PUJT valley current?

T

Tony Williams

Guest
I'm doing a job with an Onsemi 2N6027 Programmable
Unijunction transistor. Haven't seen a UJT for
years, and never used a PUJT. Forgotten everything.

The timing comps used in this circuit have a s/c
current of 400uA, so I need a valley current that
is reliably larger than that, up to 45C Tambient.

The Onsemi data sheet gives a few graphs, but don't
really pin themselves down.

Is there an algorithm that relates the value of the
valley current to Rg (and Vg, and possibly even Tamb)?

--
Tony Williams.
 
In article <3nka11ltk1sg5bqpn92cqocs8mbp9j9jmf@4ax.com>,
Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote:

That's a pretty high valley current. Can you shut it down when
the PUT triggers?
Click to expand...
No. I'm stuck with what's there atm Speff, so have
to try wriggling (probably at least until I can see
about getting an old-fashioned UJT in there).

It's a 20V supply (with about 1.5mA spare), and Vg
for the 2N6027 needs to be about 12V. A 5k+7k5 pot
down would do 12V, with about 3k source resistance.
That might give a valley current of 400uA under some
circumstances, but not reliably, especially at 45C.

What would happen if the 7k5 were replaced with a
12V zener? That would have a few hundred ohms
source-R, and the 2N6027 is guaranteed to have an
Iv of 1mA with 200R source-R.

Or is it better to replace the 5k with an 8V zener,
so that the source-R is low to the +ve supply?

Would that damage the 2N6027?

I don't know enough about PUJT's.

--
Tony Williams.
 
Tony Williams wrote:
In article <4216AB8E.7060404@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:


You're not giving us much to go on- like why can't you reduce Ct
to make the anode resistor more reasonable? -where is the output
taken from.................


It's the sawtooth generator for an ancient PWM
field-coil driver. My end of the job was originally
to do no more than replace the equally ancient 20A
pnp darlington output transistor with something more
bullet-proof.

But the front end doesn't work, so I'm having to
investigate. That was the reason for the original
PUJT question.

The (1KHz?) sawtooth is taken directly off the
capacitor with a resistor network, whose values
have to remain as-is until downstream scopings
have been done. The loaded timing network reduces
to an 18.8V source, with 45700 ohms and 0.022uF.


.......and what amplitude do you want?- what frequency and
stability?- what temperature range?


Rough calcs suggest that the raw sawtooth should be about
12v pk-pk, at about 1KHz. Frequency stability is
probably not important, but it does run in a high ambient
temperature.
Click to expand...
This topology can be adjusted to handle any situation where s/c current
exceeds reasonable valley currents- nothing is especially critical at
frequencies on the order of 1KHz, and high temperatures favor this ckt:

View in a fixed-width font such as Courier.




18.8V
,--------[27k]---------------. |
| | |
| e |
| |/ |
| ,-----------------| pnp [45.7k]
| | |\ |
| | | |
| | +--------|----------+----+
+-[5.1k]-+-[5.1k]-G |A | | |
| _\|__ | [220] |
--- \ /PUT | | |
- 12v _\_/_ | |/ |
| |K +-[5.1k]-| npn | 0.022u
| | | |\e ===C
| | | | |
| | [5.1k] | |
| | | | |
+-------------------+--------+----------+----+--0v
 
In article <421B5B56.6040508@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:

Actually- the simulation shows the capacitor discharging to
0.25V- you can see Vanode dropping abruptly to 1.1V and then a
40us reduced slope to 0.25V before the capacitor begins
recovery. So it looks pretty good.
Click to expand...
A 2N6027, (running by itself, with 20mA Ig short
circuit current available), pulls the cap down
to about 0.56V.

The negative pulse at the Gate is only about 12uS
wide.... roughly 3.2uS to discharge the cap, and
then an 8.8uS recovery to the Off state.

______ <12uS> ______9.44V
| |
| |
| /
| /
\ /
\/ __0.32V

I'm still running with the brute-force-and-ignorance
zener diode scheme. A not-bad approximation for Iv
(for the 2N6017) seems to be.

Guaranteed min Iv = 0.029*Ia + 0.041. All in mA.

Where Ia= Vg/Rg, over the range 1 mA to 50mA.

So an Iv of 0.4mA can be obtained with an Ia of 12.4mA.

Off the data sheet, the tempco of Iv looks to be about
-1.4%/C, over the range 0 to 50C.

So an Ia of 1.5*12.4mA would be required for 50C operation.

--
Tony Williams.
 
In article <4216AB8E.7060404@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:

You're not giving us much to go on- like why can't you reduce Ct
to make the anode resistor more reasonable? -where is the output
taken from.................
Click to expand...
It's the sawtooth generator for an ancient PWM
field-coil driver. My end of the job was originally
to do no more than replace the equally ancient 20A
pnp darlington output transistor with something more
bullet-proof.

But the front end doesn't work, so I'm having to
investigate. That was the reason for the original
PUJT question.

The (1KHz?) sawtooth is taken directly off the
capacitor with a resistor network, whose values
have to remain as-is until downstream scopings
have been done. The loaded timing network reduces
to an 18.8V source, with 45700 ohms and 0.022uF.

.......and what amplitude do you want?- what frequency and
stability?- what temperature range?
Click to expand...
Rough calcs suggest that the raw sawtooth should be about
12v pk-pk, at about 1KHz. Frequency stability is
probably not important, but it does run in a high ambient
temperature.


------------------+--18.8V
|
[22K]
2x 1n4148 |
,--|<|-|<|----+
| |
| [22K]
| |
| +------+
+----G |A |
| _\|__ |
| \ /PUT |
[100K] _\_/_ |
| |K |
| | ===C
| | |
-----+------+------+--0v
Click to expand...
That's a crafty one, using the Gate to get the
Anode current below Iv. Never thought of that.

The 470 and 6v2 is working well enough atm, but
that one is held in reserve. Thanks.

--
Tony Williams.
 
In article <421CA7C4.9090502@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
 
In article <4d42310fe2tonyw@ledelec.demon.co.uk>,
Tony Williams <tonyw@ledelec.demon.co.uk> wrote:
In article <421CA7C4.9090502@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
Click to expand...
So where did 3086 bytes of text disappear to?

Reposted below.

In article <421CA7C4.9090502@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:
 
On Tue, 22 Feb 2005 09:23:43 +0000 (GMT), the renowned Tony Williams
<tonyw@ledelec.demon.co.uk> wrote:

In article <42176E76.9060708@nospam.com>,
Fred Bloggs <nospam@nospam.com> wrote:

18.8V
,--------[27k]---------------. |
| | |
| e |
| |/ |
| ,-----------------| pnp [45.7k]
| | |\ |
| | | |
| | +--------|----------+----+
+-[5.1k]-+-[5.1k]-G |A | | |
| _\|__ | [220] |
--- \ /PUT | | |
- 12v _\_/_ | |/ |
| |K +-[5.1k]-| npn | 0.022u
| | | |\e ===C
| | | | |
| | [5.1k] | |
| | | | |
+-------------------+--------+----------+----+--0v

Since the pnp is const-I that 5.1k in series with
the base of the npn is probably not needed.

When using an external discharge transistor, does the
PUT get forced off when the transistor takes the A-K
voltage below the forward drop of the PUT? Similar to
commutating an SCR?
Click to expand...
Yes. You an get rid of the PNP above and drive the NPN from the
cathode of the 2N6027.



Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
 
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