Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

DC measuring

Status
Not open for further replies.

Alex18

New Member
Need help.
Which DC current can you measure according to picture a, b, c? I think its 19,8V because a diode needs 0,7V so 20,5V-0,7V=19,8V
 

Attachments

  • 1.PNG
    1.PNG
    137.6 KB · Views: 316
Last edited:
You would be measuring DC voltage [gleichspannung], not current.
(Voltage is equivalent to pressure, current is equivalent to flow).

The capacitor would charge to the peak voltage, 1.414 x the RMS rating (minus the 0.6 - 0.7V diode drop).

If the AC voltage is 20.5V RMS, the peak will be 20.5 x 1.414 ~= 29V
- 0.6 = 28.4V

In the first circuit A will be positive, in the other two B is positive.
 
20,5VeFF is in a foreign language so we do not know if it is RMS or is peak-to-peak.
Why is the transformer measured in V but the voltage across the capacitor is measured in U?
 
20,5VeFF is in a foreign language so we do not know if it is RMS or is peak-to-peak.
Why is the transformer measured in V but the voltage across the capacitor is measured in U?

Presumably it's obviously German? - and also presumably V is AC and U is DC?.

As you're well aware, mains transformers are always specified in RMS.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top