Voltage to current converters BING!
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Posts: 46,084
Thanks to your explanation, I finally understand the first stage amplifier
function. I couldn't recognise it against any of the "standard"
configurations in my book and none of the math was working. I was starting
to think I was mentally deficient or something. Its the signal coming in on
the inverting that was throwing me off. Here's my explanation of what
happens, correct me if I'm wrong.
To analyze this circuit you have to look at it as two separate circuits -
both an inverting and non-inverting amplifier.
For the non-inverting amplifier, lower the inverting input signal to zero
(or take to ground, same thing). Assuming the feedback resistor is Rf and
the signal resistor is Rin, Vo=Vin(Rf/Rin +1). If the non-inverting input
has 4 volts applied you get 4*(10,000/10,000 +1)=4*(1+1)=4x2=8 volts out.
Now look at it as an inverting amplifier with gain. Ignore the non-inverting
input voltage for the moment. Vo=-Vin(Rf/Rin). Assuming 1 volt input
thats -1 * (10,000/10,000)=-1.
Now add up the voltages 8 + -1 = 7 volts. Pull out my meter and Bingo!
Now for the 5 volt input
Non-Inverting 4 * (10,000/50,000 + 1) = 4 x 1.2 = 4.8 volts
Inverting -5 * (10,000/50,000)= -5 * 0.2 = -1 volt.
Output = 4.8 volts + -1 volt = 3.8 volts!
Since my meter is agreeing with me all along here, I think I can crack open
a brewski and celebrate [noparse]:D[/noparse] Damn if feels good to finally understand this
thing!
>
Original Message
> From: Tracy Allen [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=ZtJgdF50qic6CCYle3VZkAXG6uGFDeJ-pWv8kKpLRdKblTR6uORFMCbiy4tWXQFyznvQ2Qu9wkhKdbU]tracy@e...[/url
> Sent: September 24, 2003 11:44 AM
> To: basicstamps@yahoogroups.com
> Subject: RE: [noparse][[/noparse]basicstamps] Voltage to current converters (4-20mA
> transmitter)
>
>
> The 4-20ma circuit has an input impedance of 10 kohms, so the PWM has
> to be bufffered first. The thing about the Stamp PWM to analog is it
> needs to look into a high impedance. Here is an additional op-amp as
> a buffer:
>
> 1/2 LMC6482
> 100k |\ ;----buffer to 4-20 circuit
> Stamp P0 ---/\/\--o
|+\ |
> PWM | | >--o-/\/\-o-/\/\-;
> | ;-|-/ | 40.2k 10k |
> 0.1uf ===== | |/ | R1 R2 |
> | | | Vss
> | `
'
> Vss
>
>
> R1 and R2 can be chosen with a ratio of R2/(R1+R2)=1/5, then into
> another buffer (another 1/2 your dual LMC6482 op-amp), and then into
> the 4-20 circuit, adjusted for 0->1 volt input as it is now.
>
> Or, here is an alternative:
>
> 1/2 LMC6482
> 100k |\
> Stamp P0 ---/\/\--o
|+\
> PWM | | >--o-/\/\--- 4-20 circuit (-) input
> | ;-|-/ | 49.9k
> 0.1uf ===== | |/ |
> | | |
> | `
'
> Vss
>
> Use 49.9 kohms, right into the (-) input of the 4-20 circuit.
> http://www.elecdesign.com/Files/29/2985/Figure_02.gif (Replacing the
> 10kohm resistor that is there now). That makes the inverting gain of
> that circuit -1/5 instead of -1, so the scaling becomes 0 to 5 volts,
> as you want, instead of the original 0->1 volt.
>
> That also changes the zero point calculation, because with an
> inverting gain of -1/5, the non-inverting gain becomes 1.2 instead of
> 2.0. You still need 8.75 volts at the input to the second stage, to
> produce the 4 ma offset. So now the wiper of the "zero" pot P1 has
> to be set at about 8.75/1.2 = 7.29 volts.
>
> -- Tracy Allen
> http://www.emesystems.com
>
>
>
>
> >Actually, one more thing on this current driver. I wanted to change the
> >input range to 0-5 volts (Stamp PWM output). 0-1 would give me
> 51 steps of
> >PWM output which isn't very good for an actuator.
> >
> >I read read my opamp chapter (for the 256th time [noparse]:D[/noparse] ) and decided I can't
> >really do anything to the circuit itself to change the input range from
> >there. So the simple solution comes to mind, divide the PWM
> output to 1/5th.
> >Do the RC output as per normal but add a resistor (multi-turn
> pot) to ground
> >between the pin and the capacitor.
> >
> >Is there a better way to do this?
> >
> >
> >The simplest answer would be a voltage divider (100k and about
> 12K seems to
> >do the trick).
> >
> >>
Original Message
> >> From: PatM [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=KLN79x24HGSozbwOZDSSvqRNV54JE92PqmAHOZXm5Ip2yVZbX_omfM-5aZZc6JktxUPbbexrLA]pmeloy@s...[/url
> >> Sent: September 23, 2003 11:26 PM
> >> To: basicstamps@yahoogroups.com
> >> Subject: RE: [noparse][[/noparse]basicstamps] Voltage to current converters (4-20mA
> >> transmitter)
> >>
> >>
> >> Received the LMC6482s today so I set up the circuit again. I set
> >> P2 to 62.5
> >> ohms and the output of the first amplifier stage to Vcc - 0.25V.
> >> Slapped on
> >> the ammeter and got a reading of 3.87mA! A bit of twiddling
> with the two
> >> pots and voila, 4-20mA current driver! Just to be sure I measured the
> >> current shorted to ground and then through a 470 ohm resistor.
> Exactly the
> >> same readings both times!
> >>
> >> Thank you for the help Tracy Allen, you're a great boon to this group!
> >>
> >> Tomorrow I start on the PID section of the industrial controls
> pdf on the
> >> parallax site. I've read through it already and understand about
> >> 98% of what
> >> I read. Have to see it in action to be sure though.
> >>
> >>
> >>
> >> To UNSUBSCRIBE, just send mail to:
> >> basicstamps-unsubscribe@yahoogroups.com
> >> from the same email address that you subscribed. Text in the
> >> Subject and Body of the message will be ignored.
> >>
> >>
> >> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
> >>
> >>
> >>
> >
> >
> >
> >To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@yahoogroups.com
> >from the same email address that you subscribed. Text in the
> >Subject and Body of the message will be ignored.
> >
> >
> >Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the
> Subject and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
function. I couldn't recognise it against any of the "standard"
configurations in my book and none of the math was working. I was starting
to think I was mentally deficient or something. Its the signal coming in on
the inverting that was throwing me off. Here's my explanation of what
happens, correct me if I'm wrong.
To analyze this circuit you have to look at it as two separate circuits -
both an inverting and non-inverting amplifier.
For the non-inverting amplifier, lower the inverting input signal to zero
(or take to ground, same thing). Assuming the feedback resistor is Rf and
the signal resistor is Rin, Vo=Vin(Rf/Rin +1). If the non-inverting input
has 4 volts applied you get 4*(10,000/10,000 +1)=4*(1+1)=4x2=8 volts out.
Now look at it as an inverting amplifier with gain. Ignore the non-inverting
input voltage for the moment. Vo=-Vin(Rf/Rin). Assuming 1 volt input
thats -1 * (10,000/10,000)=-1.
Now add up the voltages 8 + -1 = 7 volts. Pull out my meter and Bingo!
Now for the 5 volt input
Non-Inverting 4 * (10,000/50,000 + 1) = 4 x 1.2 = 4.8 volts
Inverting -5 * (10,000/50,000)= -5 * 0.2 = -1 volt.
Output = 4.8 volts + -1 volt = 3.8 volts!
Since my meter is agreeing with me all along here, I think I can crack open
a brewski and celebrate [noparse]:D[/noparse] Damn if feels good to finally understand this
thing!
>
Original Message
> From: Tracy Allen [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=ZtJgdF50qic6CCYle3VZkAXG6uGFDeJ-pWv8kKpLRdKblTR6uORFMCbiy4tWXQFyznvQ2Qu9wkhKdbU]tracy@e...[/url
> Sent: September 24, 2003 11:44 AM
> To: basicstamps@yahoogroups.com
> Subject: RE: [noparse][[/noparse]basicstamps] Voltage to current converters (4-20mA
> transmitter)
>
>
> The 4-20ma circuit has an input impedance of 10 kohms, so the PWM has
> to be bufffered first. The thing about the Stamp PWM to analog is it
> needs to look into a high impedance. Here is an additional op-amp as
> a buffer:
>
> 1/2 LMC6482
> 100k |\ ;----buffer to 4-20 circuit
> Stamp P0 ---/\/\--o
|+\ |
> PWM | | >--o-/\/\-o-/\/\-;
> | ;-|-/ | 40.2k 10k |
> 0.1uf ===== | |/ | R1 R2 |
> | | | Vss
> | `
'
> Vss
>
>
> R1 and R2 can be chosen with a ratio of R2/(R1+R2)=1/5, then into
> another buffer (another 1/2 your dual LMC6482 op-amp), and then into
> the 4-20 circuit, adjusted for 0->1 volt input as it is now.
>
> Or, here is an alternative:
>
> 1/2 LMC6482
> 100k |\
> Stamp P0 ---/\/\--o
|+\
> PWM | | >--o-/\/\--- 4-20 circuit (-) input
> | ;-|-/ | 49.9k
> 0.1uf ===== | |/ |
> | | |
> | `
'
> Vss
>
> Use 49.9 kohms, right into the (-) input of the 4-20 circuit.
> http://www.elecdesign.com/Files/29/2985/Figure_02.gif (Replacing the
> 10kohm resistor that is there now). That makes the inverting gain of
> that circuit -1/5 instead of -1, so the scaling becomes 0 to 5 volts,
> as you want, instead of the original 0->1 volt.
>
> That also changes the zero point calculation, because with an
> inverting gain of -1/5, the non-inverting gain becomes 1.2 instead of
> 2.0. You still need 8.75 volts at the input to the second stage, to
> produce the 4 ma offset. So now the wiper of the "zero" pot P1 has
> to be set at about 8.75/1.2 = 7.29 volts.
>
> -- Tracy Allen
> http://www.emesystems.com
>
>
>
>
> >Actually, one more thing on this current driver. I wanted to change the
> >input range to 0-5 volts (Stamp PWM output). 0-1 would give me
> 51 steps of
> >PWM output which isn't very good for an actuator.
> >
> >I read read my opamp chapter (for the 256th time [noparse]:D[/noparse] ) and decided I can't
> >really do anything to the circuit itself to change the input range from
> >there. So the simple solution comes to mind, divide the PWM
> output to 1/5th.
> >Do the RC output as per normal but add a resistor (multi-turn
> pot) to ground
> >between the pin and the capacitor.
> >
> >Is there a better way to do this?
> >
> >
> >The simplest answer would be a voltage divider (100k and about
> 12K seems to
> >do the trick).
> >
> >>
Original Message
> >> From: PatM [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=KLN79x24HGSozbwOZDSSvqRNV54JE92PqmAHOZXm5Ip2yVZbX_omfM-5aZZc6JktxUPbbexrLA]pmeloy@s...[/url
> >> Sent: September 23, 2003 11:26 PM
> >> To: basicstamps@yahoogroups.com
> >> Subject: RE: [noparse][[/noparse]basicstamps] Voltage to current converters (4-20mA
> >> transmitter)
> >>
> >>
> >> Received the LMC6482s today so I set up the circuit again. I set
> >> P2 to 62.5
> >> ohms and the output of the first amplifier stage to Vcc - 0.25V.
> >> Slapped on
> >> the ammeter and got a reading of 3.87mA! A bit of twiddling
> with the two
> >> pots and voila, 4-20mA current driver! Just to be sure I measured the
> >> current shorted to ground and then through a 470 ohm resistor.
> Exactly the
> >> same readings both times!
> >>
> >> Thank you for the help Tracy Allen, you're a great boon to this group!
> >>
> >> Tomorrow I start on the PID section of the industrial controls
> pdf on the
> >> parallax site. I've read through it already and understand about
> >> 98% of what
> >> I read. Have to see it in action to be sure though.
> >>
> >>
> >>
> >> To UNSUBSCRIBE, just send mail to:
> >> basicstamps-unsubscribe@yahoogroups.com
> >> from the same email address that you subscribed. Text in the
> >> Subject and Body of the message will be ignored.
> >>
> >>
> >> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
> >>
> >>
> >>
> >
> >
> >
> >To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@yahoogroups.com
> >from the same email address that you subscribed. Text in the
> >Subject and Body of the message will be ignored.
> >
> >
> >Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the
> Subject and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
Comments
>To analyze this circuit you have to look at it as two separate circuits -
>both an inverting and non-inverting amplifier.
Vout = V1*(1+Rf/Ri) - V2*(Rf/Ri)
non-inverting inverting
V1 from"zero" pot V2 "signal" from buffer
>
>For the non-inverting amplifier, lower the inverting input signal to zero
>(or take to ground, same thing). Assuming the feedback resistor is Rf and
>the signal resistor is Rin, Vo=Vin(Rf/Rin +1). If the non-inverting input
>has 4 volts applied you get 4*(10,000/10,000 +1)=4*(1+1)=4x2=8 volts out.
>
>Now look at it as an inverting amplifier with gain. Ignore the non-inverting
>input voltage for the moment. Vo=-Vin(Rf/Rin). Assuming 1 volt input
>thats -1 * (10,000/10,000)=-1.
>
>Now add up the voltages 8 + -1 = 7 volts. Pull out my meter and Bingo!
>
>Now for the 5 volt input
>
>Non-Inverting 4 * (10,000/50,000 + 1) = 4 x 1.2 = 4.8 volts
>
>Inverting -5 * (10,000/50,000)= -5 * 0.2 = -1 volt.
>
>Output = 4.8 volts + -1 volt = 3.8 volts!
>
>Since my meter is agreeing with me all along here, I think I can crack open
>a brewski and celebrate [noparse]:D[/noparse] Damn if feels good to finally understand this
>thing!
Cheers with the brewski!
-- Tracy