what should be the output voltage of a 8v zener if input is 37 VDC ? thanks

what should be the output voltage of a 8v zener if input voltage is 37 VDC ?


thanks

How is the diode connected to the 37v?
Can you provide a schematic?

Alex

If reasonable current is allowed through the zener using a series resistance, the voltage across the zener will be 8V.

According to the datasheet of the device you have to select the Zener current and the calculate the series resistance for (37-8=29) for 29 volts. For example if you select 20mA then (37-8)/.02=1,450 ohms. You can use rounded figure 1K5 ohms resistor in series. The output should be 8 volts.

Hi,

The selected current for the zener is equal:
Iz_max = 5mA + maximum load current (in mA)
When the load draws its maximum current from the terminals of the zener, the remaining 5mA (see datasheet for minimum current) keeps the zener voltage at about 8V.
The worst case is when the load is removed completely, hence Iz = Iz_max. In this case, the zener diode has to dissipate about:
Pz_max = Vz * Iz_max.
Note that at relatively high current the zener voltage Vz will increase a bit (due to the internal dynamic series resistance) so Vz could become 8.5V (see datasheet).
In any case, Pz_max should be kept smaller than Pz_rated (like 400mW for small zeners).
Of course, if this worst case (no load) cannot happen (load always connected), only the added 5mA will heat the zener.
In practice, the input DC (as your 37V) is likely not regulated, so one needs to analyse the design twice:
For the lowest input DC, the zener current should not become lower than its minimum (as 5mA or see datasheet) when full load is applied at the zener diode.
For the highest input DC, the zener current should not become large so that Pz < Pz_rated (worst case if the load is removed).

Kerim

You may see first part of this
https://www.edaboard.com/threads/232191/#post990303

---------- Post added at 01:13 ---------- Previous post was at 00:41 ----------

If your current need is between 200mA and 350mA

Your voltage drop across series resistance is 29Volt
Zener diode current 5mA(min)
Series resistance should be 75.32ohm(calculated value)
Power dissipation will be 1.48 W(calculated)
As far as I recall the nearest available rating above the calculated value is 5W

papunblg said:

You may see first part of this
https://www.edaboard.com/threads/232191/#post990303

---------- Post added at 01:13 ---------- Previous post was at 00:41 ----------

If your current need is between 200mA and 350mA

Your voltage drop across series resistance is 29Volt
Zener diode current 5mA(min)
Series resistance should be 75.32ohm(calculated value)
Power dissipation will be 1.48 W(calculated)
As far as I recall the nearest available rating above the calculated value is 5W

Click to expand...

I don't understand, how did you calculate the 75.32 ohm

29V/75.32ohm = 385mA

but why 385mA

If the load seek max 350mA and the zener needs min 5mA,

you don't want 29V/355mA = 81.69ohm instead of the 75.32ohm series resistance

I mean where did you find the 385mA, did you allow more current not to be too borderline?

thanks

Gauthier

Surely you are right Mr.Gauthier

Value of series resistance R_s=(E_DC(min)-E_z)/1.1 I_L(max)

When R_s is known, maximum power dissipation is calculated as,

P_D=[(E_DC(max)-E_z)/R_s)-I_L(min)]E_z


Note:- For Practical application add 10% safety buffer with maximum current. Thats why 1.1 x I_L(max)is used.

See: 1.Practical Electronic Power Supply ch-8/page117
2. Zener Diode Voltage Regulator - Electric Circuit