary brain fart... was thinking the pnp base resistor in series between two transistors.

Like this (R3 is opto current limit resistor):

Hi Mike,
What program do you use to draw your schematics?
My name is Robert.

 

[KeepItSimpleStupid]

Active low is used a lot because nothing bad happens when a board is disconnected. Take a motor controller with a DIRECTION and NOT ENABLE input.
Low causes the motor to move. Disconnect the signal and the motor doesn't move.

Now say your driving the chip with a CPU. The pins are normally tri-stated or configured as inputs. So, what happens, again nothing.

Some problems do arise if a standard TTL gate is used. When disconnected it may be the same as floating high. This can cause problems with some circuits.

With the ULN2003 or a simple transistor, you get DISCONNECTED=OFF, LOW=OFF and HIGH = ON.

 

[crosslakeguy]

Active low is used a lot because nothing bad happens when a board is disconnected. Take a motor controller with a DIRECTION and NOT ENABLE input.
Low causes the motor to move. Disconnect the signal and the motor doesn't move.

Now say your driving the chip with a CPU. The pins are normally tri-stated or configured as inputs. So, what happens, again nothing.

Some problems do arise if a standard TTL gate is used. When disconnected it may be the same as floating high. This can cause problems with some circuits.

With the ULN2003 or a simple transistor, you get DISCONNECTED=OFF, LOW=OFF and HIGH = ON.

Active low is used a lot because nothing bad happens when a board is disconnected. Take a motor controller with a DIRECTION and NOT ENABLE input.
Low causes the motor to move. Disconnect the signal and the motor doesn't move.

Now say your driving the chip with a CPU. The pins are normally tri-stated or configured as inputs. So, what happens, again nothing.

Some problems do arise if a standard TTL gate is used. When disconnected it may be the same as floating high. This can cause problems with some circuits.

With the ULN2003 or a simple transistor, you get DISCONNECTED=OFF, LOW=OFF and HIGH = ON.

motor moves, with NOTHING connected!!!! then power opto, it stops. I would be happy to send this ILL designed controller to someone if they feel the need to be irritated. p.s my controller is active High then (-) low or visa versa +5v - low (-)

just to clarify this post started because (and still do) was looking for a multi-line driver capable of inputing +5 low current-------output +5v at least 37ma+

 

[KeepItSimpleStupid]

Use a high side driver: https://www.st.com/web/catalog/sense_power/FM1965/SC1037/PF63388 This one is active low too, so you would need an inverter as well.

 

[crosslakeguy]

Do you have a schematic of the cheep chinese motor driver board?

I do not have a schematic

Use a high side driver: https://www.st.com/web/catalog/sense_power/FM1965/SC1037/PF63388 This one is active low too, so you would need an inverter as well.

looks as low current?

 

[KeepItSimpleStupid]

3 channel; +-5 to +-30 uA is what I get. There are lots of other "high side drivers" out there.

 

[crosslakeguy]

3 channel; +-5 to +-30 uA is what I get. There are lots of other "high side drivers" out there.

I smell smoke........once you let it out you cannot get it back in there......

 

[Mike odom]

Hi Mike,
What program do you use to draw your schematics?
My name is Robert.

Robert, Hi,
I use PADS Logic for schematic entry and PowerPCB for pcb layout.

 

[rc3po]

Robert, Hi,
I use PADS Logic for schematic entry and PowerPCB for pcb layout.

Thanks Mike.
I need to find one that is quick and easy to use. I don't want to waste all day drawing a schematic.
I tried to post a picture of a schematic that I drew by hand, but it won't work for some reason.

 

[rc3po]

Cool!! Somehow I got it to work...
I've since changed the circuit and gave each LED it's own 470 ohm resistor.

 

[Mike odom]

The suggestion from post #5 is a simple emitter follower, and is all that's required, the 0.7V loss is meaningless for driving an LED - I've no idea what you mean by 'dual follower'?

then perhaps you should read post #5 again, in it's entirety... especially look at the LAST sentence copied here.

A simple solution would be to use a small NPN transistor (such as a 2N2222) as a emitter follower buffer but that reduces the output voltage by about 0.7V. If that's a problem then you could use an NPN driving a PNP.

An NPN driving a PNP.... this is the point where I asked Crutchow "wouldn't the PNP also drop .7V?" Then he responded, "not if they were both followers the BE drops would cancel", hence the "DUAL FOLLOWER". And then my posted schematic was trying to figure out what he meant or how they would be connected... Note the post doesn't say "HERE IS THE BEST SOLUTION AVAILABLE". Or presented as a solution at all...

The example high side drivers in my tutorial would do it without the loss, as the link already posted.

sorry but you are wrong. The high side driver you present in your post does drop the .7V. Any time you switch a transistor like that, it's going to drop .7V emitter to collector.

I never took it out of context, it was an exceptionally poor 'solution'.

And that is EXACTLY how you took it out of context... it was never presented as a 'design solution'... I was just trying to figure out what Crutchow was talking about by using the dual follower (NPN feeding PNP)...