Reply by Harold Hallikainen●July 24, 20092009-07-24
<*>[Attachment(s) from Harold Hallikainen included below]
> Hi,
> I got the circuit from a lighting desk, so can't explain why certain
> things have been put in, but its good for discussion.
>
> Here is the circuit as i have found it, the earlier one i adapted slightly
>
>
http://f1.grp.yahoofs.com/v1/gNppSg6xNS0X-VGoRhwoeS68XZrvLSLtQvrBSMQQixPTQmYuqckEIklzfP_Qb17cMR2BaA7VhQXHnIvB0BPchQRwuG7T/PWM2.jpg
>
> Does the make anymore sense?
> I think the diode on the output could be to stop damage if wired
> incorrectly also?
>
> I am now wondering what sort of cycle time this system must have to
> service 24 channels and to keep the lights at the correct level? I know
> digital is quick, but it sometimes make you wonder!
>
> Cheers,
> Paul
I think the circuit suffers from "brute force engineering" where parts
kept being added until it worked. It could be simplified considerably.
I'm
sure the diode in the output is for "pile on".
When originally designing the MTX-DE48, I thought of having the PWM output
of the PIC generate an analog voltage, then demux that out to sample and
hold capacitors. It occurred to me that I could put the R of the RC low
pass before the demux and the C after the demux to have the C serve as
both a the C of the low pass filter and as the hold capacitor. So, the
output section looks like the attached (assuming the attachment gets
through). As described before, there's a series R in front of the demux
(between the pin marked PWM and the PIC PWM output). There's also a pot
to
adjust the PWM amplitude to allow the output voltage to be adjusted as
desired.
Hi,
I think the diodes may be to avoid the op amps having to give an output
right down to 0 volts The first diode will add about 0.6 volts to the level and
then the second one will subtract about the same value. If this is the case then
the technique should have been applied on the first op amp. A better way would
be to use a negative supply to the op amps or use op amps rated to work right
down to the negative rail. By the way I could not view the original link.
Les.
--- In p..., "Paul Laverick" wrote: >
> Hi,
> I got the circuit from a lighting desk, so can't explain why certain
things have been put in, but its good for discussion.
>
> Here is the circuit as i have found it, the earlier one i adapted slightly
>
>
http://f1.grp.yahoofs.com/v1/gNppSg6xNS0X-VGoRhwoeS68XZrvLSLtQvrBSMQQixPTQmYuqckEIklzfP_Qb17cMR2BaA7VhQXHnIvB0BPchQRwuG7T/PWM2.jpg
>
> Does the make anymore sense?
> I think the diode on the output could be to stop damage if wired incorrectly
also?
>
> I am now wondering what sort of cycle time this system must have to service 24
channels and to keep the lights at the correct level? I know digital is quick,
but it sometimes make you wonder!
>
> Cheers,
> Paul
> From: Harold Hallikainen
> Sent: Friday, July 24, 2009 4:56 PM
> To: p...
> Subject: Re: [piclist] PWM to 0-10V Output
> OK, that worked this time. I think the circuit can be simplified a bit.
>
> 1. I don't know that you need three sections of RC filtering. If the
PWM
> frequency is quite a ways above the "signal" frequency (how fast you want
> the output to move), a single RC may be sufficient.
>
> 2. I don't think the voltage follower is necessary. The input
resistance
> of the non-inverting amplifier that follows it will be very high, so it
> will not load the RC.
>
> 3. The diode in the feedback of the second stage is interesting. This is
> apparently to compensate for the diode in the output of the third stage.
> The diode on the third stage is probably to allow the paralleling of
> multiple outputs, allowing a "pile on" or "highest takes precedence",
> common in lighting control. You could get the same effect by just taking
> the output from the cathode of the diode in the second stage.
>
> 4. The 2.7k and 4.7nF will have no effect (other than increasing the
> possibility the second stage will oscillate due to capacitive loading).
> These components are in parallel with a near zero ohm output of the second
> stage, so they have no effect on the voltage at the output of the second
> stage.
>
> So, I'd simplify this a bit. I'd use a single RC driving the circuit
of
> stage number two, and take the output from the cathode of the diode of
> stage number two. This is the approach taken in these products:
>
> http://www.dovesystems.com/products/MTX-DE%208.html
> http://www.dovesystems.com/products/MTX-DE%2048.html
>
> Harold
> (formerly designer at Dove Systems (http://www.dovesystems.com)
>
> --
> FCC Rules Updated Daily at http://www.hallikainen.com - Advertising
> opportunities available!
>
Reply by Paul Laverick●July 24, 20092009-07-24
Hi,
I got the circuit from a lighting desk, so can't explain why certain things
have been put in, but its good for discussion.
Here is the circuit as i have found it, the earlier one i adapted slightly
Does the make anymore sense?
I think the diode on the output could be to stop damage if wired incorrectly
also?
I am now wondering what sort of cycle time this system must have to service 24
channels and to keep the lights at the correct level? I know digital is quick,
but it sometimes make you wonder!
Cheers,
Paul
From: Harold Hallikainen
Sent: Friday, July 24, 2009 4:56 PM
To: p...
Subject: Re: [piclist] PWM to 0-10V Output
OK, that worked this time. I think the circuit can be simplified a bit.
1. I don't know that you need three sections of RC filtering. If the PWM
frequency is quite a ways above the "signal" frequency (how fast you want
the output to move), a single RC may be sufficient.
2. I don't think the voltage follower is necessary. The input resistance
of the non-inverting amplifier that follows it will be very high, so it
will not load the RC.
3. The diode in the feedback of the second stage is interesting. This is
apparently to compensate for the diode in the output of the third stage.
The diode on the third stage is probably to allow the paralleling of
multiple outputs, allowing a "pile on" or "highest takes precedence",
common in lighting control. You could get the same effect by just taking
the output from the cathode of the diode in the second stage.
4. The 2.7k and 4.7nF will have no effect (other than increasing the
possibility the second stage will oscillate due to capacitive loading).
These components are in parallel with a near zero ohm output of the second
stage, so they have no effect on the voltage at the output of the second
stage.
So, I'd simplify this a bit. I'd use a single RC driving the circuit
of
stage number two, and take the output from the cathode of the diode of
stage number two. This is the approach taken in these products:
Reply by Harold Hallikainen●July 24, 20092009-07-24
OK, that worked this time. I think the circuit can be simplified a bit.
1. I don't know that you need three sections of RC filtering. If the PWM
frequency is quite a ways above the "signal" frequency (how fast you want
the output to move), a single RC may be sufficient.
2. I don't think the voltage follower is necessary. The input resistance
of the non-inverting amplifier that follows it will be very high, so it
will not load the RC.
3. The diode in the feedback of the second stage is interesting. This is
apparently to compensate for the diode in the output of the third stage.
The diode on the third stage is probably to allow the paralleling of
multiple outputs, allowing a "pile on" or "highest takes precedence",
common in lighting control. You could get the same effect by just taking
the output from the cathode of the diode in the second stage.
4. The 2.7k and 4.7nF will have no effect (other than increasing the
possibility the second stage will oscillate due to capacitive loading).
These components are in parallel with a near zero ohm output of the second
stage, so they have no effect on the voltage at the output of the second
stage.
So, I'd simplify this a bit. I'd use a single RC driving the circuit
of
stage number two, and take the output from the cathode of the diode of
stage number two. This is the approach taken in these products:
You mention a multiplexer or demux. Years ago I designed a DMX to 48
channel decoder. The PWM output of the PIC went to a pot, which was used
to set the output voltage, then to a series resistor, then to a series of
demultiplexers. Each output had a capacitor to ground as a holding
capacitor. This then drove a non-inverting amplifier to provide the
required gain. As I recall, we could output up to +36V. In sync with the
PWM, the demux would be enabled on a certain channel, then about 10 pulses
would be output with the appropriate duty cycle. The demuxes would then be
disabled, the address lines set to point to the next capacitor, and the
appropriate demux enabled. The next 10 pulses would set the voltage on
that capacitor. This would be repeated for 48 channels, then go back to
the start.
You mention a multiplexer or demux. Years ago I designed a DMX to 48
channel decoder. The PWM output of the PIC went to a pot, which was used
to set the output voltage, then to a series resistor, then to a series of
demultiplexers. Each output had a capacitor to ground as a holding
capacitor. This then drove a non-inverting amplifier to provide the
required gain. As I recall, we could output up to +36V. In sync with the
PWM, the demux would be enabled on a certain channel, then about 10 pulses
would be output with the appropriate duty cycle. The demuxes would then be
disabled, the address lines set to point to the next capacitor, and the
appropriate demux enabled. The next 10 pulses would set the voltage on
that capacitor. This would be repeated for 48 channels, then go back to
the start.
----- Original Message -----
From: Paul
To: p...
Sent: Friday, July 24, 2009 2:40 PM
Subject: [piclist] PWM to 0-10V Output
I have recently come across a circuit diagram:
Now the purpose of this circuit, is to convert a 5V PWM Signal to a 0-10V
Output.
The details of the PWM are unknown. So it could be a normal type of PWM,
with the mark to space ratio changing, but seen as it is a PIC that is
supplying this signal, it could just be a oscilator.
Just to add, i've removed part of the circuit that lies just before the
last
OP-AMP. It was a multiplexer IC. I've mentioned this because this one PWM
signal services 24 channels. So i'm guessing, that each channel is getting
a
pulse of voltage, depending on the frequency or ratio of the signal, in a
continious circle. So the device that uses this 0-10V can't be to fussy
about pulsing?
Anyone got any thoughts on this please?
---------------------------
It's actually a low-pass filter and a DC amplifier, giving a variable DC
voltage from 0V - 10V.
Leon
Reply by Paul Laverick●July 24, 20092009-07-24
Having done a search on the net, there is quite a bit about this method.
I've looked at the circuit, and found this:
1 - The first 3 RC networks are filters, now is there any way of figuring out
what frequency they are chopping? This could help in getting the PWM
frequency
2 - The first OP-AMP i would say is just a buffer
3 - The second been a voltage gain, in this case x2, which makes sense of PWM
been 0-5V and needing 0-10V
Probably best to make this circuit up and have a play, don't think my
simulator likes the filters...
Paul
From: Paul
Sent: Friday, July 24, 2009 2:40 PM
To: p...
Subject: [piclist] PWM to 0-10V Output
I have recently come across a circuit diagram:
Now the purpose of this circuit, is to convert a 5V PWM Signal to a 0-10V
Output.
The details of the PWM are unknown. So it could be a normal type of PWM, with
the mark to space ratio changing, but seen as it is a PIC that is supplying this
signal, it could just be a oscilator.
Just to add, i've removed part of the circuit that lies just before the
last OP-AMP. It was a multiplexer IC. I've mentioned this because this one
PWM signal services 24 channels. So i'm guessing, that each channel is
getting a pulse of voltage, depending on the frequency or ratio of the signal,
in a continious circle. So the device that uses this 0-10V can't be to
fussy about pulsing?
Anyone got any thoughts on this please?
Thanks
Paul
Now the purpose of this circuit, is to convert a 5V PWM Signal to a 0-10V
Output.
The details of the PWM are unknown. So it could be a normal type of PWM, with
the mark to space ratio changing, but seen as it is a PIC that is supplying this
signal, it could just be a oscilator.
Just to add, i've removed part of the circuit that lies just before the
last OP-AMP. It was a multiplexer IC. I've mentioned this because this one
PWM signal services 24 channels. So i'm guessing, that each channel is
getting a pulse of voltage, depending on the frequency or ratio of the signal,
in a continious circle. So the device that uses this 0-10V can't be to
fussy about pulsing?
Anyone got any thoughts on this please?
Thanks
Paul