RC4194 Dual Tracking Voltage Regulators

Thread Starter

WSUBG

Joined Jun 12, 2021
26
Hi,
Could someone help me understand how the voltage output is controlled on this device. Particularly I need to understand the purpose of the red bobble pins. The data sheet is too advanced and does not have good explanation of the pins purpose. RC4194 pdf, RC4194 description, RC4194 datasheets, RC4194 view ::: ALLDATASHEET ::: For instance if I wanted the device to output +-4.7 volts with +-12 volt input how would I design the circuit. Also when looking at the package diagram does it show the top or the PCB mount side of view?

1627712448726.png
 

Ian0

Joined Aug 7, 2020
3,245
The data sheet is too advanced and does not have good explanation of the pins purpose
That's a matter of opinion, but. . .

The two COMP pins are where the frequency compensation capacitors for the output amplifiers connect. See the "compensation" section for how to use them

Ro and Rset are involved in setting the output voltage. It's in the section called "RC4194 Switchable Power Supply".
Rset is used to establish a precision current source. The precision current source is connected to Ro. That produces a voltage = Ro * Iset which is the output voltage.
If you stick with Rset = 71.5k you get Vout = 2.5 * Ro (If Ro is in kΩ). If you put 2 x 36k in series you get 2.517* Ro which will probably be near enough.

Bal doesn't get much of an explanation - but it seems it for adjusting the balance between the negative and positive outputs (Figure 7 gives the only clue)

Are you sure the RC4194 doesn't belong in a museum?
 

Thread Starter

WSUBG

Joined Jun 12, 2021
26
That's a matter of opinion, but. . .

The two COMP pins are where the frequency compensation capacitors for the output amplifiers connect. See the "compensation" section for how to use them

Ro and Rset are involved in setting the output voltage. It's in the section called "RC4194 Switchable Power Supply".
Rset is used to establish a precision current source. The precision current source is connected to Ro. That produces a voltage = Ro * Iset which is the output voltage.
If you stick with Rset = 71.5k you get Vout = 2.5 * Ro (If Ro is in kΩ). If you put 2 x 36k in series you get 2.517* Ro which will probably be near enough.

Bal doesn't get much of an explanation - but it seems it for adjusting the balance between the negative and positive outputs (Figure 7 gives the only clue)

Are you sure the RC4194 doesn't belong in a museum?
I am working on 1977 first electronic Singer sewing machine Athena 2000.
my measurements of the debug test points indicate that RC4194 could be faulty. TP 3 is indicating 12.64V and traces directly to the pin 6 of the RC4194. There are some RC4194 available for purchase but I need to test the device for correct operation before I go through the trouble of installing it into PCB.
1627714247890.png
 

Ian0

Joined Aug 7, 2020
3,245
The positive side seems to have gone short, but the negative side is still working.
That's a bit unusual - usually if something's dead, it's all dead. Before you give up, and replace it, look for external shorts between positive input and positive output. I bet it won't be easy to desolder!
 

Thread Starter

WSUBG

Joined Jun 12, 2021
26
That's a matter of opinion, but. . .

The two COMP pins are where the frequency compensation capacitors for the output amplifiers connect. See the "compensation" section for how to use them

Ro and Rset are involved in setting the output voltage. It's in the section called "RC4194 Switchable Power Supply".
Rset is used to establish a precision current source. The precision current source is connected to Ro. That produces a voltage = Ro * Iset which is the output voltage.
If you stick with Rset = 71.5k you get Vout = 2.5 * Ro (If Ro is in kΩ). If you put 2 x 36k in series you get 2.517* Ro which will probably be near enough.

Bal doesn't get much of an explanation - but it seems it for adjusting the balance between the negative and positive outputs (Figure 7 gives the only clue)

Are you sure the RC4194 doesn't belong in a museum?
I apologize but I do not understand logic of provided explanation. Most likely it is the format in which it is presented. I suspect that in your explanation you are referring to
1627718048695.png
of the Electrical Characteristics table. But from your calculations I do not see how output becomes +-4.7v. I cannot visualize this " If you put 2 x 36k in series you get 2.517* Ro which will probably be near enough." What is 36k, why *2 and how is 2.517* Ro gives +-4.7v? Again my apology for not caching on more quickly.
 

Ian0

Joined Aug 7, 2020
3,245
Output scale factor = 2.5kΩ/V, for for 4.7V you use 2.5kΩ*4.7 = 11.75kΩ (or 12kΩ would give 4.8V, because 11.75kΩ is a non-stadard value)
Similarly 71.5kΩ is a non-standard value. The nearest standard values (68kΩ and 75kΩ) would introduce some error but 2 x 36kΩ would give you 72kΩ which would be accurate within 0.7%, and it would make the scale factor 2.517 instead of 2.50

(Though, if you wanted 4.7V with standard values, replacing the 71.5kΩ with 91kΩ and using Ro=15kΩ, you should get 4.71V)
 

Thread Starter

WSUBG

Joined Jun 12, 2021
26
Output scale factor = 2.5kΩ/V, for for 4.7V you use 2.5kΩ*4.7 = 11.75kΩ (or 12kΩ would give 4.8V, because 11.75kΩ is a non-stadard value)
Similarly 71.5kΩ is a non-standard value. The nearest standard values (68kΩ and 75kΩ) would introduce some error but 2 x 36kΩ would give you 72kΩ which would be accurate within 0.7%, and it would make the scale factor 2.517 instead of 2.50

(Though, if you wanted 4.7V with standard values, replacing the 71.5kΩ with 91kΩ and using Ro=15kΩ, you should get 4.71V)
Is this how you calculated the change in the scaling factor based on the 72kΩ?
1627725034216.png
I understand your explanation now I think, thank you.
this relationship did not make sense to me 1627725795533.pngbecause 2.5 was without units and units were in the table and I couldn't crosslink the two.
So Ro is solely responsible for controlling the output and Rset is responsible for scaling factor. Changing the Rset will affect the scaling factor proportionally.
 
Last edited:

Ian0

Joined Aug 7, 2020
3,245
Is this how you calculated the change in the scaling factor based on the 72kΩ?
View attachment 244793
I understand your explanation now I think, thank you.
this relationship did not make sense to me View attachment 244794because 2.5 was without units and units were in the table and I couldn't crosslink the two.
So Ro is solely responsible for controlling the output and Rset is responsible for scaling factor. Changing the Rset will affect the scaling factor proportionally.
That's it.
 
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