# Step Down Switching Regulator

#### Ndjs

Joined Sep 26, 2012
18
Hi Guys,

I am trying to figure out from the maths of a Step-down Switching regulator. The attached image is from the data sheet

1. How the output voltage is calculated?
2. How would I know the peak current which this is designed.?
I think from the data sheet Rsc=0.33/Ipk therefore Ipk=0.33/0.33 = 1 amp (i maybe way off hear)

#### PackratKing

Joined Jul 13, 2008
847
this information should be in the datasheet ?

#### Ndjs

Joined Sep 26, 2012
18
The data sheet only provides the details of the chip it self. If i wanted a Vout of say 20 Volts instead of the 10 what would I need to change for this to occur. ?

#### THE_RB

Joined Feb 11, 2008
5,438
Output voltage is set by the voltage divider R1:R2 ratio, so that there is 1.25v across R2 when the output is at the right voltage.

So you need to change R1 to change the 10v output.

For 20v output you would have 1.25v on R2, and 20-1.25v (18.75v) on R1.

#### bountyhunter

Joined Sep 7, 2009
2,512
The 78S40 chip is so old, it's data sheet may have gone through some "cut downs" which is what they do with old products that they don't want to waste a lot of catalog pages on.

#### bountyhunter

Joined Sep 7, 2009
2,512
Current limit (from data sheet)

The current limit modifies the ON time. The current limit is
activated when a 300 mV potential appears between lead
13 (VCC) and lead 14 (Ipk). This potential is intended to re-
sult when designed for peak current flows through RSC.

When the peak current is reached the current limit is
turned on. The current limit circuitry provides for a quick
end to ON time and the immediate start of OFF time.

#### ian field

Joined Oct 27, 2012
6,539
Hi Guys,

I am trying to figure out from the maths of a Step-down Switching regulator. The attached image is from the data sheet

1. How the output voltage is calculated?
2. How would I know the peak current which this is designed.?
I think from the data sheet Rsc=0.33/Ipk therefore Ipk=0.33/0.33 = 1 amp (i maybe way off hear)

That looks similar to the MC34063 or possibly the 78S something or other -there are various calculator apps you can download as EXE files or some you can enter values in boxes online.

#### ian field

Joined Oct 27, 2012
6,539
The 78S40 chip is so old, it's data sheet may have gone through some "cut downs" which is what they do with old products that they don't want to waste a lot of catalog pages on.
IIRC - the MC34063 is a cut down version (I think, minus the op-amp bit) There is a fairly comprehensive appnote from Motorola/On-Semi and various calculator apps you can use online or download as EXE.

#### Ndjs

Joined Sep 26, 2012
18
Output voltage is set by the voltage divider R1:R2 ratio, so that there is 1.25v across R2 when the output is at the right voltage.

So you need to change R1 to change the 10v output.

For 20v output you would have 1.25v on R2, and 20-1.25v (18.75v) on R1.

If the output voltage is set by the R1:R2 and R2 = 1.25V that means 8.75 across R2.

R1 = 12k (volt drop across is 1.25V)
R2 = 85k This has to be 8.75 but is not the case

R1 = 12k = 1.25Vdrop
R2= 85k = (85/12)x1.25 = 8.85v

Therfore output voltage is 1.25+8.85 = 10.1V.

I think That is correct

#### THE_RB

Joined Feb 11, 2008
5,438
If the output voltage is set by the R1:R2 and R2 = 1.25V that means 8.75 across R2.

R1 = 12k (volt drop across is 1.25V)
R2 = 85k This has to be 8.75 but is not the case

R1 = 12k = 1.25Vdrop
R2= 85k = (85/12)x1.25 = 8.85v

Therfore output voltage is 1.25+8.85 = 10.1V.
I think That is correct
Looks like you understand it correctly.

Standard resistor values mean that it may not be a perfect 10v or 20v output, and there will probably be a small amount of voltage "sag" too making the real output a fraction less than the calculated value.